SPA 3e_ Teachers Edition _ Ch 6

6 

Sampling Distributions 

Please read the Introduction to the Teacher’s Edition. 

It will help prepare you for teaching this course, as it 

includes a lot of helpful information and advice. 

The Big Picture 

This chapter focuses on sampling distributions. A sampling 

distribution describes the possible values of a statistic such 

as the sample mean x or the sample proportion p^ and how 

often they occur. Three characteristics of sampling distributions 

will be examined in detail: center, variability, and shape. 

These are the same three characteristics used in Chapter 1 

to describe distributions of quantitative data. The mean and 

standard deviation measure the center and variability of 

sampling distributions. The shapes of sampling distributions 

will be described with the same terms in use since Chapter 1: 

skewed left, skewed right, symmetric, mound-shaped, and a 

new term from Chapter 5: approximately normal. 

The variables examined in this chapter are examples of 

random variables, which were introduced in Chapter 5. 

The sample count X is a binomial random variable and 

is therefore discrete. The sample proportion p^ is closely 

related to the sample count X. Finally, the sample mean x 

is a continuous random variable. 

The process of making a conclusion about a population 

based on the data in a sample is called statistical inference. 

Chapter 6 lays the foundation for the statistical inference 

techniques learned in Chapters 7–10. Chapter 6 describes 

the sampling distribution of a sample statistic when certain 

characteristics are known about a population. In future 

chapters, we will test claims and estimate population 

parameters using what we have learned about these sampling 

distributions, even when population characteristics 

are unknown. 

The sampling distribution of p^ , the sample proportion, 

and x, the sample mean, are of particular importance for 

Chapters 7–9. In those chapters, the values of unknown 

population proportions and population means will be estimated, 

and claims about them will be tested. Furthermore, 

estimates will be made and claims will be tested about the 

difference in two proportions and the difference in two 

means. 

Pacing and Assignment Guide 

Day Lesson Learning Targets/Classroom Activities Suggested Assignment 

1 Ch. 6 Introduction Lesson 6.1 Activity: A penny for your thoughts? None 

2 6.1 What Is a Sampling 

Distribution? 

• Distinguish between a parameter and a statistic. 

• Create a sampling distribution using all possible samples from a 

small population. 

• Use the sampling distribution of a statistic to evaluate a claim 

about a parameter. 

1–15 odd, 19 

3 6.2 Introduction Lesson 6.2 Activity: How many craft sticks are in the bag? None 

4 6.2 Sampling 

Distributions: Center 

and Variability 

5 6.3 Sampling 

Distribution of a Sample 

Count (The Normal 

Approximation to the 

Binomial Distribution) 

• Determine if a statistic is an unbiased estimator of a population 

parameter. 

• Describe the relationship between sample size and the variability 

of a statistic. 

• Calculate the mean and the standard deviation of the sampling 

distribution of a sample count and interpret the standard 

deviation. 

• Determine if the sampling distribution of a sample count is 

approximately normal. 

• If appropriate, use the normal approximation to the binomial 

distribution to calculate probabilities involving a sample count. 

1–15 odd, 19 

1–15 odd, 19 

6-2 

C H A P T E R 6 • Sampling Distributions 

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6 Flex Day Consider giving Quiz 6A: Lessons 6.1–6.3, showing one or more of 

the online videos listed in the Additional Chapter 6 Resources, or 

re-teaching (if needed). 

Optional assignment: 

6.1 Ex12, 6.2 Ex10, 6.2 Ex14, 

6.3 Ex14 

7 6.4 The Sampling 

Distribution of the 

Sample Proportion 

• Calculate the mean and standard deviation of the sampling 

distribution of a sample proportion p^ and interpret the 

standard deviation. 

• Determine if the sampling distribution of p^ is approximately 

normal. 

• If appropriate, use a normal distribution to calculate probabilities 

involving p^ . 

1–15 odd, 19 

8 6.5 The Sampling 

Distribution of the 

Sample Mean 

• Find the mean and standard deviation of the sampling distribution 

of a sample mean x and interpret the standard deviation. 

• Use a normal distribution to calculate probabilities involving x 

when sampling from a normal population. 

1–15 odd, 19 

9 6.6 The Central Limit 

Theorem 

• Determine if the sampling distribution of x is approximately 

normal when sampling from a non-normal population. 

• If appropriate, use a normal distribution to calculate probabilities 

involving x. 

1–15 odd, 19 

10 Flex Day This is a great day to have students work in groups on the STATS 

applied! at the end of Lesson 6.6. Also consider showing one 

or more of the online videos listed in the Additional Chapter 6 

Resources, giving Quiz 6B: Lessons 6.4–6.6, or re-teaching (if 

needed). 

Optional assignment: 

6.4 Ex14, 6.5 Ex10, 6.5 Ex12, 

6.6 Ex10, 6.6 Ex12 

11 Ch. 6 Review Ch. 6 Practice Test Ch. 6 Review Exercises 1–6 

12 Ch. 6 Test Ch. 6 Test 

To save time, it is possible to skip Lessons 6.2 and 6.3 

without losing much continuity with future chapters. However, 

the other lessons in this chapter are crucial to understanding 

much of the remainder of the course. 

Four kinds of exercises end each lesson: Mastering Concepts 

and Skills, Applying the Concepts, Extending the Concepts, 

and Recycle and Review. They have been written so 

that assigning the odd-numbered exercises provides appropriate 

practice. Mastering Concepts and Skills exercises 

address a single learning target, while Applying the Concepts 

exercises address two or more learning targets. Recycle 

and Review exercises reinforce concepts learned earlier. For 

exceptional or motivated students, Extending the Concepts 

exercises are a good way to differentiate instruction. 

The even-numbered exercises form a pair with the preceding 

odd exercise in the Mastering Concepts and Skills 

and Applying the Concepts sections so that another full 

assignment can be created from the even-numbered exercises 

of these types. Consider using the even-numbered 

exercises to spiral into future lessons, for re-teaching, or 

as additional practice for students. The answers to the 

odd-numbered exercises appear in the back of the student 

textbook, while the answers to the even-numbered exercises 

do not. The answers to all of the Chapter Review 

Exercises and the Chapter Test are in the back of the 

student textbook so that students can check their own 

progress. 

C H A P T E R 6 • Sampling Distributions 6-3 


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Promoting Good Habits and Skills 

Chapter 6 is about the characteristics of sampling distributions. 

The sampling distribution of the sample proportion 

and sample mean will play a key role in Chapters 7–9, so 

understanding them is very important for future success. 

Here are some important habits to develop in your students 

as you teach Chapter 6: 

1. Pay attention to the vocabulary: Understanding the differences 

among population, parameter, sample, and statistic is 

absolutely vital to understanding this and future chapters. 

Also, make sure your students understand the difference 

between the distribution of the population, the distribution 

of a single sample, and the sampling distribution. They are 

not the same! 

2. Emphasize symbols, not formulas: The symbols used in 

this chapter are quite standard in all of statistics. While we 

don’t want the symbols to overwhelm students, they are 

an integral part of basic statistical practice. The symbols 

n, p, p^ , m, s, and x will be used extensively in Chapters 

7–9. Being comfortable with them now will be of great 

help later. On the other hand, don’t have students memorize 

the formulas for the mean and standard deviations of 

the sampling distributions in this chapter. Having students 

understand the symbols is far more important than having 

them memorize the formulas. 

3. Look for the underlying variable: If your students can 

recognize the underlying variable and classify it as categorical 

or quantitative, they can tell which sampling distribution 

is appropriate. Categorical variables (like color of 

a Reese’s Pieces ® candy) lead to sample counts and sample 

proportions. Quantitative variables (like the year a penny 

was manufactured) lead to sample means. Developing this 

skill now will pay dividends in future chapters as well! 

4. Think back to the simulations: Because sampling distributions 

are very abstract, simulating the sampling process can 

provide insight for students. If students have trouble understanding 

the different distributions in sampling situations, 

have them think back to concrete simulations like the 

“A penny for your thoughts?” activity in Lesson 6.1 and 

computer simulations with software like the SPA applets. 

Physical simulations are so important for student understanding 

that if you were to do only one activity in the entire 

chapter, it should be the penny activity. 

5. Watch the conditions: Students will often pay little attention 

to the conditions about sampling distributions. Have 

them focus on the Large Counts condition and the Normal/ 

Large Sample condition because these concepts will be used 

repeatedly in future chapters. If students don’t pay attention 

to them now, they will have a more difficult time in the 

future. 

Lesson-by-Lesson Content 

Overview 

Lesson 6.1 What Is a Sampling Distribution? 

A large collection of individuals is called a population. 

A subset of that population is called a sample. A number 

that measures some characteristic of a population is 

called a parameter, while a numerical measure of some 

characteristic of a sample is called a statistic. Statistics 

vary from sample to sample. The distribution of values 

taken on by a statistic from every possible sample of a 

given size is called the sampling distribution of that statistic. 

We can evaluate claims about a parameter by calculating 

probabilities from the sampling distribution of 

the corresponding statistic. 

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Lesson 6.2 Sampling Distributions: 

Center and Variability 

If the mean of the sampling distribution of a statistic is equal 

to the corresponding population parameter, the statistic is 

said to be an unbiased estimator of the population parameter. 

Otherwise, the statistic is a biased estimator of the population 

parameter. The standard deviation of a sampling distribution 

of a statistic measures the variability of a statistic. The smaller 

the standard deviation, the more precise the estimate of the 

parameter. The variability of the sampling distribution of a 

statistic will decrease as sample size increases. 

Lesson 6.3 The Sampling Distribution of a Sample 

Count (The Normal Approximation to 

the Binomial) 

Let the random variable X be the count of successes in a sample 

of size n, where p is the probability of a success on a single 

trial. The sampling distribution of X will have mean m X = np 

and standard deviation s X = "np(1 − p). The Large Counts 

condition states that the sampling distribution of X will have 

an approximately normal distribution whenever np ≥ 10 

and n(1 2 p) ≥ 10. When the sampling distribution of X 

is approximately normal, probabilities involving the sample 

count X may be approximated by a normal distribution. 

Lesson 6.4 The Sampling Distribution of a Sample 

Proportion 

Let the random variable p^ be the proportion of successes in 

a sample size of size n, where p is the proportion of successes 

in the population. The sampling distribution of p^ will have 

p(1 − p) 

mean m p^ = p and standard deviation s p^ = . The 

Å n 

Large Counts condition states that the sampling distribution 

of p^ will have an approximately normal distribution whenever 

np ≥ 10 and n(1 2 p) ≥ 10. When the sampling distribution 

of p^ is approximately normal, probabilities involving 

the sample proportion p^ may be approximated by a normal 

distribution. 

Lesson 6.5 The Sampling Distribution of a 

Sample Mean 

Let the random variable x be the sample mean in a sample 

of size n from a population with mean m and standard 

deviation s. The sampling distribution of x will have mean 

m x = m and standard deviation s x = s . If the population 

"n 

is normal, then the sampling distribution of x will be normal. 

When the sampling distribution of x is exactly normal, 

probabilities involving the sample mean x may be calculated 

using a normal distribution. 

Lesson 6.6 The Central Limit Theorem 

The Central Limit Theorem states that when sampling from 

a non-normal population, the sampling distribution of x is 

approximately normal when the sample size is large. As a rule 

of thumb, when sampling from non-normal populations, we 

will consider the sampling distribution of x to be approximately 

normal when n ≥ 30. When the sampling distribution 

of x is approximately normal, probabilities involving the sample 

mean x may be approximated by a normal distribution. 



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Chapter 6 Resources 

SPA Applets 

highschool.bfwpub.com/spa3e 

• The Normal Approximation to the Binomial applet 

allows students to view a binomial probability 

distribution with parameters n and p and a normal 

probability distribution with the same mean and 

standard deviation superimposed on the binomial. 

Sliders allow students to easily change the values of 

n and p. 

• The Probability applet computes probabilities for normal 

distributions, which are used frequently in Chapter 6. 

Teacher’s Resource Materials 

The following resources can be found by clicking on the 

links in the Teacher’s e-Book (TE-book), logging into 

LaunchPad (password required) highschool.bfwpub.com 

/launchpad/spa3e, or opening the Teacher’s Resource Flash 

Drive (TRFD). 

• Chapter Videos 

• Chapter 6 Overview video (for teachers) 

• Lesson Overview videos for Lessons 6.1–6.3 and 

Lessons 6.4–6.6 (for teachers but fine to share with 

students, if desired) 

• Worked Example videos for every example 

(for students and teachers) 

• Chapter 6 Review Exercise videos (for students and 

teachers) 

• Alternate Examples 

All of the Chapter 6 Alternate Examples are provided 

in a Word document. Use these as additional examples 

in class, as the basis for assessments, or as additional 

practice for students. 

• Lesson App Handout 

All of the Chapter 6 Lesson Apps are provided in PDF 

format. The Lesson Apps assess each learning target 

in the lesson. Print these for use as exit tickets or as a 

performance task for individuals or groups of students. 

Each Lesson App can also be used as formative assessment. 

• Teacher’s Resource Material Documents 

• Lesson 6.1 Activity Overview for teachers 

• Lesson 6.1 Activity Handout [Use with Lesson 6.1.] 

• Lesson 6.2 Activity Fathom file [Use with Lesson 6.2.] 

• Chapter 6 Activity: Sampling Movies [Use after 

Lesson 6.4.] 

• Sampling Distributions Summary Chart [Use with 

Lesson 6.5.] 

• Chapter 6 Activity: Sampling Movies (The Sequel) 

[Use after Lesson 6.6.] 

• Chapter 6 Learning Targets Grid 

• Lecture Presentation Slides—one prepared PowerPoint 

presentation per lesson (for teachers) 

• Chapter Quizzes and Tests 

• Quiz 6A: Lessons 6.1–6.3 

• Quiz 6B: Lessons 6.4–6.6 

• Chapter 6 Test 

• Chapter 6 Answers to Quizzes and Tests 

• Full Solutions to Exercises—the worked solutions file for 

each lesson and end-of-chapter exercises and test are 

provided. 

• Chapter Data Files 

• Additional Chapter Resources 

We have created a list of third-party videos and other 

resources to support the content in this chapter. The 

Word document includes clickable URLs to help you 

access this external content. (Note: All of the URLs 

were live when this book was published.) 

6-6 



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Notes 



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PD Chapter 6 Overview 

Watch the chapter overview video for 

guidance from the authors on teaching 

the content in this chapter. Find it in the 

Teacher’s Resource Materials by clicking 

on the link in the TE-book, logging into 

the Teacher’s Resource site highschool 

.bfwpub.com/launchpad/spa3e, 

or accessing it on the TRFD. 

6 

Sampling 

Distributions 

Lesson 6.1 What Is a Sampling Distribution? 400 

Lesson 6.2 Sampling Distributions: Center and Variability 409 

Lesson 6.3 The Sampling Distribution of a Sample Count 

(The Normal Approximation to the Binomial) 417 

Lesson 6.4 The Sampling Distribution of a Sample Proportion 424 

Lesson 6.5 The Sampling Distribution of a Sample Mean 432 

Lesson 6.6 The Central Limit Theorem 439 

Chapter 6 Main Points 445 

Chapter 6 Review Exercises 447 

Chapter 6 Practice Test 448 

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Steve Gorton and Gary Ombler/Getty Images 

STATS applied! 

How can we build “greener” batteries? 

Kids love getting toys for their birthdays, especially electronic ones that have flashing lights 

and make loud noises. But these devices require lots of power and can drain batteries quickly. 

Battery manufacturers are constantly searching for ways to build longer-lasting batteries. 

When the manufacturing process is working correctly, AA batteries from a particular 

company should last an average of 17 hours, with a standard deviation of 0.8 hours. Also, 

at least 73% of the batteries should last 16.5 hours or more. 

Quality-control inspectors select a random sample of 50 batteries during each hour of 

production and then drain them under conditions that mimic normal use. The graph and 

summary statistics describe the distribution of the lifetimes (in hours) of the batteries from 

one sample of 50 AA batteries. 

Frequency 

12 

10 

8 

6 

4 

2 

15.0 15.5 16.0 

16.5 17.0 17.5 18.0 18.5 

Lifetime (h) 

n Mean SD min Q 1 med Q 3 max 

50 16.718 0.66 15.46 16.31 16.7 17.28 17.98 

Do these data suggest that the production process isn’t working properly? Or is it safe 

for plant managers to send out all the batteries produced in this hour for sale? 

Teaching Tip: 


The STATS applied! feature is designed 

to appeal to students and preview 

interesting questions that statistics 

can answer. To answer this STATS 

applied!, students must make inferences 

about a population from sample data. 

Knowledge of the sampling distribution 

of the sample proportion p^ and the 

sampling distribution of the sample 

mean x are needed to answer this 

question, although students won’t 

understand why until the end of the 

chapter. However, students should 

understand that this question is 

about quality control, an important 

statistical application for manufacturing 

businesses. 

We’ll revisit STATS applied! at the end of the chapter, so you can use what you have learned to help 

answer these questions. 

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399 


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PD LESSONS 6.1–6.3 Overview 

Watch the Lessons 6.1–6.3 overview 

video for guidance on teaching the 

content in these lessons. Find it in the 

Teacher’s Resource Materials by clicking 


the Teacher’s Resource site, or accessing 

it on the TRFD. 

Lesson 6.1 

What is a Sampling 

Distribution? 

L e A r n i n g T A r g e T S 

d Distinguish between a parameter and a statistic. 

d Create a sampling distribution using all possible samples from a small 

population. 

d Use the sampling distribution of a statistic to evaluate a claim about a 

parameter. 

Learning Target Key 

The problems in the test bank are 

keyed to the learning targets using 

these numbers: 

d 6.1.1 

d 6.1.2 

d 6.1.3 

BELL RINGER 

What is the difference between random 

sampling and random assignment when 

collecting data? What inferences can be 

made in each case? Discuss your answers 

with a partner. 

AcT iviT y 

A penny for your thoughts? 

In this activity, your class will investigate how the 

mean year x and the proportion of pennies from the 

2000s p^ vary from sample to sample, using a large 

population of pennies of various ages. 1 

1. Have each member of the class randomly select 1 

penny from the population and record the year of 

the penny with an “X” on the dotplot provided by 

your teacher. Return the penny to the population. 

Repeat this process until at least 100 pennies have 

been selected and recorded. This graph gives you 

an idea of what the population distribution of 

penny years looks like. 

2. Have each member of the class take an SRS of 5 

pennies from the population and note the year on 

each penny. 

• Record the average year of these 5 pennies 

with an “x” on a new class dotplot. Make 

sure this dotplot is on the same scale as the 

dotplot in Step 1 above. 

• Record the proportion of pennies from 

the 2000s with a “p^ ” on a different dotplot 

provided by your teacher. 

Return the pennies to the population. Repeat 

this process until there are at least 100 x's and 

100 p^'s. 

3. Repeat Step 2 with SRSs of size n 5 20. Make sure 

these dotplots are on the same scale as the corresponding 

dotplots from Step 2 above. 

4. Compare the distribution of X (year of penny) 

with the two distributions of x (mean year). 

How are the distributions similar? How are they 

different? What effect does sample size seem to 

have on the shape, center, and variability of the 

distribution of x ? 

5. Compare the two distributions of p^ . How are the 

distributions similar? How are they different? What 

effect does sample size seem to have on the shape, 

center, and variability of the distribution of p^ ? 

Teaching Tip 

This activity is the most important in 

the whole chapter and worth the time 

it takes because it introduces students 

to sampling distributions by simulating 

repeated sampling from a population. 

We recommend spending an entire class 

period on this activity as the introduction 

to this chapter. As an alternative, 

consider starting it in Chapter 4 or 

Chapter 5 and do a little each day, as 

explained in the Activity Overview 

document. 

400 


Activity Overview 

Time: 40–50 minutes 

Materials: Large chart paper and markers, 

dot stickers, or bingo daubers to make a 

dotplot. Alternatively, you can make a class 

dotplot on the whiteboard. You will also need 

a population of pennies. You need a minimum 

of 600 pennies, but 1000 or more is ideal. 

Teaching Advice: See the Lesson 6.1 Activity 

overview and activity handout. 

To estimate the mean income of U.S. residents with a college degree, the Current 

Population Survey (CPS) selected a random sample of more than 60,000 people with 

at least a bachelor’s degree. The mean income in the sample was $69,609. 2 How close 

is this estimate to the mean income for all members of the population? To find out 

how an estimate varies from sample to sample, we want to gain some understanding 

of sampling distributions. 

TRM Lesson 6.1 Activity Overview 

for Teachers 

TRM Lesson 6.1 Activity Handout 

A detailed Activity Overview document with 

sample graphs for teachers, as well as an 

Activity Handout for students, is available for 

this important activity. Consider giving the 

handout to your students so they don’t look 

ahead in their books for ideas and hints. You 

can find these resources by clicking on the 

link in the TE-book, logging into the Teacher’s 

Resource site, or accessing them on the TRFD. 


400 



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L E S S O N 6.1 • What Is a Sampling Distribution? 401 

Parameters and Statistics 

For the sample of college graduates contacted by the CPS, the mean income was 

x 5 $69,609. The number $69,609 is a statistic because it describes this one sample. 

The population that the researchers want to draw conclusions about is all U.S. college 

graduates. In this case, the parameter of interest is the mean income m of the population 

of all college graduates. 

DEFINITION Statistic, Parameter 

A statistic is a number that describes some characteristic of a sample. 

A parameter is a number that describes some characteristic of the population. 

Because we can’t examine the entire population, the value of a parameter is usually 

unknown. To estimate the value of the parameter, we use a statistic calculated using 

data from a random sample of the population. 

Remember s and p: statistics come from samples, and parameters come from 

populations. The notation we use should reflect this distinction. For example, we 

write m (the Greek letter mu) for the population mean and x for the sample mean. 

The table lists some additional examples of statistics and their corresponding 

parameters. 

Teaching Tip 

Remind students that we have seen a 

“hat” before in this course. The estimated 

value of y is denoted y^ . Likewise, the 

estimated value of p is denoted p^ . 

TRM chapter 6 Alternate Examples 

You can find the Alternate Examples for 

this chapter in Microsoft Word format by 

clicking the link in the TE-book, logging 

into the Teacher’s Resource site, or 

accessing it on the TRFD. 

Alternate Example 

Lesson 6.1 

Sample statistic 

Population parameter 

x (the sample mean) estimates m (the population mean) 

p^ (the sample proportion) estimates p (the population proportion) 

s (the sample SD) estimates s (the population SD) 

How are teens different from turkeys? 

Parameters and statistics 

PROBLEM: Identify the population, the parameter, 

the sample, and the statistic in each of the following 

settings: 

(a) A Pew Research Center poll asked 1102 12- to 

17-year-olds in the United States if they have a cell 

phone. Of the respondents, 71% said “Yes.” 3 

(b) Tom is roasting a large turkey breast for a holiday 

meal. He wants to be sure that the turkey is 

safe to eat, which requires a minimum internal 

temperature of 165°F. Tom uses a thermometer 

to measure the temperature of the turkey breast 

at four randomly chosen points. The minimum 

reading he gets is 170°F. 

e XAMPLe 

SOLUTION: 

(a) Population: all 12- to 17-year-olds in the United 

States. Parameter: p 5 the proportion of all 12- to 

17-year-olds with cell phones. Sample: the 1102 

12- to 17-year-olds contacted. Statistic: the sample 

proportion with a cell phone, p^ 5 0.71. 

(b) Population: all possible locations in the turkey breast. 

Parameter: the true minimum temperature in all possible 

locations. Sample: the four randomly chosen locations. 

Statistic: the sample minimum, 170°F. 

FOR PRACTICE TRY EXERCISE 1. 

Pictures of coworkers? 

Parameters and statistics 

PROBLEM: Identify the population, 

parameter, sample, and statistic in each 

of the following settings: 

(a) A professional photographer is 

interested in the average number of 

photographs she took per day last year. 

She randomly selected 10 days from the 

year and recorded the number of photographs 

she took on each of the 10 days. 

The average number of photographs she 

took on those 10 days is 831.2 photos. 

(b) A Pew Research Center Poll asked a 

random sample of U.S. adults 18 or older 

whether they prefer to have a male 

coworker, a female coworker, or 

whether it doesn’t matter. Of the 2002 

respondents, 77% said it “doesn’t matter.” 

SOLUTION: 


Teaching Tip 

18/08/16 4:58 PM 

Point out the phrase “who would say” in the 

solution to part (b) of the alternate example. 

It is not correct to say that the parameter is 

“the true proportion of all U.S. adults 18 or 

older who said it ‘doesn’t matter.’” The true 

proportion who said it doesn’t matter is 77%, 

which is the statistic (the sample proportion). 

Tell students to be careful about using the 

past tense to describe parameters! 

(a) Population: all days last year. 

Parameter: m, the true average 

number of photographs per day the 

photographer took over all days last year. 

Sample: the 10 randomly chosen days. 

Statistic: the sample mean number of 

photographs per day, x 5 831.2 photos. 

(b) Population: all U.S. adults 18 or older. 

Parameter: p 5 the true proportion of all 

U.S. adults 18 or older who would say it 

“doesn’t matter.” 

Sample: the 2002 U.S. adults 18 or older 

who participated in the survey. 

Statistic: the sample proportion who said 

it “doesn’t matter,” p^ 5 0.77. 



11/01/17 3:53 PM

402 


Common Error 

The phrase “sampling distribution” 

sounds similar to “distribution of a 

sample,” but they mean very different 

things. In the “A penny for your 

thoughts?” activity, the distribution of 

a sample is distribution of year for the 

5 (or 20) pennies in a student’s hand. 

The dotplots of the sample means and 

sample proportions created by the class 

are examples of sampling distributions. 

While some parameters and statistics have special symbols (such as p for the population 

proportion and p^ for the sample proportion), many parameters and statistics 

do not have their own symbol. To distinguish between a parameter and statistic, use 

descriptors such as “true” minimum and “sample” minimum as we did in the turkey 

example. 

Sampling Distributions 

In the Penny for Your Thoughts Activity, you encountered sampling variability— 

meaning that different random samples of the same size from the same population 

produce different values of a statistic. The statistics that come from these samples 

form a sampling distribution. 

DEFINITION Sampling distribution 

The sampling distribution of a statistic is the distribution of values taken by the statistic 

in all possible samples of the same size from the same population. 


Disproportionate males? 

Sampling distributions 

PROBLEM: There are six employees in 

a small company, Atsuko, Bernadette, 

Carlos, Dandre, Easton, and Freddie. 

Atsuko and Bernadette are female 

and the others are male. List all 15 

possible SRSs of size n 5 4, calculate the 

proportion of males for each sample, and 

display the sampling distribution of the 

sample proportion on a dotplot. 

SOLUTION: 

Sample 1: A, B, C, D p^ 5 0.50 

Sample 2: A, B, C, E p^ 5 0.50 

Sample 3: A, B, C, F p^ 5 0.50 

Sample 4: A, B, D, E p^ 5 0.50 

Sample 5: A, B, D, F p^ 5 0.50 

Sample 6: A, B, E, F p^ 5 0.50 

Sample 7: A, C, D, E p^ 5 0.75 

Sample 8: A, C, D, F p^ 5 0.75 

Sample 9: A, C, E, F p^ 5 0.75 

Sample 10: A, D, E, F p^ 5 0.75 

Sample 11: B, C, D, E p^ 5 0.75 

Sample 12: B, C, D, F p^ 5 0.75 

Sample 13: B, C, E, F p^ 5 0.75 

Sample 14: B, D, E, F p^ 5 0.75 

Sample 15: C, D, E, F p^ 5 1.00 

a 

e XAMPLe 

Just how tall are their sons? 

Sampling distributions 

Remember that a distribution describes the possible values of a variable and how 

often these values occur. The easiest way to picture a distribution is with a graph, such 

as a dotplot or histogram. 

PROBLEM: John and Carol have four grown sons. Their heights (in inches) are 71, 75, 72, and 

68. List all 6 possible SRSs of size n 5 2, calculate the mean height for each sample, and display 

the sampling distribution of the sample mean on a dotplot. 

SOLUTION: 

Sample 1: 71, 75 x 5 73 Sample 4: 75, 72 x 5 73.5 

Sample 2: 71, 72 x 5 71.5 Sample 5: 75, 68 x 5 71.5 

Sample 3: 71, 68 x 5 69.5 Sample 6: 72, 68 x 5 70 

FigUre 6.1 Dotplot 

showing the sampling 

distribution of the 

sample range of height 

for SRSs of size n 5 2. 


69 70 71 72 73 74 

Sample mean height (in.) 


Every statistic has its own sampling distribution. For example, Figure 6.1 shows 

the sampling distribution of the sample range of height for SRSs of size n 5 2 from 

John and Carol’s four sons. 

Sample 1: 71, 75 sample range 5 4 Sample 4: 75, 72 sample range 5 3 



d d 

d d d d 

0 1 2 3 4 5 6 7 8 

Sample range of height (in.) 


d 

d 

d 

d 

d 

d 

d 

d 

0.5 0.6 0.7 0.8 0.9 1.0 

Sample proportion of men 

402 



11/01/17 3:53 PM


Be specific when you use the word “distribution.” There are three different types of distributions 

in this setting: 

1. The distribution of height in the population (the four heights): 

d d d d 

67 68 69 70 71 72 73 74 75 76 

Height (in.) 

2. The distribution of height in a particular sample (two of the heights): 

d 

67 68 69 70 71 72 73 74 75 76 

Height (in.) 

3. The sampling distribution of the sample range for all possible samples (the six 

sample ranges): 

d d 

d d d d 

0 1 2 3 4 5 6 7 8 

Sample range of height (in.) 

Notice that the first two distributions consist of heights (data values), while the third 

distribution consists of ranges (statistics). Lesson: Always use “the distribution of __” 

and never just “the distribution.” 

d 

cAutIOn 

! 

Common Error 

Emphasize the difference in the three 

distributions shown here. It will be 

difficult, but important, for students to 

do this. Ask students to describe what 

the leftmost dot represents in each 

graph. In graph 1, it represents the 

height of a son (in the population of 

four sons) who is 68 inches tall. In graph 

2, it represents the height of a son (in 

a sample of two sons) who is 71 inches 

tall. In graph 3, it represents the sample 

range of heights for the sample of the 

two sons who are 71 and 72 inches 

tall. Each dot in dotplot 3 represents a 

statistic from a sample, not a value from 

a single individual. 

Lesson 6.1 

Using Sampling Distributions to Evaluate Claims 

Sampling distributions are the foundation for the methods of statistical inference you 

will learn about in Chapters 7–10. Knowing the sampling distribution of a statistic 

will help us know how much the statistic tends to vary from its corresponding parameter 

and what values of the statistic should be considered unusual. 

How long will we bead doing the homework? 

Evaluating a claim 

PROBLEM: At the beginning of class, Mrs. Chauvet shows her 

class a box filled with black and white beads. She claims that 

the proportion of black beads in the box is p 5 0.50. To determine 

the number of homework exercises she will assign that 

evening, she invites a student to select an SRS of n 5 30 beads 

from the box. The number of black beads selected will be the 

number of homework exercises assigned. When the student 

selects 19 black beads (p^ 5 19/30 5 0.63), the students groan 

and suggest that Mrs. Chauvet included more than 50% black 

beads in the box. 

To determine if a sample proportion of p^ 5 0.63 provides convincing evidence that Mrs. Chauvet 

cheated, the class simulated 100 SRSs of size n 5 30, assuming that she was telling the truth. That is, 

they sampled from a population with 50% black beads. For each sample, they recorded the sample 

proportion of black beads. The results of the simulation are shown on the next page. 

e XAMPLe 

© Monalyn Gracia/Corbis 



What’s in the box? 

Evaluating a claim 

PROBLEM: At the end of class, Mr. Osters 

allows one student to select a ticket from 

a shoebox without looking. The tickets are 

labeled either “Homework pass” or “Try 

again.” Once a ticket is drawn, it is replaced 

for the next drawing and the tickets are 

mixed thoroughly. Mr. Osters claims that 

the proportion of homework passes in 

the shoebox is p 5 0.25. At the end of the 

first quarter, one student noted that only 

6 students won in 50 drawings ( p^ 5 0.12). 

The students were suspicious that less than 

25% of the tickets in the box are homework 

passes. 

18/08/16 4:58 PM 

To determine if a sample proportion of 

p^ 5 0.12 provides convincing evidence that 

the true proportion of homework passes is 

less than 25%, the class simulated 100 SRSs of 

size n 5 50, assuming that 25% of the tickets 

were homework passes. For each sample, they 

recorded the sample proportion of homework 

passes. Here are the results of the simulation: 

d d dddddd d 

d d d d d d 

d d d d d d 

d 

d d d d d d d 

d d d d d d d 

d d d d d d d 

d d d d d d d d 

d d d d d d d d d d 

d d d d d d d d d d d 


d 

d d d d d d d d d d d d 

d d d d 

0.10 0.15 0.20 0.25 0.30 0.35 0.40 

Sample proportion of 

homework passes 

(a) There is one dot on the graph at p^ 5 0.38 

Explain what this dot represents. 

(b) Would it be unusual to get a sample 

proportion of 0.12 or less in a sample of 

size 50 when p 5 0.25? Explain. 

(c) Based on your answer to part (b), is 

there convincing evidence that Mr. Osters 

lied about the contents of the shoebox? 

SOLUTION: 

(a) In one SRS of size n 5 50, 38% of the 

tickets were homework passes. 

(b) Yes; in the 100 trials of the 

simulation, only 2 of the SRSs included 

12% or fewer homework passes. 

(c) Yes; because the probability from part 

(b) is small—only 0.02—it is not plausible 

that the proportion of homework passes in 

the shoebox is p 5 0.25 and the students 

got a sample proportion of p^ 5 0.12 by 

chance alone. 



11/01/17 3:53 PM

404 


FYI 

Sampling distributions of statistics from 

most real populations are extremely 

large and therefore difficult to imagine. 

For example, if we were to sample 6 U.S. 

senators from the population of 100 

current senators, there are 1,192,052,400 

possible samples. In this case, a sampling 

distribution would have 1,192,052,400 

different dots in its dotplot! When 

sampling distributions are very large, 

we use simulations to create a good 

approximation. 

SOLUTION: 

(a) 

(b) 

(c) 

(a) There is one dot on the graph at p^ 5 0.77. Explain 

what this dot represents. 

(b) Would it be unusual to get a sample proportion 

of 0.63 or higher in a sample of size 30 when 

d d dddddddddd d ddddddd d dddddddddddd p 5 0.50? Explain. 

d 

d 

d 

d 

d d d d d d d d d 




d d d d d d d d d d d d d 

d d d d d d d d d d d d d d 

0.2 0.3 0.4 0.5 0.6 0.7 0.8 

Sample proportion of black beads 

In one simulated SRS of size n 5 30, 77% of the beads 

were black. 

No. In the 100 simulated samples, 9 of the SRSs included 

at least 63% black beads. 

No. Because the probability from part (b) isn’t that small, 

it is plausible that the proportion of black beads in the box 

is p 5 0.50 and the student got a sample proportion of 

(c) Based on your answer to part (b), is there convincing 

evidence that Mrs. Chauvet lied about the 

contents of the box? 

Notice that 9 of the 

100 simulated SRSs 

d d dd resulted in a sample 

d d 

d d proportion of 0.63 

d d d 

d d d 

or higher. 

d d d 

d d d 

d d d d 

d d d d 

d d d d 

d d d d d 







p^ 5 0.63 by chance alone. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 

Sample proportion of black beads 


We used 100 simulated samples to produce the dotplot of sample proportions in 

this example. Because it doesn’t include all possible samples of size 30, it is only an 

approximation of the actual sampling distribution of p^ . Thankfully, the simulated 

sampling distribution should be a good approximation as long as we use a large number 

of samples in the simulation. 

L e SSon APP 6. 1 

How cold is it inside the cabin? 

During the winter months, outside temperatures at the Starneses' cabin in 

Colorado can stay well below freezing (32°F, or 0°C) for weeks at a time. To 

prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The 

manufacturer claims that the thermostat allows variation in home temperature 

that follows a normal distribution with s 5 3°F. To test this claim, Mrs. Starnes 

programs her digital thermostat to take an SRS of n 5 10 readings during a 

24-hour period. The standard deviation of the results is s x 5 5°F. 

Quasarphoto/Getty Images 

TRM Chapter 6 Lesson App Handout 

All of the Chapter 6 Lesson Apps can 

be found by clicking on the link in the 

TE-book, logging into the Teacher’s 

Resource site, or accessing this resource 

on the TRFD. The Lesson Apps assess all 

learning targets in the lesson, so they 

are excellent resources to gauge student 

understanding. Use them as a formative 

evaluation at the end of each lesson to 

help you and your students understand 

exactly which learning targets are 

challenging and which are not. 

1. Identify the population, the parameter, the sample, and the statistic in this context. 

Suppose the thermostat is working properly and that the temperatures in the cabin vary according to a normal 

distribution with mean m 5 50°F and standard deviation s 5 3°F. The dotplot shows the distribution of the sample 

standard deviation in 100 simulated SRSs of size n 5 10 from this distribution. 

Lesson App 


Answers 

1. Population: All possible times during 

the 24-hour period. Parameter: s 5 the 

true standard deviation of temperature 

readings at all possible times during the 

24-hour period. Sample: The SRS of 10 times. 

Statistic: The sample standard deviation of 

temperature readings, S x 5 5°F. 

2. Yes; in the 100 simulated samples, 0 of 

the SRSs had a sample standard deviation 

of 5°F or higher. Based on the simulation, 

P(s x > 5) = 0∙100 = 0. 

3. Yes; because the probability from Question 

2 is small, it is not plausible that the true 

standard deviation is s 5 3°F and 

Mrs. Starnes got the sample standard 

deviation of S x 5 5°F by chance alone. 


404 



11/01/17 3:53 PM



d 

2. Would it be unusual to get a sample standard 

d d ddd deviation of s x 5 5°F or higher in a sample of size 

n 5 10 when s 5 3°F? Explain. 

d dddd d d 

dd dddd ddddd d d d 

d dd ddddddddddddd dd d 

d dd ddddddddddddddddd dd 

ddddddddddddddddddddddddddddddddd 

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 

Sample standard deviation of temperature (°F) 

Lesson 6.1 

3. Based on your answer to Question 2, is there convincing 

evidence that the thermometer is more 

variable than the manufacturer claims? Explain. 

WhAT DiD y o U LeA rn? 

LEARNINg TARgET EXAMPLES EXERCISES 

Distinguish between a parameter and a statistic. p. 401 1–4 

Create a sampling distribution using all possible samples from a small 

population. 

Use the sampling distribution of a statistic to evaluate a claim about a 

parameter. 

Exercises 

Mastering Concepts and Skills 

For Exercises 1–4, identify the population, the parameter, 

the sample, and the statistic in each setting. 

1. Smoking and height 

(a) From a large group of people who signed a card 

saying they intended to quit smoking, a random 

sample of 1000 people was selected. It turned out 

that 210 (21%) of the sampled individuals had not 

smoked over the past 6 months. 

(b) A pediatrician wants to know the 75th percentile 

for the distribution of heights of 10-year-old boys, 

so she selects a sample of 50 10-year-old male 

patients and calculates that the 75th percentile in 

the sample is 56 inches. 

2. Unemployment and gas prices 

(a) Each month, the Current Population Survey 

interviews a random sample of individuals in 

about 60,000 U.S. households. One of its goals 

is to estimate the national unemployment rate. In 

January 2015, 5.7% of those interviewed were 

unemployed. 

(b) How much do gasoline prices vary in a large 

city? To find out, a reporter records the price per 

gallon of regular unleaded gasoline at a random 

sample of 10 gas stations in the city on the same day. 

pg 401 

Lesson 6.1 

p. 402 5–8 

p. 403 9–12 

The range (Maximum – Minimum) of the prices in 

the sample is 25 cents. 

3. Tea and screening 

(a) On Tuesday, the bottles of iced tea filled in a plant were 

supposed to contain an average of 20 ounces of iced 

tea. Quality-control inspectors sampled 50 bottles at 

random from the day’s production. These bottles contained 

an average of 19.6 ounces of iced tea. 

(b) On a New York–Denver flight, 8% of the 125 passengers 

were selected for random security screening 

before boarding. According to the Transportation 

Security Administration, 10% of passengers at this airport 

are supposed to be chosen for random screening. 

4. Bearings and thermostats 

(a) A production run of ball bearings is supposed to 

have a mean diameter of 2.5000 centimeters. An 

inspector chooses a random sample of 100 bearings 

from the container and calculates a mean diameter 

of 2.5009 centimeters. 

(b) During the winter months, Mrs. Starnes sets the thermostat 

at 50°F to prevent the pipes from freezing in 

her cabin. She wants to know how low the interior 

temperature gets. A digital thermometer records the 

indoor temperature at 20 randomly chosen times 

during a given day. The minimum reading is 38°F. 

Teaching Tip 

18/08/16 4:59 PM 

In part (a) of Exercise 3, the value of the 

parameter is not necessarily 20. It is a target 

the company is hoping to achieve, but it may 

not be the actual average number of ounces 

in the bottles. Likewise, in part (b), 10% may 

not be the true proportion of all passengers 

selected for a security screening, so it is not 

necessarily the actual parameter value. 

TRM Full Solutions to Lesson 6.1 

Exercises 

You can find the full solutions for this 

lesson by clicking on the link in the 



on the TRFD. 

Answers to Lesson 6.1 Exercises 

1. (a) Population: All people who 

signed a card saying that they intend to 

quit smoking. Parameter: p 5 the true 

proportion of the population who quit 

smoking. Sample: A random sample 

of 1000 people who signed the cards. 

Statistic: The proportion of the sample 

who quit smoking; p^ 5 0.21. 

(b) Population: All 10-year-old boys. 

Parameter: The true 75th percentile of all 

10-year-old boys. Sample: Sample of 50 

patients. Statistic: The 75th percentile of 

the sample, 56 inches. 

2. (a) Population: Individuals in 

U.S. households. Parameter: p 5 true 

proportion of the U.S. population who 

are unemployed. Sample: A random 

sample of individuals from 60,000 U.S. 

households. Statistic: The proportion of 

the sample who were unemployed; 

p^ 5 0.057. 

(b) Population: All gasoline stations in 

a large city. Parameter: True range of 

gas prices at all gasoline stations in the 

city. Sample: A random sample of 10 gas 

stations in the city. Statistic: The range 

of prices in the sample; sample range 5 

25 cents. 

3. (a) Population: All bottles of iced tea 

filled in a plant on Tuesday. Parameter: 

m 5 the true mean amount of tea in the 

population. Sample: A random sample of 

50 bottles. Statistic: The mean amount of 

tea in the sample; x 5 19.6 ounces. 

(b) Population: All passengers in the 

airport. Parameter: p 5 the true proportion 

of the population who are chosen for 

random screening. Sample: The 125 

passengers on a New York-to-Denver flight. 

Statistic: The proportion of the sample 

selected for security screening; p^ 5 0.08. 

4. (a) Population: All ball bearings in 

the production run. Parameter: m 5 the 

true mean diameter in the population. 

Sample: A random sample of 100 

bearings. Statistic: The mean diameter in 

the sample; x 5 2.5009 cm. 

(b) Population: All possible times during 

the given day. Parameter: The true minimum 

temperature during the 24-hour period. 

Sample: The 20 randomly chosen times 

during the day. Statistic: The minimum 

temperature in the sample 5 38°F. 

Lesson 6.1 



11/01/17 3:53 PM

406 


Exercises 5–8 refer to the following population of 2 

Teaching Tip 

male students and 3 female students, along with their 

quiz scores: 

Exercises 5–8 are very important because 

Abigail 10 Bobby 5 Carlos 10 DeAnna 7 Emily 9 

students can create (and see) an entire 

5. Sample means List all 10 possible SRSs of size 

sampling distribution. Don’t skip these 

pg 402 n 5 2, calculate the mean quiz score for each sample, 

and display the sampling distribution of the sample 

exercises! 

mean on a dotplot. 

6. Sample ranges List all 10 possible SRSs of size 

5. 

n 5 3, calculate the range of quiz scores for each 

sample, and display the sampling distribution of 

Sample #1: Abigail (10), Bobby (5) x 5 7.5 

the sample range on a dotplot. 

Sample #2: Abigail (10), Carlos (10) x 5 10 

7. Sample proportions List all 10 possible SRSs of size 

n 5 2, calculate the proportion of females for each 

Sample #3: Abigail (10), DeAnna (7) x 5 8.5 


Sample #4: Abigail (10), Emily (9) x 5 9.5 

the sample proportion on a dotplot. 

8. Sample medians List all 10 possible SRSs of size 

Sample #5: Bobby (5), Carlos (10) x 5 7.5 

n 5 3, calculate the median quiz score for each 

Sample #6: Bobby (5), DeAnna (7) x 5 6 


the sample median on a dotplot. 

Sample #7: Bobby (5), Emily (9) x 5 7 

9. Who does their homework? A school newspaper 

Sample #8: Carlos (10), DeAnna (7) x 5 8.5 

pg 403 article claims that 60% of the students at a large 

high school completed their assigned homework 

Sample #9: Carlos (10), Emily (9) x 5 9.5 

last week. Some statistics students want to investigate 

if this claim is true, so they choose an SRS of 

Sample #10: DeAnna (7), Emily (9) x 5 8 

100 students from the school to interview. When 

d d d 

they found that only 45 of the 100 students completed 

their assigned homework last week, they 

d d d d d d d 

6 6.5 7 7.5 8 8.5 9 9.5 10 

suspected that the proportion of all students who 

Sample mean quiz score 

completed their assigned homework last week is 

less than the 60% claimed by the newspaper. 

6. 

To determine if a sample proportion of p^ 5 0.45 

provides convincing evidence that the true proportion 

is less than p 5 0.60, the class simulated 250 

Sample #1: Abigail (10), range 5 5 

Bobby (5), Carlos (10) 

SRSs of size n 5 100 from a population in which 

p 5 0.60. Here are the results of the simulation. 


d 

d 

Bobby (5), DeAnna (7) 

d 


d d d d 

d d 

d d d 

Bobby (5), Emily (9) 

dddd d d d d 

d d d d 

ddd 

d d d d d 

d d d d d d 



d d d 


d d d 

d d d 

Carlos (10), DeAnna (7) 

d d d 

d d d dddd d d d d d 

d d d 

d d 

d 

d d d d 

d d d d d d 

d d d d d d d 


d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d 





Carlos (10), Emily (9) 

dd dddddddddd dddddddddddddddd d dddddd d ddddddd 

d d d d d 



d dddddd d d dd d d d 


0.45 0.50 0.55 0.60 0.65 0.70 0.75 

DeAnna (7), Emily (9) 

Sample proportion of students who 

completed homework 

Sample #7: Bobby (5), range 5 5 


(a) There is one dot on the graph at 0.73. Explain what 

this dot represents. 


(b) Would it be surprising to get a sample proportion 

of 0.45 or less in an SRS of size 100 when 



p 5 0.60? Explain. 


Sample #10: Carlos (10), range 5 3 


d 

d dd 

d d d d 

1 1.5 2 2.5 3 3.5 4 4.5 5 


Sample range of quiz score 

8. 

Sample #1: Abigail (10), median 5 10 

7. 

Bobby (5), Carlos (10) 

Sample #1: Abigail, Bobby p^ 5 0.50 Sample #2: Abigail (10), median 5 7 

Sample #2: Abigail, Carlos 

Bobby (5), DeAnna (7) 

p^ 5 0.50 


Sample #3: Abigail, DeAnna p^ 5 1 

Bobby (5), Emily (9) 

Sample #4: Abigail, Emily p^ 5 1 


Sample #5: Bobby, Carlos p^ 5 0 



Sample #6: Bobby, DeAnna p^ 5 0.50 Carlos (10), Emily (9) 

Sample #7: Bobby, Emily p^ 5 0.50 Sample #6: Abigail (10), median 5 9 

Sample #8: Carlos, DeAnna p^ 5 0.50 DeAnna (7), Emily (9) 

Sample #9: Carlos, Emily Sample #7: Bobby (5), median 5 7 

p^ 5 0.50 


Sample #10: DeAnna, Emily p^ 5 1 

Sample #8: Bobby (5), median 5 9 


Sample #9: Bobby (5), median 5 7 

d d ddddd d dd DeAnna (7), Emily (9) 

0 0.5 1 

Sample #10: Carlos (10), median 5 9 

Sample proportion of females 



evidence that the proportion of all students 

who completed their assigned homework last week 

is less than p 5 0.60? Explain. 

10. First-serve percentage One important aspect of 

a tennis player’s effectiveness is her first-serve 

percentage—the proportion of the time the first of 

her two attempts to serve the ball to her opponent 

is successful. For her first three years on the 

tennis team, Shruti’s first-serve percentage is 53%. 

Hoping to improve, Shruti works over the summer 

with a coach who specializes in serves. In her 

first match of the next season, Shruti’s first serve 

is successful 42 times in 60 attempts, a first-serve 

percentage of 70%. 

Suppose we treat Shruti’s first 60 attempts 

as an SRS of her serves after working with the 

new coach. To determine if a sample proportion 

of p^ 5 0.70 provides convincing evidence that 

the true proportion is greater than p 5 0.53, we 

simulate 200 SRSs of size n 5 60 from a population 

in which p 5 0.53. Here are the results of 

the simulation. 

d 

d 

d 

d 

d 

d 

d d 

d d 

d d d d 

d d d d 

d d d d 

d ddddddddd d d d d 

d d d d 

d d d d 

d d 

dd 

d d d d 

d d d d d d d 


d 

d d ddd d d d d d d d d 


d d 

dd d d d d d d d d d d ddddddddddddddddd 

0.37 0.40 0.43 0.47 0.50 0.53 0.57 0.60 0.63 0.67 0.70 

Sample proportion of successful first serves 

d 

d 

d 

d 

d 

d 

d 

7 7.5 8 8.5 9 9.5 10 

Sample median of quiz scores 

9. (a) In one SRS of size n 5 100, 73% of the 

students did all their assigned homework. 

(b) Yes; in the 250 simulated samples, 0 

of the SRSs had a sample proportion of 

0.45 or lower. Based on the simulation, 

P( p^ ≤ 0.45) = 0∙250 = 0. 


(b) is small, it is not plausible that the true 

proportion is p 5 0.60 and the statistics 

students got a sample proportion of p^ 5 0.45 

by chance alone. 

d d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d d d 

d d d d d 



(b) Would it be surprising to get a sample proportion 

of 0.70 or more in an SRS of size 60 when 

p 5 0.53? Explain. 


evidence that Shruti’s first-serve percentage has 

improved since working with the new coach? Explain. 

11. Are we taller? According to the National Center 

for Health Statistics, the distribution of heights for 

16-year-old females is modeled well by a normal 

distribution with mean m 5 64 inches and standard 

deviation s 5 2.5 inches. To see if this distribution 

applies at their high school, a statistics class takes 

an SRS of 20 of the 300 16-year-old females at the 

school and measures their heights. When they calculate 

a sample mean of 64.7 inches, they wonder 

if the population of 16-year-old girls at their school 

has a mean height greater than 64 inches. 

To determine if a sample mean of x 5 64.7 inches 

provides convincing evidence that the average 

height of 16-year-old girls at the school is taller 

Answers 10–11 are on page 407 


406 



11/01/17 3:53 PM


than 64 inches, the class simulated 200 SRSs of size 

n 5 20 from a normal population with mean 

m 5 64 inches and standard deviation s 5 2.5 

inches. Here are the results of the simulation. 

d 

d 

d 

d d 

d d 

d d 

d d d d 

d d d d 

d d d d 

d d d d 

d d d d d 

d d d d d d 






d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d d d d d dddd 

62.5 63.0 63.5 64.0 64.5 65.0 65.5 66.0 

Sample mean height (in.) 



(b) Would it be unusual to get a sample mean of 64.7 

or more in a sample of size 20 when m 5 64? 

Explain. 


evidence that the mean height of the population 

of 16-year-old girls at this school is greater than 64 

inches? Explain. 

12. Relying on bathroom scales A manufacturer of 

bathroom scales says that when a 150-pound 

weight is placed on a scale produced in the factory, 

the weight indicated by the scale is normally 

distributed with a mean of 150 pounds and a 

standard deviation of 2 pounds. A consumeradvocacy 

group acquires an SRS of 12 scales from 

the manufacturer and places a 150-pound weight 

on each one. The group gets a mean weight of 

149.1 pounds, which makes them suspect that 

the scales underestimate the true weight. To test 

this, they use a computer to simulate 200 samples 

of 12 scales from a population with a mean of 

150 pounds and standard deviation of 2 pounds. 

Here is a dotplot of the means from these 200 

samples. 

d 

d 

d d 

d 

d 

d 

d 

d d 

d d d 

d d d d 

d d d d d d d 







d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d d d d d 

d d d d d d d d d d d d d d d d d d d d d d d d dd d d 

148.0 148.5 149.0 149.5 150.0 150.5 151.0 151.5 

Sample mean weight (lb) 

(a) There is one dot on the graph at 151.2. Explain 

what this dot represents. 

d 

(b) Would it be unusual to get a sample mean of 149.1 or 

less in a sample of size 12 when m 5 150? Explain. 


evidence that the scales produced by this manufacturer 

underestimate true weight? Explain. 

Applying the Concepts 

13. Instant winners A fast-food restaurant promotes 

certain food items by giving a game piece with 

each item. Advertisements proclaim that “25% of 

the game pieces are Instant Winners!” To test this 

claim, a frequent diner collects 20 game pieces and 

gets only 3 instant winners. 

(a) Identify the population, the parameter, the sample, 

and the statistic in this context. 

Suppose the advertisements are correct and 

p 5 0.25. The dotplot shows the distribution of the 

sample proportion of instant winners in 100 simulated 

SRSs of size n 5 20. 

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0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 

Sample proportion of instant winners 

(b) Would it be unusual to get a sample proportion of 

p^ 5 3/20 5 0.15 or less in a sample of size n 5 20 

when p 5 0.25? Explain. 


evidence that fewer than 25% of all game 

pieces are instant winners? Explain. 

14. Puny guppies? A large pet store that specializes 

in tropical fish has several thousand guppies. The 

store claims that the lengths of its guppies are 

approximately normally distributed with a mean of 

5 centimeters and a standard deviation of 0.5 centimeter. 

You come to the store and buy 10 randomly 

selected guppies and find that the mean length of 

your 10 guppies is only 4.8 centimeters. 

(a) Identify the population, the parameter, the sample, 

and the statistic in this context. 

Suppose the store’s description of the lengths of 

its guppies is true. The dotplot on the next page 

shows the distribution of sample means from 200 

simulated SRSs of size n 5 10 from a normally distributed 

population with m 5 5 centimeters and 

s 5 0.5 centimeter. 

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11. (a) In one SRS of size n 5 20, the 

mean height was 62.5 inches. 

(b) No, in the 200 simulated samples, 

23 of the SRSs had a mean of 64.7 

or more. Based on the simulation, 

P( x ≥ 64.7) = 23∙200 = 0.115. 

(c) No; because the probability from 

part (b) isn’t small, it is plausible that the 

true mean is m 5 64 and the class got 

a sample mean of x 5 64.7 by chance 

alone. 

12. (a) In one SRS of size n 5 12, the 

mean weight was 151.2 pounds. 

(b) No; in the 200 simulated samples, 

22 of the SRSs had a mean of 

149.1 or less. Based on the simulation, 

P( x ≤ 149.1) = 22∙200 = 0.11. 



true mean is m 5 150 and the group got 

a sample mean of x 5 149.1 by chance 

alone. 

13. (a) Population: All game pieces. 

Parameter: p 5 the true proportion of 

the population that are instant winners. 

Sample: The 20 game pieces collected by 

the frequent diner. Statistic: The proportion 

of the sample that are instant winners; 

p^ 5 3/20 5 0.15. 

(b) No; in the 100 simulated samples, 

18 of the SRSs had a sample proportion 

of 0.15 or lower. Based on the simulation, 

P( p^ ≤ 0.15) = 18∙100 = 0.18. 


part (b) isn’t small, it is plausible that 

the true proportion is p 5 0.25 and the 

frequent diner got a sample proportion 

of p^ 5 0.15 by chance alone. 

Lesson 6.1 


Answers continued 

10. (a) In one SRS of size n 5 60, 36.7% of 

the first serves were successful. 

(b) Yes; in the 200 simulated samples, only 

1 of the SRSs had a sample proportion of 

0.70 or higher. Based on the simulation, 

P( p^ ≥ 0.70) = 1∙200 = 0.005. 


(b) is small, it is not plausible that the true 

proportion is still p 5 0.53 and the player got 

a sample proportion of p^ 5 0.70 by chance 

alone. 

18/08/16 4:59 PM 



11/01/17 3:53 PM

408 


14. (a) Population: All guppies at the 

pet store. Parameter: m 5 the true mean 

length of the population. Sample: A 

random sample of 10 guppies. Statistic: The 

mean length of the sample; x 5 4.8 cm. 

(b) No; in the 200 simulated 

samples, 21 of the SRSs had a mean 

of 4.8 or less. Based on the simulation, 

P( x ≤ 4.8) = 21∙200 = 0.105. 



true mean is m 5 5 and I got a sample 

mean of x 5 4.8 by chance alone. 

15. (a) The distribution of heights for 

16-year-old females is approximately 

normal with a mean of m 5 64 inches 

and standard deviation of s 5 2.5 inches. 

56.5 59 61.5 64 66.5 

Height (in.) 

69 71.5 

(b) Answers will vary. This is the distribution 

of one possible sample. 

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55 60 65 70 

Height (in.) 

16. (a) The distribution of measured 

weights for all scales is approximately 

normal with a mean of m 5 150 pounds 

and standard deviation of s 5 2 pounds. 

144 146 148 150 152 

Weight (lb) 

154 156 

(b) Answers will vary. This is the 

distribution of one possible sample. 

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146 147 148 149 150 151 152 153 154 155 

Weight (lb) 

17. (a) In 10 cases of taking a random 

sample of size n 5 50 from each high 

school, the difference in proportions of 

students with Internet access at home 

is 0%. This means the proportion of 

students with Internet access was the 

same for each high school in 10 pairs of 

simulated samples 

(b) Yes; in the 100 pairs of simulated 

samples, 0 of the pairs had a 

difference in proportions of 0.20 

or higher. Based on the simulation, 

P(p^ N − p^ S ≥ 0.20) = 0∙100 = 0. 

(c) Yes; because the probability from 

part (b) is small, it is not plausible that 

the true difference in proportions is 

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4.6 4.8 5.0 5.2 5.4 

Sample mean length (cm) 

(b) Would it be unusual to get a sample mean of x = 4.8 

centimeters or less in a sample of size n 5 10 from 

this population? Explain. 


evidence that the mean length of guppies at this 

store is less than 5 centimeters? Explain. 

15. More tall girls Refer to Exercise 11. 

(a) Make a graph of the population distribution of 

heights for 16-year-old females. 

(b) Sketch a possible dotplot of the distribution of sample 

data for an SRS of size 20 from this population. 

16. More bathroom scales Refer to Exercise 12. 

(a) Make a graph of the population distribution of 

weights, assuming the manufacturer’s claim is correct. 

(b) Sketch a possible dotplot of the distribution of sample 

data for an SRS of size 12 from this population. 

Extending the Concepts 

17. Difference of proportions A school superintendent 

believes that the proportion of North High School 

students with Internet access at home is greater 

than the proportion of South High School students 

with Internet access at home. To investigate, she 

selects SRSs of size n 5 50 from each school and 

finds p^ N 5 46/50 5 0.92 and p^ S 5 36/50 5 0.72. 

To determine if a difference in proportions of 

0.20 provides convincing evidence that North High 

School has a greater proportion of students with 

Internet access at home, we simulated two random 

samples of size n 5 50 from populations with the 

same proportion of students with Internet access. 

Then, we subtracted the sample proportions. Here 

are the results from repeating this process 100 times. 

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p N 2 p S 5 0 and the superintendent 

got a sample difference of proportion of 

p^ N − p^ S = 0.20 by chance alone. 

18. (a) 300 C 25 = 1.95 × 10 36 different 

possible samples of 25 tomatoes. 

(b) Due to the extremely large number of 

possible samples, it is not practical to examine 

the complete sampling distribution of means 

for samples of size 25. 

19. (a) z = x − m 

s ; 

28,000 − 23,300 

0.67 = ; 

s 

s = 7014.93 

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(a) There are ten dots at 0. Explain what these dots 

represent. 

(b) Would it be unusual to get a difference in sample 

proportions of at least 0.20 when there is no difference 

in the population proportions? Explain. 


evidence that North High School has a greater 

proportion of students with Internet access at 

home? Explain. 

Recycle and Review 

18. Sampling tomatoes (4.8, 6.1) Zach runs a roadside 

stand during the summer, selling produce from his 

farm. On a single day in mid-August, he harvests 

300 tomatoes. Suppose Zach wants to take a simple 

random sample of 25 tomatoes from the day’s 

pick to estimate mean weight. 

(a) How many possible sets of 25 tomatoes could 

be sampled from the 300 tomatoes in the day’s 

crop? 

(b) What does this say about the practicality of examining 

the complete sampling distribution of the sample 

mean for samples of size 25 from this population? 

19. College debt (5.7) A report published by the Federal 

Reserve Bank of New York in 2012 reported the 

results of a nationwide study of college student 

debt. Researchers found that the average student 

loan balance per borrower is $23,300. They also 

reported that about one-quarter of borrowers owe 

more than $28,000. 4 

(a) Assuming that the distribution of student loan 

balances is approximately normal, estimate the 

standard deviation of the distribution of student 

loan balances. 

(b) Assuming that the distribution of student loan 

balances is approximately normal, use your answer 

to part (a) to estimate the proportion of borrowers 

who owe more than $54,000. 

(c) In fact, the report states that about 10% of borrowers 

owe more than $54,000. What does this 

fact indicate about the shape of the distribution of 

student loan balances? 

(d) The report also states that the median student loan 

balance is $12,800. Does this fact support your 

conclusion in part (c)? Explain. 

54,000 − 23,300 

(b) z = 

≈ 4.38; 

7014.93 

P(X ≥ 54,000)≈ P(Z ≥ 4.38) ≈ 0 

Using technology: Applet/normalcdf(lower: 

54000, upper:100000, mean:23300, 

SD:7014.93) 5 0.000006 

(c) If the distribution of loan balances is 

approximately normal, then we would expect 

almost no one to have a balance that large. 

Because 10% of borrowers owe more than 

$54,000, we can conclude that the distribution 

of loan balances isn’t normal and is rightskewed. 

(d) Yes; because the mean ($23,300) is so 

much larger than the median ($12,800), we can 

conclude that the distribution of loan balances 

is skewed to the right. 


408 



11/01/17 3:54 PM

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Lesson 6.2 

Sampling Distributions: 

center and variability 


d Determine if a statistic is an unbiased estimator of a population parameter. 

d Describe the relationship between sample size and the variability of a 

statistic. 





d 6.2.1 

d 6.2.2 

Lesson 6.2 

AcT iviT y 

How many craft sticks are in the bag? 

In this activity, you will create a statistic for estimating 

the total number of craft sticks in a bag (N). The 

sticks are numbered 1, 2, 3, . . . , N. Near the end of the 

activity, your teacher will select a random sample of 

n 5 7 sticks and read the number on each stick to the 

class. The team that has the best estimate for the total 

number of sticks will win a prize. 

1. Form teams of three or four students. As a team, 

spend about 10 minutes brainstorming different 

ways to estimate the total number of sticks. Try to 

come up with at least three different statistics. 

2. Before your teacher provides the sample of sticks, 

use simulation to investigate the sampling distribution 

of each statistic. For the simulation, assume 

that there are N 5 100 sticks in the bag and that 

you will be selecting samples of size n 5 7. 

j Using your TI-83/84 calculator, select an SRS of 

size 7 using the command RandIntNoRep(lower: 

1,upper:100,n:7). [With older OS, use the command 

RandInt(lower:1,upper:100,n:7) and verify 

that there are no repeated numbers. If there are 

repeats, press ENTER to get a new sample.] 

j For each sample, calculate the value of each of 

your three statistics. 

Unbiased Estimators 

j 

j 

graph these values on a set of dotplots like those 

shown here. 

Perform as many trials of the simulation as possible. 

Statistic 3 Statistic 2 Statistic 1 

60 

70 80 90 100 110 120 130 140 

Estimated total 

3. Based on the simulated sampling distributions, 

which of your statistics is likely to produce the 

best estimate? Discuss as a team. 

4. Your teacher will now draw a sample of n 5 7 

sticks from the bag. On a piece of paper, write 

the names of your group members, your group’s 

estimate for the number of sticks in the bag (a 

number), and the statistic you used to calculate 

your estimate (a formula). 

In the craft sticks activity, the goal was to estimate the maximum value in a population, 

with the assumption that the members of the population are numbered 1, 2, . . . , N. 

Two possible statistics that might be used to estimate N are the sample maximum (max) 

and twice the sample median (2 3 median). 

Assuming that the population has N 5 100 members and we use SRSs of size n 5 7, 

Figure 6.2 shows the simulated sampling distributions of the sample maximum and 

twice the sample median. 

409 

Bell Ringer 

Suppose you wish to estimate the 

average (mean) height of all students at 

your school by taking a random sample 

of students and calculating the average 

height of the sample. Would you expect 

the sample mean to be closer to the 

true average height from a sample of 

4 students or 40 students? 

The best statistics are centered at 100 

with low variability. A pre-made Fathom 

file is included in the Teacher’s Resource 

Materials. If you don’t have Fathom, 

the figure shows simulated sampling 

distributions for several commonly used 

statistics using 200 random samples. The 

statistics are 

• TwiceMean 5 2 · sample mean 

• TwiceMedian 5 2 · sample median 

• Max 5 sample maximum 

• MeanPlusMed 5 mean 1 median 

• SumQuartiles 5 Q 1 1 Q 3 

• TwiceIQR 5 2 · IQR 

• MeanPlus2SD 5 sample mean 1 

2 · sample standard deviation 

• Partition 5 (8/7) · sample maximum 

TwiceMean 

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Materials: Graphing calculators or an Internetconnected 

device for each student or group of 

students, a prize for the winning team, and a 

population of craft sticks. We recommend using 

at least 100 sticks (but not exactly 100). Use an 

opaque container so students can’t use their 

eyes to estimate the total. 

Teaching Advice: The point of this activity is to 

illustrate the concepts of bias and variability of 

statistics. This is a long activity and should be 

done during an entire class period. 

Start by selecting one or two sticks to 

make the contents of the bag less abstract. 

Be patient in Step 1! Teams will struggle. 

18/08/16 4:59 PM 

Let them keep at it. Once one statistic is 

proposed, usually more will follow. 

During Step 1, rotate around the room to 

give suggestions to teams that are struggling. 

Consider giving away one or more of the 

methods used later in the lesson, such as the 

sample maximum, twice the sample median, 

or twice the sample mean. These are all 

decent statistics but not the best. 

During Step 2, students can use their 

calculators or a random generator on the 

Internet (like random.org) to generate 7 

integers without repeats. Groups should 

generate at least 20 samples. If you have the 

ability to simulate the sampling distributions 

for a number of statistics using Fathom or 

other software, consider offering another 

prize for the team with the best statistic. 

TwiceMedian 

Max 

MeanPlusMed 

SumQuartiles 

TwiceIQR 

MeanPlus2SD 

Partition 

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Answers: 

1. Answers will vary by student group. 

2. Dotplots will vary. 

3. Answers will vary. The best statistics 

are centered at 100 with low 

variability. 

4. Answers will vary by student group. 

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L E S S O N 6.2 • Sampling Distributions: Center and Variability 409 


11/01/17 3:54 PM

410 


TRM Lesson 6.2 Activity 

fAthom File 

If you are familiar with Fathom software, 

you can use the pre-made Fathom file to 

simulate statistics in Step 2 of the Lesson 

6.2 activity. Click on the link in the TE-book, 

log into the Teacher’s Resource site, or 

access this resource on the TRFD. 

Teaching Tip 

If you need to save time, Lessons 6.2 

and/or 6.3 can be skipped without losing 

much continuity in future chapters. 

However, the other lessons in this 

chapter are crucial to understanding 

much of the remainder of the course. 

FigUre 6.2 Simulated 

sampling distributions 

of the sample maximum 

and twice the sample 

median for samples of 

size n 5 7 from a population 

with N 5 100. 

Sample 

maximum 

twice sample 

median 

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These simulated sampling distributions look quite different. The sampling distribution 

of the sample maximum is skewed left, while the sampling distribution of 

twice the sample median is roughly symmetric. 

The values of the sample maximum are consistently less than the population maximum 

N. However, the values of twice the sample median aren’t consistently less than 

or consistently greater than the population maximum N. It appears that twice the 

sample median might be an unbiased estimator of the population maximum, while 

the sample maximum is clearly biased. 

DEFINITION unbiased estimator 

A statistic used to estimate a parameter is an unbiased estimator if the mean of its sampling 

distribution is equal to the value of the parameter being estimated. 

cAutIOn 

! 

The use of the word “bias” here is consistent with its use in Chapter 3. The design 

of a statistical study shows bias if it would consistently underestimate or consistently 

overestimate the value you want to know when you repeat the study many times. 

Recall the Federalist Papers activity (page 188) in which the estimates were consistently 

too large when students were allowed to choose the words in the sample. Don’t 

trust an estimate that comes from a biased sampling method. 


What is the mean-ing of bias? 

Unbiased estimators 

PROBLEM: The dotplot displays 

simulated sampling distributions of two 

statistics that can be used to estimate 

the mean of a population distribution. 

The simulated sampling distributions are 

based on 1000 SRSs of size n 5 5, and 

the population mean m 5 40. The mean 

of each distribution is indicated by a blue 

line segment. 

a 

e XAMPLe 

Why do we divide by n 2 1? 

Unbiased estimators 

PROBLEM: In Chapter 1, you learned to calculate the 

standard deviation of sample data using the formula 

∑ (x i − x ) 2 

s x = Å n − 1 

What if you divided by n instead of n 2 1? Let’s 

simulate the sampling distributions of two statistics 

that can be used to estimate the variance of a 

distribution, where the variance is the square of the 

standard deviation (variance 5 standard deviation 2 ). 

∑ (x i − x ) 2 

Statistic 1: 

Statistic 2: ∑ (x i − x ) 2 

n − 1 

n 

These simulated sampling distributions are based 

on 1000 SRSs of size n 5 3 from a population with 

variance 5 25. The mean of each distribution is 

indicated by a blue line segment. 

Is either of these statistics an unbiased estimator of 

the population variance? Explain your reasoning. 

Statistic 2 Statistic 1 

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Is either of these statistics an unbiased estimator of the population mean? Explain your 

reasoning. 

SOLUTION: Statistic 1 appears to be unbiased because the mean of its sampling 

distribution is very close to 40, the value of the population mean. Statistic 2 appears to be 

biased because the mean of its sampling distribution is about 44, which is clearly greater 

than 40, the value of the population mean. 

FYI 

For a sample size of 7, the best estimator of 

N in a population like the one in the activity 

is (8/7) · sample maximum 2 1. For any 

sample size n, the best estimator is 

(n 1 1)/n · sample maximum 2 1. 

FYI 

The definition of an unbiased estimator given 

here is based on the mean of a sampling 

distribution. If the median of the sampling 

distribution of a statistic is equal to the value 

of the parameter being estimated, it is also 

considered unbiased. 


410 



11/01/17 3:54 PM


Statistic 1 

Statistic 2 

0 20 40 60 80 100 120 140 160 180 200 

Estimated variance 

SOLUTION: 

Statistic 1 appears to be unbiased because the mean of 

its sampling distribution is very close to 25, the value of 

the population variance. Statistic 2 appears to be biased 

because the mean of its sampling distribution is clearly less 

than 25, the value of the population variance. 


Teaching Tip: 

Differentiate 

Students who have trouble with 

mathematical notation might be 

intimidated by the formulas in the 

preceding example. Tell these students 

to ignore the formulas and focus on 

the big idea that there are two slightly 

different ways to calculate the standard 

deviation. 

Lesson 6.2 

We divide by n 2 1 when calculating the sample variance so it will be an unbiased 

estimator of the population variance. If we divided by n instead, our estimates would 

be consistently too small. Likewise, it is better to divide by n 2 1 instead of n when 

calculating the standard deviation for a distribution of sample data. 

Sampling Variability 

Another possible statistic that could be used in the craft sticks activity is twice the 

sample mean. Figure 6.3 shows the simulated sampling distributions of twice the 

sample mean and twice the sample median. 

Twice sample 

mean 

Twice sample 

median 

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Both statistics appear to be unbiased estimators because the mean of each sampling 

distribution is around 100. However, the sampling distribution of twice the 

sample mean (standard deviation ≈ 22) is less variable than the sampling distribution 

of twice the sample median (standard deviation ≈ 34). In general, we prefer statistics 

that are less variable because they produce estimates that tend to be closer to the value 

of the parameter. 

For some parameters, there is an obvious choice for a statistic. For example, to estimate 

the proportion of successes in a population p, we use the proportion of successes 

in the sample, p^ . Fortunately, p^ is an unbiased estimator of p. And as we learned in 

Lesson 3.3, we can reduce the variability of an estimate by increasing the sample size. 

Figure 6.4 on the next page shows the simulated sampling distributions for p^ 5 

the proportion of students in the sample who take the bus to school when taking SRSs 

of size n 5 10 and SRSs of size n 5 50 from a population in which the proportion of 

all students who take the bus to school is p 5 0.70. 


sampling distributions of 

twice the sample mean 

and twice the sample 

median for samples 

of size n 5 7 from a 

population with N 5 100. 

Common Error 

Some students think that bias is about 

the shape of a sampling distribution. 

These students think that a statistic is 

unbiased if its sampling distribution 

is symmetric and/or mound-shaped. 

Remind them that bias is about the 

center of the sampling distribution. 


18/08/16 5:00 PM 



11/01/17 3:54 PM

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412 


Teaching Tip 

Have students look closely at the bottom 

dotplot in Figure 6.4 and the dotplot 

in the next example. Ask them if this 

simulated sampling distribution has 

a familiar shape. Later in this chapter, 

students will learn that certain statistics 

have approximately normal sampling 

distributions. 


Can you hand me that wrench? 

Sampling variability 

PROBLEM: A local auto parts store has 

records of the daily sales of hand tools 

(in dollars) over the last several years. 

To estimate the average daily sales, 

a manager selects 10 days and finds 

the sample mean daily sales. Here is a 

simulated sampling distribution of x, 

the sample mean daily sales of hand 

tools (in dollars) for 1000 samples of 

size n 5 10. 

a 



the sample proportion p^ 

for samples of size n 5 10 

and samples of size 

n 5 50 from a population 

with p 5 0.7. 

e XAMPLe 

n = 10 

n = 50 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

ddddddddddddddddddd 

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 

Sample proportion who take the bus 

As expected, both simulated sampling distributions have means near p 5 0.70. 

Also, the sampling distribution of p^ is much more variable when the sample size 

is n 5 10, compared with n 5 50. In a small sample, it is plausible that the sample 

proportion could be much smaller or much larger than the parameter, just by chance. 

However, when the sample size gets bigger, we expect the sample proportion to be 

fairly close to the value of the parameter. 

The lifetime of batteries: Hours or days? 

Sampling variability 

Decreasing sampling variability 

The sampling distribution of any statistic will have less variability when the sample size is larger. 

PROBLEM: For quality control, workers at a battery factory regularly 

select random samples of batteries to estimate the mean lifetime. Here is a 

simulated sampling distribution of x, the sample mean lifetime (in hours) for 

1000 random samples of size n 5 100 from a population of AAA batteries. 

(a) What would happen to the sampling distribution of the sample 

mean x if the sample size were n 5 50 instead? Justify. 

(b) What is the practical consequence of this change in sample size? 

d 

d 

d dddd 

d 

d 

ddd 

dd dd 

d ddddddd 

d ddd ddd 

dd 

d 

ddd d 

d d ddd 

ddd dd 

d d dd ddd dd 

dd d 

dd dd ddd ddd 

d d 

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dd d dd d dd 

d dd dd d dd d dd d d d dd dddd dd d dd 

d dddd dd 

d dd d ddd 

d d d d 

dd d ddd ddd d dddd dd 

ddd 

dddd d 

d d d 

d 

d d ddd d ddd d d d d dd dd ddd 

d dd ddd 

ddd 

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dd 

d dd d 

dd 

dd d d dd 

d dddd ddd dd dddddddddd ddddddddd 

d 

ddd 

ddd 

d dddd dddd d ddd 

dddd d dddd ddd 

ddd 

ddd ddd 

d dddd d d d dddd d dd dd 

d dddd d 

ddd ddd d ddd dddd dddd dddd ddd d 

ddd ddd 

dddd ddd dd dd 

d d d d ddd 

dddd dddddd ddd dddd d 

dd 

d ddd 

ddd d dddd d dddd ddd d 

ddd 

d d dddd dd 

dd ddd 

d dd 

ddd d d dd 

dd 

d d d ddd 

d d 

ddddd 

ddddd ddd d ddd 

d ddd ddd ddd ddd 

ddd 

dddd 

dd ddd 

ddd 

ddd dd dd ddd dddddd dddd dd 

dd ddd d d dd d 

dd dd 

d ddd 

ddd dd d dddddd d 

dd d ddddd 

d 

ddd dd dd 

dd 

ddd d 

ddd d dd 

d dd d 

ddd ddd 

d d 

ddd 

d ddd d 

d d 

ddd ddd ddd ddd d d dddddd ddd ddd ddd ddd ddd d d 

dd d 

ddd d 

dd dd dddd 

d 

d 

ddd d 

ddd ddd dd 

d 

d 

d d dd 

d 

d 

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d d d 

d 

d 

d dd ddd ddd d 

d d d dd dd 

100 120 140 160 180 200 220 

Sample mean daily sales of 

hand tools ($) 

(a) What would happen to the sampling 

distribution of the sample mean x if the 

sample size were n 5 30 instead? Justify 

your answer. 

(b) What is the practical consequence of 

this change in sample size? 

SOLUTION: 

(a) The sampling distribution of the 

sample mean x will be less variable 

because the sample size is larger. 

(b) The estimated mean daily sales of 

hand tools will typically be closer to the 

true mean daily sales of hand tools. In 

other words, the estimate will be more 

precise. 

d 

SOLUTION: 

(a) The sampling distribution of the sample mean x will be more variable because the 

sample size is smaller. 

44 46 48 50 52 54 56 58 60 62 

Sample mean lifetime (h) 

(b) The estimated mean lifetime will typically be farther away from the true mean lifetime. In other words, the 

estimate will be less precise. 



Putting It All Together: Center and Variability 

We can think of the true value of the population parameter as the bullseye on a target 

and of the sample statistic as an arrow fired at the target. Both bias and variability 

describe what happens when we take many shots at the target. 

• Bias means that our aim is off and we consistently miss the bullseye in the same 

direction. That is, our sample values do not center on the population value. 


412 



11/01/17 3:54 PM


• High variability means that repeated shots are widely scattered on the target. In 

other words, repeated samples do not give very similar results. 

Figure 6.5 shows this target illustration of bias and variability. Notice that low variability 

(shots are close together) can accompany high bias (shots are consistently away 

from the bullseye in one direction). And low or no bias (shots center on the bullseye) 

can accompany high variability (shots are widely scattered). Ideally, we’d like our 

estimates to be accurate (unbiased) and precise (have low variability). 

d 

ddd dd dd 

High bias, low variability 

(a) 

d 

d 

d 

d d 

Low bias, high variability 

(b) 

d 

d 

d 

d 

d 

d 

High bias, high variability 

(c) 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

FigUre 6.5 Bias and variability. (a) High bias, low variability. (b) Low bias, high variability. 

(c) High bias, high variability. (d) The ideal: no bias, low variability. 


How many tanks does the enemy have? 

During World War II, the Allies captured many german 

tanks. Each tank had a serial number on it. Allied 

commanders wanted to know how many tanks the 

germans had so that they could allocate their forces 

appropriately. They sent the serial numbers of the 

captured tanks to a group of mathematicians in 

Washington, D.C., and asked for an estimate of the 

total number of german tanks N. 

Here are simulated sampling distributions for three 

statistics that the mathematicians considered, using 

samples of size n 5 7. The blue line marks N, the total 

number of german tanks. The shorter red line segments 

mark the mean of each simulated sampling distribution. 

Statistic 3 Statistic 2 Statistic 1 

d 

d 

d 

d 

d 

d dd 

d 

d 

dd 

d 

d ddd 

dd ddd d 

dd dd d d 

d 

d 

ddd 

d d 

d d d 

dd d 

ddd 

dd 

d dddddd d 

ddd 

d 

d 

dd 

d d 

ddd ddd d 

d 

d 

d 

N 


dd d 

ddd 

The ideal: no bias, low variability 

(d) 

1. Do any of these statistics appear to be unbiased? 

Justify. 

2. Which of these statistics do you think is best? 

Explain your reasoning. 

3. Explain how the Allies could get a more precise 

estimate of the number of german tanks using the 

statistic you chose in Question 2. 

© Bettmann/Corbis 

Teaching Tip: 

Differentiate 

The target illustration in Figure 6.5 

will be the best way for some students 

to understand the ideas of bias and 

variability in sample statistics. Compare 

the statistics in the screenshot in the 

Activity Overview at the start of this 

lesson to the figure. The Max statistic 

corresponds to target (a), the TwiceIQR 

statistic corresponds to target (b), and 

the Partition statistic corresponds to 

target (d). 

There is no statistic from the 

screenshot in the Activity Overview at 

the start of this lesson that corresponds 

to target (c). Challenge your top students 

to draw a dotplot of a hypothetical 

statistic that would correspond to target 

(c). They should use the same scale as the 

one in the screenshot. 

Teaching Tip 

In the Lesson App, Statistic 1 is sample 

min 1 sample max, Statistic 2 is sample 

mean 1 3SD, Statistic 3 is sample max· 

(n 1 1)/n. The Allies used a statistic 

similar to Statistic 3 because it was 

unbiased and had low variability! Thus, 

their estimate of the number of tanks 

would be “on target” and was unlikely to 

be far from the true value. 

Lesson App 

Answers 

Lesson 6.2 


18/08/16 5:00 PM 

1. Statistics 1 and 3 both appear to be 

unbiased because the mean of each 

sampling distribution is very close to N, 

the value of the population maximum. 

Statistic 2 appears to be biased because 

the mean of its sampling distribution 

is clearly more than N, the value of the 

population maximum. 

2. Statistic 3; while both Statistics 1 and 

3 are unbiased, Statistic 3 appears to 

have less variability. 

3. The Allies could get a more precise 

estimate of the number of German tanks 

by capturing more tanks (increasing the 

sample size). This way, the estimated 

number of tanks would typically be 

closer to the true number of tanks 

(more precise). 



11/01/17 3:54 PM

414 


Lesson 6.2 



Determine if a statistic is an unbiased estimator of a population 

parameter. 

p. 410 1–4 

Describe the relationship between sample size and the variability of a 

statistic. 

p. 412 5–8 

TRM full Solutions to Lesson 6.2 

Exercises 


lesson by clicking on the link in the 



on the TRFD. 

Answers to Lesson 6.2 

Exercises 

1. Yes; the mean of the sampling 

distribution is very close to 22.96, 

the value of the population median. 

2. No; the mean of the sampling 

distribution is clearly more than 0.20, 

the value of the population minimum. 

3. No; the mean of the sampling 

distribution is clearly less than 153.53, the 

value of the population range. Population 

range 5 max 2 min 5 153.73 2 0.20 

5 153.53. 

Exercises 


Exercises 1–4 refer to the following setting. The manager 

of a grocery store records the total amount spent 

(in dollars) for each customer who makes a purchase 

at his store during a week. The values in the table 

summarize the distribution of amount spent for this 

population: 

N mean SD Min Q 1 med Q 3 Max 

749 29.85 24.63 0.20 12.29 22.96 39.93 153.73 

1. Is the median unbiased? To investigate if the sample 

pg 410 median is an unbiased estimator of the population 

median, 1000 SRSs of size n 5 10 were selected from 

the population described. The sample median for 

each of these samples was recorded on the dotplot. 

The mean of the simulated sampling distribution is 

indicated by an orange line segment. Does the sample 

median appear to be an unbiased estimator of the 

population median? Explain your reasoning. 

d 

d d 

d 

d 

d 

d 

d d 

d 

d 

Lesson 6.2 

d 

d 

d 

d 

d 

d 

d 

ddddd 

dd d 

0 10 20 30 40 50 60 70 

Sample median 

2. Is the minimum unbiased? To investigate if the sample 

minimum is an unbiased estimator of the population 

minimum, 1000 SRSs of size n 5 10 were 

selected from the population described. The sample 

minimum for each of these samples was recorded 

on the dotplot. The mean of the simulated sampling 

distribution is indicated by an orange line segment. 

Does the sample minimum appear to be an unbiased 

estimator of the population minimum? Explain your 

reasoning. 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d d 

d d 

d 

d 

d d d d d d d 

0 5 10 15 20 25 

Sample minimum 

3. Is the range unbiased? To investigate if the sample 

range is an unbiased estimator of the population 

range, 1000 SRSs of size n 5 10 were selected from 

the population described. The sample range for 

each of these samples was recorded on the dotplot. 

The mean of the simulated sampling distribution 

is indicated by an orange line segment. Does the 

sample range appear to be an unbiased estimator 

of the population range? Explain your reasoning. 

d 

d d 

d 

d 

d 

dd 

d 

d 

d 

d 

d 

d 

d d 

d 

d 

d 

20 40 60 80 100 120 140 160 

Sample range 

d 

d 

d 

d 

d d 

d 

dd d d 

d 

d d d 


Teaching Tip 

Exercise 19 on p. 416 previews Lesson 6.3, 

which is based on binomial random variables. 

Make sure your students do this exercise. 


414 



11/01/17 3:54 PM


4. Is the IQR unbiased? To investigate if the sample 

IQR is an unbiased estimator of the population 

IQR, 1000 SRSs of size n 5 10 were selected from 

the population described. The sample IQR for each 

of these samples was recorded on the dotplot. The 

mean of the simulated sampling distribution is indicated 

by an orange line segment. Does the sample 

IQR appear to be an unbiased estimator of the 

population IQR? Explain your reasoning. 

dd 

0 

d 

d 

d 

d 

d 

d 

20 

d 

d 

d 

d 

d 

dd 

d 

d 

d d 

d 

40 60 

Sample IQR 

d d d 

d 

d 

d d dd d d 

80 

100 

5. More about medians Refer to Exercise 1. 

(a) What would happen to the sampling distribution of 

pg 412 the sample median if the sample size were n 5 50 

instead? Justify. 

(b) What is the practical consequence of this change in 

sample size? 

6. More about minimums Refer to Exercise 2. 

(a) What would happen to the sampling distribution of 

the sample minimum if the sample size were n 5 50 



sample size? 

7. More about ranges Refer to Exercise 3. 

(a) What would happen to the sampling distribution 

of the sample range if the sample size were n 5 5 



sample size? 

8. More about IQRs Refer to Exercise 4. 

(a) What would happen to the sampling distribution 

of the sample IQR if the sample size were n 5 5 



sample size? 


9. Cholesterol in teens A study of the health of teenagers 

plans to measure the blood cholesterol levels of 

an SRS of 13- to 16-year-olds. The researchers will 

report the mean x from their sample as an estimate 

of the mean cholesterol level m in this population. 

Explain to someone who knows little about statistics 

what it means to say that x is an unbiased 

estimator of m. 

10. Predict the election A polling organization plans to 

ask a random sample of likely voters who they will 

vote for in an upcoming election. The researchers 

will report the sample proportion p^ that favors the 

incumbent as an estimate of the population proportion 

p that favors the incumbent. Explain to 

someone who knows little about statistics what it 

means to say that p^ is an unbiased estimator of p. 

11. Sampling more teens Refer to Exercise 9. The sample 

mean x is an unbiased estimator of the population 

mean m no matter what size SRS the study chooses. 

Explain to someone who knows nothing about statistics 

why a large random sample will give more reliable 

results than a small random sample. 

12. Sampling more voters Refer to Exercise 10. The 

sample proportion p^ is an unbiased estimator of 

the population proportion p no matter what size 

random sample the polling organization chooses. 

Explain to someone who knows nothing about statistics 

why a large random sample will give more 

trustworthy results than a small random sample. 

13. Housing prices In a residential neighborhood, the 

median value of a house is $200,000. For which of 

the following sample sizes, n 5 10 or n 5 100, is 

the sample median most likely to be greater than 

$250,000? Explain. 

14. Houses with basements In a particular city, 74% 

of houses have basements. For which of the following 

sample sizes, n 5 10 or n 5 100, is the sample 

proportion of houses with a basement more likely 

to be greater than 0.70? Explain. 

15. Bias and variability The histograms show sampling 

distributions for four different statistics intended to 

estimate the same parameter. 

(i) 

(ii) 

(iii) 

(iv) 





10. If we chose many random samples 

and calculated the sample proportion 

p^ for each sample, the distribution of p^ 

would be centered at the value of p. In 

other words, when we use p^ to estimate 

p, we will not consistently underestimate 

p or consistently overestimate p. 

11. A larger random sample will provide 

more information about the population 

and, therefore, more precise results. 

The variability of the distribution of x 

decreases as the sample size increases. 

12. A larger random sample will provide 

more information about the population 

and, therefore, more precise results. 

The variability of the distribution of p^ 

decreases as the sample size increases. 

13. n 5 10; the sampling distribution 

of the sample median will be more 

variable with n 5 10 than with n 5 100. 

Because the distribution is more variable, 

it is more likely to get a sample median 

($250,000) that is far away from the true 

median ($200,000). 

14. n 5 100; the sampling distribution 

of the sample proportion will be less 

variable with n 5 100 than with n 5 10. 

Because the distribution is less variable, 

it is less likely to get a sample proportion 

that is far away (less than 0.70) from 

the true proportion (0.74). This makes it 

more likely for the sample proportion to 

be above 0.70 with n 5 100. 

Teaching Tip 

Don’t skip Exercise 15! It’s a wonderful 

way to assess the two learning targets 

from this Lesson. Consider having a 

short class discussion on it after students 

have had a chance to try the exercise for 

themselves. 

Lesson 6.2 


4. Yes; the mean of the sampling distribution is 

very close to 27.64, the value of the population 

IQR. Population IQR 5 Q 3 2 Q 1 5 39.93 2 12.29 

5 27.64. 

5. (a) It will be less variable because the 

sample size is larger. 

(b) The estimated median amount spent will 

typically be closer to the true median amount 

spent. In other words, the estimate will be 

more precise. 

6. (a) It will be less variable because the 

sample size is larger. 

(b) The estimated minimum amount spent 

will typically be closer to the true minimum 

amount spent. In other words, the estimate 

will be more precise. 

18/08/16 5:00 PM 

7. (a) It will be more variable because the 


(b) The estimated range amount spent 

will typically be farther from the true range 

amount spent. In other words, the estimate 

will be less precise. 

8. (a) It will be more variable because the 


(b) The estimated IQR amount spent will typically 

be farther from the true IQR amount spent. 

In other words, the estimate will be less precise. 

9. If we chose many SRSs and calculated the 

sample mean x for each sample, the distribution 

of x would be centered at the value of m. In 

other words, when we use x to estimate m, 

we will not consistently underestimate m or 

consistently overestimate m. 



11/01/17 3:54 PM

416 


15. (a) Statistics (ii) and (iii) both appear 

to be unbiased because the mean of each 

sampling distribution is very close to the 

value of the population parameter. 

(b) Statistic (ii); while both statistics 

(ii) and (iii) are unbiased, statistic (ii) has 

lower variability. 

16. (a) 

10 + 5 + 10 + 7 + 9 41 

m = 

= 

5 

5 = 8.2 

(b) 

Sample #1: Abigail (10), x 5 7.5 

Bobby (5) 

Sample #2: Abigail (10), x 5 10 

Carlos (10) 


DeAnna (7) 


Emily (9) 

Sample #5: Bobby (5), x 5 7.5 

Carlos (10) 

Sample #6: Bobby (5), x 5 6 

DeAnna (7) 

Sample #7: Bobby (5), x 5 7 

Emily (9) 

Sample #8: Carlos (10), x 5 8.5 

DeAnna (7) 

Sample #9: Carlos (10), x 5 9.5 

Emily (9) 

Sample #10: DeAnna (7), x 5 8 

Emily (9) 

(c) 

d d d 

d 

d d d d d 

6 6.5 7 7.5 8 8.5 9 9.5 10 

Sample mean quiz score 

m x = 

7.5 + 10 + 8.5 + 9.5 + 7.5 + 

6 + 7 + 8.5 + 9.5 + 8 

10 

= 82 

10 = 8.2. 

Yes, the sample mean is an unbiased 

estimator of the population mean. The 

mean of the sampling distribution is equal 

to 8.2, which is the value of the population 

mean. 

(a) Which statistics are unbiased estimators? Justify 

your answer. 

(b) Which statistic does the best job of estimating the 

parameter? Explain. 


16. More about means In the Exercises for Lesson 6.1, 

you were introduced to the following population of 

2 male students and 3 female students, along with 

their quiz scores: 


(a) Calculate the mean quiz score for the entire population. 

(b) List all 10 possible SRSs of size n 5 2, calculate 

the mean quiz score for each sample, and display 

the sampling distribution of the sample mean in a 

dotplot. 

(c) Calculate the mean of the sampling distribution 

from part (b). Is the sample mean an unbiased estimator 

of the population mean? Explain. 

17. More about proportions In the Exercises for 

Lesson 6.1, you were introduced to the following 

population of 2 male students and 3 female 

students, along with their quiz scores: 


(a) Calculate the proportion of females in the entire 

population. 

(b) List all 10 possible SRSs of size n 5 2, calculate the 

proportion of females for each sample, and display 

the sampling distribution of the sample proportion 

in a dotplot. 

(c) Calculate the mean of the sampling distribution 

from part (b). Is the sample proportion an unbiased 

estimator of the population proportion? Explain. 


18. Students and housing (4.3, 4.4) There are 104 

students in Professor Negroponte’s statistics class, 

49 males and 55 females. Sixty of the students live in 

the dorms and the rest live off campus. Twenty of the 

males live off-campus. Choose a student at random 

from this class. Let Event M 5 the student is male and 

Event D 5 the student lives in the dorms. 

(a) Construct a Venn diagram to represent the outcomes 

of this chance process using the events M 

and D. 

(b) Find each of the following probabilities and interpret 

them in context. 

(i) P(M c D) (ii) P(M C d D) (iii) P(D k M) 

19. Students and homework (5.3, 5.4) Refer to Exercise 

18. At the beginning of each day that Professor 

Negroponte’s class meets, he randomly selects a 

member of the class to present the solution to a homework 

problem. Suppose the class meets 40 times during 

the semester and the selections are made with 

replacement. Let X 5 the number of times a female 

student is selected to present a solution. 

(a) Is X a binomial random variable? Justify your 

answer. 

(b) Calculate the mean and standard deviation of X. 

(c) For the first 10 meetings of the class, Professor Negroponte 

selects only 1 female student to solve a problem. 

Is there convincing evidence that his selection 

process is not really random? Support your answer 

with an appropriate probability calculation. 

17. (a) p = 3 5 = 0.6 

(b) 

Sample #1: Abigail, Bobby p^ 5 0.5 

Sample #2: Abigail, Carlos p^ 5 0.5 

Sample #3: Abigail, DeAnna p^ 5 1 

Sample #4: Abigail, Emily p^ 5 1 

Sample #5: Bobby, Carlos p^ 5 0 

Sample #6: Bobby, DeAnna p^ 5 0.5 

Sample #7: Bobby, Emily p^ 5 0.5 

Sample #8: Carlos, DeAnna p^ 5 0.5 

Sample #9: Carlos, Emily p^ 5 0.5 

Sample #10: DeAnna, Emily p^ 5 1 

d 

d 

d d dd d d 

0 0.5 1 

Sample proportion of female 


(c) 

0.5 + 0.5 + 1 + 1 + 0 + 

0.5 + 0.5 + 0.5 + 0.5 + 1 

m p^ = 

= 6 

10 

10 = 0.6. 

Yes, the sample proportion is an unbiased 

estimator of the population proportion. The 

mean of the sampling distribution is equal 

to 0.6, which is the value of the population 

proportion. 

18. (a) 

Male 20 29 Dorms 31 

24 

20 + 29 + 31 

(b) (i) P(M c D) = P(M or D) = 

104 

= 80 = 0.769. There is about a 77% chance 

104 

that a randomly selected student is a male or 

lives in the dorm. 

(ii) P(M C d D) = P(M C 31 

and D)= 

104 = 0.298. 

There is about a 30% chance that a randomly 

selected student is not a male and lives in the 

dorm. 

P(D and M) 

(iii) P(D 0 M) = = 29∙104 

P(M) 49∙104 = 29 

49 

= 0.592. There is about a 59% chance that a 

randomly selected student lives in the dorm, 

given that the student is a male. 

Answer 19 is on page 417 


416 



11/01/17 3:54 PM


Lesson 6.3 

The Sampling Distribution 

of a Sample count 

(The normal Approximation 

to the Binomial) 


d Calculate the mean and the standard deviation of the sampling distribution of 

a sample count and interpret the standard deviation. 

d Determine if the sampling distribution of a sample count is approximately 

normal. 

d If appropriate, use the normal approximation to the binomial distribution to 

calculate probabilities involving a sample count. 

In many cases, we are interested in the number of successes X in a random sample 

from some population. For example, X 5 the number of defective flash drives in a 

random sample of 10 flash drives or X 5 the number of Democrats in a random 

sample of 1000 registered voters. To do probability calculations involving X, we want 

an understanding of the sampling distribution of the sample count X. 

DEFINITION Sampling distribution of the sample count X 

The sampling distribution of the sample count X describes the distribution of values 

taken by the sample count X in all possible samples of the same size from the same 

population. 

The sampling distribution of X is closely related to the binomial distributions that 

you learned about in Lessons 5.3 and 5.4. 

Suppose that a supplier inspects an SRS of 10 flash drives from a shipment of 

10,000 flash drives in which 200 are defective. Let X 5 the number of bad flash drives 

in the sample. This is not quite a binomial setting. Because we are sampling without 

replacement, the independence condition is violated. The conditional probability that 

the second flash drive chosen is bad changes when we know whether the first is good 

or bad: P(second is bad | first is good) 5 200/9999 5 0.0200 but P(second is bad | first 

is bad) 5 199/9999 5 0.0199. These probabilities are very close because removing 

1 flash drive from a shipment of 10,000 changes the makeup of the remaining 9999 

flash drives very little. The distribution of X is very close to the binomial distribution 

with n 5 10 and p 5 0.02. 


19. (a) Yes. Binary? “Success” 5 female is 

selected. “Failure” 5 male is selected. 

Independent? Knowing whether or not one 

randomly selected student is a female tells you 

nothing about whether or not another randomly 

selected student is a female. Number? n 5 40. 

Same probability? p = 55 

104 = 0.529 

(b) m X = np = 40(0.529) = 21.16 

s X = "np(1 − p) = "40(0.529)(1 − 0.529) 

= "40(0.529)(0.471) = "9.97 = 3.16 

(c) P(X = 0) = 10 C 0 (0.529) 0 (1 − 0.529) 10 

= 1(0.529) 0 (0.471) 10 = 0.0005 

P(X = 1) = 10C 1 (0.529) 1 (1 − 0.529) 9 

= 10(0.529) 1 (0.471) 9 = 0.006 

417 

18/08/16 5:00 PM 

P(X ≤ 1) = P(X = 0) + P(X = 1) 

= 0.0005 + 0.006 = 0.0065 

If the professor were to randomly choose 

students for the first 10 meetings, there is 

less than a 1% chance that he would select 

1 female or fewer purely by chance. Because 

this is unlikely, we have convincing evidence 

that his selection process is not really random. 

Teaching Tip 

To save time, Lessons 6.2 and/or 6.3 

can be skipped without losing much 

continuity in future chapters. However, 

the other lessons in this chapter are 

crucial to understanding much of the 

remainder of the course. 





d 6.3.1 

d 6.3.2 

d 6.3.3 

BELL RINGER 

According to the manufacturer, the 

true proportion of blue M&M’S® milk 

chocolate candies is 0.24. If Mrs. Gallas 

takes a random sample of 50 candies, 

how many should she expect to be blue? 

If she repeatedly takes random samples 

of 50 candies, will she always get the 

same number of blue candies? Why or 

why not? 

FYI 

When sampling with replacement from 

a finite population or sampling from an 

infinite population, the distribution of 

the sample count is exactly binomial. 

However, these types of sampling 

methods are not common in practice. It 

is far more common to sample without 

replacement from a finite population. 

In this case, the sample count is usually 

approximately binomial. 

Lesson 6.3 

L E S S O N 6.3 • The Sampling Distribution of a Sample Count 417 


11/01/17 3:54 PM

418 


Teaching Tip 

The rule of thumb mentioned here is 

sometimes called the 10% condition. As 

long as the sample size is less than 10% 

of the population size, the sample count 

will have a distribution that is close 

enough to a binomial distribution that 

binomial probabilities will be reasonably 

accurate. 

Teaching Tip 

Remind students that the mean is also 

called the expected value! 

In practice, we can ignore the violation of the independence condition caused by 

sampling without replacement whenever the sample size is relatively small compared 

to the population size. Specifically, we can assume that the sampling distribution of a 

sample count X is approximately binomial when the sample size is less than 10% of 

the population size. 


Because the sampling distribution of a sample count X is approximately binomial 

when the sample is a small fraction of the population, we can use the formulas from 

Lesson 5.4 to calculate the mean and standard deviation of X. 

How to Calculate μ x and σ x for a Binomial Distribution 

Suppose X is the number of successes in a random sample of size n from a large population 

with proportion of successes p. Then: 

• The mean of the sampling distribution of X is m X = np. 

• The standard deviation of the sampling distribution of X is s X = "np(1 − p). 

The formula for the mean is always correct, even if we are sampling without 

replacement. However, the formula for the standard deviation is not appropriate to 

use when the sample size is more than 10% of the population size. 


Rubber ducky, are you the one? 

Mean and SD of the sampling 

distribution of X 

PROBLEM: A popular carnival game 

has players choose a rubber duck from 

a small pool and look under the duck. 

If a special mark is on the duck, the 

player wins a prize. Suppose that a pool 

has 5000 ducks, 800 of which have the 

special mark. One generous father pays 

for his children to choose 20 rubber 

ducks. Let X 5 the number of ducks with 

the mark in the sample of 20. 

(a) Calculate the mean and standard 

deviation of the sampling distribution of X. 

(b) Interpret the standard deviation 

from part (a). 

a 

e XAMPLe 

How many flash drives are defective? 

Mean and SD of the sampling distribution of X 

PROBLEM: Two percent of the flash drives in a shipment of 10,000 flash drives are defective. An 

inspector randomly selects 10 flash drives from the shipment and records X 5 the number of 

defective flash drives in the sample. 

(a) Calculate the mean and standard deviation of the sampling distribution of X. 

(b) Interpret the standard deviation from part (a). 

SOLUTION: 

(a) m x = 10(0.02) = 0.2 flash drives 

s x = "10(0.02)(1 − 0.02) = 0.44 f lash drives 

(b) If the inspector took many samples of size 10, the number of 

defective flash drives would typically vary by about 0.44 from 

the mean of 0.2. 

Because the sample size is less than 10% of the 

population size, the distribution of X is approximately 

binomial with n 5 10 and p 5 0.2. The mean is m X = np 

and the standard deviation is s X = "np(1 − p). 


SOLUTION: 

(a) m X = 20a 800 b = 20(0.16) = 3.2 

5000 

ducks with the mark 

s X = "20(0.16)(1 − 0.16) = 1.64 ducks 

with the mark 

(b) If the children took many samples 

of 20 ducks, the number of ducks with 

the mark would typically vary by about 

1.64 from the mean of 3.2. 


Teaching Tip 

The interpretation of the mean and standard 

deviation in the preceding example is the 

same as in past chapters. Connect the ideas of 

past lessons to the current one! 


418 



11/01/17 3:55 PM

L E S S O N 6.3 • The Sampling Distribution of a Sample Count 

419 

Shape 

As you learned in Lesson 5.3, the shape of a binomial distribution can be skewed to 

the right, skewed to the left, or roughly symmetric. The histogram in Figure 6.6 shows 

the sampling distribution of X 5 the number of defective flash drives from the previous 

example. It is clearly skewed to the right. 

Probability 

.90 

.72 

.54 

.36 

FigUre 6.6 Probability 

histogram of X 5 the 

number of defective flash 

drives in a sample of size 


in which p 5 0.02. 

Teaching Tip 

Remind students that the mean of a 

distribution is the balancing point of its 

histogram. 

Lesson 6.3 


.18 

.00 

0 1 2 3 4 5 6 7 8 9 10 

Number of defective flash drives 

The following activity explores the shape of the sampling distribution of a sample 

count for various combinations of n and p. 

AcT iviT y 

Simulating with the Normal Approximation to Binomial Distributions applet 

In this activity, you will explore the shape of the sampling distribution of a sample count X using an applet from 

the book’s website. 

1. Launch the Normal Approximation to the Binomial Distributions 

applet at highschool.bfwpub.com/spa3e. You 

.90 

will see a histogram with a normal curve overlaid. 

.72 

2. Using the sliders, set the number of trials to n 5 10 and the 

probability of success to p 5 0.02. Hint: You can also use 

.54 

the arrow keys on your computer’s keyboard to move the 

sliders. The normal curve has the same mean and standard 

deviation as the histogram, but it doesn’t model the 

.36 

histogram very well. 

.18 

3. Use the slider (or the arrow keys) to gradually 

change the probability from p 5 0.02 to p 5 1.00 

while keeping the number of trials the same. 

Does the normal curve fit well when p is close to 

0? Close to 0.5? Close to 1? 

4. Keep the number of trials set to n 5 10 and 

change the probability to p 5 0.1. Use the slider 

(or the arrow keys) to gradually increase the 

sample size from n 5 10 to n 5 100. Does the 

Probability 

.00 

0 1 2 

0.2 

3 4 5 6 7 8 9 10 

normal curve fit the histogram better when n is 

smaller or larger? 

5. Under what conditions will the distribution 

of X be approximately normal? Under what 

conditions will the distribution of X not be 

approximately normal? 

18/08/16 5:00 PM 



Materials: An Internet-connected device 

for each student or group of students 

Teaching Advice: This activity 

helps students understand that the 

distribution of a sample count (a 

binomial distribution) is sometimes 

approximately normal, but not always 

approximately normal. The binomial 

distribution is never exactly a normal 

distribution. 

If you don’t have enough devices 

for each student, students can work in 

groups or you can demonstrate the applet 

to the entire class. Showing the applet 

as a demonstration also saves time, but 

doesn’t engage students as much. 

After students have had time to discuss 

their answers, be sure to emphasize the 

following points with the class: 

• The binomial distribution is never 

exactly normal. 

• The binomial distribution is perfectly 

symmetric when p 5 0.5. 

• The binomial distribution is more 

approximately normal as n increases. 

• Therefore, the criterion for deciding 

when a binomial distribution is “close” 

to normal should include the values 

of both n and p. 

Answers: 

1. Students should launch the applet. 

2. Students should input the specified 

values for n and p. 

3. The normal curve doesn’t 

approximate the binomial 

distribution very well. 

4. The normal curve fits the binomial 

best when p 5 0.5. It doesn’t fit well 

at all when p is close to 0 or 1. 

5. Student answers will vary. The normal 

curve fits better when for values of p 

near 0.5 and larger values of n. 



11/01/17 3:55 PM

420 


As you learned in the activity, the shape of the sampling distribution of X will be 

approximately normal when the sample size is large enough. You also learned that 

“large enough” depends on the value of p. The farther p is from 0.5, the larger the 

sample size needs to be, as shown in Figure 6.7. 

0.330 

0.234 

0.148 

Probability 

0.165 

Probability 

0.117 

Probability 

0.074 

Teaching Tip 

Make sure students understand that 

both np and n(1 2 p) must be checked to 

see if they are at least 10. 

FYI 

The Large Counts condition presented 

here is one rule of thumb for 

ensuring that a binomial distribution 

is approximately normal. Another 

criterion is np ≥ 5 and n(1 2 p) ≥ 5. We 

recommend the Large Counts condition 

stated here. 

0.000 

0 8 10 

(a) 

n = 10, p = 0.8 

0.000 

0.000 

0 16 20 

0 40 50 

(b) n = 20, p = 0.8 

(c) 

n = 50, p = 0.8 

FigUre 6.7 Histograms of the sampling distribution of a sample count X with (a) n 5 10 and p 5 0.8, 

(b) n 5 20 and p 5 0.8, and (c) n 5 50 and p 5 0.8. As n increases, the shape of the sampling distribution 

gets closer and closer to normal. 

In practice, the sampling distribution of a sample count will have an approximately 

normal distribution when the Large Counts condition is met. 

DEFINITION the Large counts condition 

Suppose X is the number of successes in a random sample of size n from a population 

with proportion of successes p. The Large Counts condition says that the distribution of 

X will be approximately normal when 

np ≥ 10 and n(1 2 p) ≥10 

This condition is called “large counts” because np is the expected (mean) count of 

successes and n(1 2 p) is the expected (mean) count of failures. 


Spend or save? 

Shape of the sampling distribution of a 

sample count 

PROBLEM: Suppose that 24% of all 

Americans have more debt on their credit 

cards than they have money in their 

savings accounts. Let X 5 the number of 

Americans with more debt than savings 

in a random sample of 40 Americans. 

Would it be appropriate to use a normal 

distribution to model the sampling 

distribution of X ? Justify your answer. 

SOLUTION: 

a 

e XAMPLe 

How many teens have debit cards? 

Shape of the sampling distribution of a sample count 

PROBLEM: Suppose that 12% of teens in a large city 

have a debit card. Let X 5 the number of teens with a 

debit card in a random sample of 500 teens from this 

city. Would it be appropriate to use a normal distribution 

to model the sampling distribution of X? Justify 

your answer. 

SOLUTION: 

Because np 5 500(0.12) 5 60 ≥ 10 and n (1 2 p ) 5 500 

(1 2 0.12) 5 440 ≥ 10, the sampling distribution of X is 


Check the Large Counts condition to determine if X will 

have an approximately normal distribution. 

In this context, 60 is the expected (mean) count of teens 

with a debit card and 440 is the expected (mean) count 

of teens without a debit card. 


Because np 5 40(0.24) 5 9.6 < 10, 

the sampling distribution of X is not 

approximately normal. Although 

n(1 2 p) 5 40(1 2 0.24) 5 30.4 ≥ 10, 

we have still not met the Large Counts 

condition. 



420 



11/01/17 3:55 PM


421 

Finding Probabilities Involving X 

When the Large Counts condition is met, we can use a normal distribution to calculate 

probabilities involving X 5 the number of successes in a random sample of 

size n. 

Is it fun to shop anymore? 

Probabilities involving X 

PROBLEM: Sample surveys show that fewer people 

enjoy shopping than in the past. A survey asked a 

nationwide random sample of 2500 adults if they 

agreed or disagreed with the statement “I like 

buying new clothes, but shopping is often frustrating 

and time-consuming.” 5 Suppose that exactly 

60% of all adult U.S. residents would say “Agree” if 

asked the same question. Calculate the probability 

that at least 1520 members of the sample would say 

“Agree.” 

SOLUTION: 

• Mean: m X 5 2500(0.60) 5 1500 

• SD: s X = "2500(0.6)(1 − 0.6) 5 24.49 

• Shape: Approximately normal because 

np 5 2500 (0.60) 5 1500 ≥ 10 and 

n (1 2 p ) 5 2500 (1 2 0.6) 5 1000 ≥ 10 

1426.53 1451.02 1475.51 1500 1524.49 1548.98 1573.47 

1520 

Sample count who would say “Agree” 

1520 − 1500 

Using Table A: Z = = 0.82 

24.49 

P (Z ≥ 0.82) 5 1 2 0.7939 5 0.2061 

Using technology : Applet/normalcdf (lower:1520, 

upper:100000, mean:1500, SD: 24.49) 5 0.2071 

e XAMPLe 

Let X 5 the number who would say “Agree.” The 

sampling distribution of X is approximately binomial 

with n 5 2500 and p 5 0.60. 

To use a normal approximation to calculate probabilities 

involving X, we need to know the mean, standard 

deviation, and shape of the sampling distribution of X. 

Recall that the mean is m X = np and the standard deviation 

is s X = "np(1 − p). 

1. Draw a normal distribution. 

2. Perform calculations. 

(i) Standardize the boundary value and use Table A to 

find the desired probability; or 

(ii) Use technology. 


Chris Hondros/Getty Images 


The most romantic dinner? 

Probabilities involving X 

PROBLEM: Suppose that 23% of adult 

Americans would say the most romantic 

Valentine’s dinner option is preparing 

a home-cooked meal together. If you 

interviewed a random sample of 800 adult 

Americans, what is the probability that 

165 or fewer would say that this is the 

most romantic dinner option? 

SOLUTION: 

Let X 5 the count of adult Americans in 

the sample who would choose this option. 

• Mean: np 5 800(0.23) 5 184 

• SD: "np(1 − p)= "800(0.23)(1 − 0.23) 

= 11.90 


np 5 800(0.23) 5 184 ≥ 10 and 

n(1 2 p) 5 800(1 2 0.23) 5 616 ≥ 10. 

165 

Lesson 6.3 


Teaching Tip 

In the “Is it fun to shop anymore?” example, 

the probability can be computed as a 

binomial distribution with the Probability 

applet or with a graphing calculator 

command: 1 2 binomcdf(2500, 0.6, 1519). 

The answer is 0.213, which is very close to the 

value using the normal approximation (0.207). 

Make sure your students understand that 

probabilities using the normal approximation 

will be close as long as the Large Counts 

condition is met. 

Teaching Tip 

18/08/16 5:01 PM 

Students might ask why we go to the 

trouble of using a normal approximation to 

calculate probabilities when one could just 

use the binomial distribution to calculate the 

probability. They’re not wrong! The reason is 

that using a single distribution (the normal 

distribution) in the following chapters will 

make our work much simpler. 

148.3 160.2 172.1 184.0 195.9 207.8 219.7 

Sample count who would 

choose this option 

165 − 184 

Using Table A: z = = −1.60 

11.9 

P(Z ≤ 21.60) 5 0.0548 

Using technology: Applet/normalcdf 

(lower:2100000, upper:165, mean:184, 

SD: 11.90) 5 0.0552 



11/01/17 3:55 PM

422 


Lesson App 

Answers 

1. m X = np = 1500(0.12) = 180; s X = 

"np(1−p) ="1500(0.12)(1 − 0.12) 

= 12.59. If many samples of size 1500 

were taken, the number of American 

adults who identify themselves as black 

would typically vary by about 12.59 from 

the mean of 180. 

2. Because np = 1500(0.12) = 180 $ 10 

and n(1 − p) = 1500(1 − 0.12) = 1320 

$ 10, the sampling distribution of X is 


155 − 180 

3. z = ≈ −1.99; 

12.59 

205 − 180 

z = ≈ 1.99 

12.59 

P(155 ≤ X ≤ 205) ≈ P(−1.99 ≤ Z 

# 1.99) = 0.9767 − 0.0233 = 0.9543 



SD:12.59) 5 0.9529 

4. There’s about a 95% chance that the 

number of randomly selected American 

adults (out of an SRS of 1500) who will 

identify themselves as black is between 

155 and 205. So if our sample had 

fewer than 155 American adults who 

identify themselves as black, we would 

suspect that black Americans are being 

underrepresented in the sample. This 

same approach could be used to check 

for undercoverage using other variables 

known about the population. 

TRM Quiz 6A: Lessons 6.1–6.3 

You can find a prepared quiz for Lessons 

6.1–6.3 by clicking on the link in the 



on the TRFD. 


How can we check for bias in a survey? 

One way of checking the effect of undercoverage, 

nonresponse, and other sources of bias in a sample 

survey is to compare the sample with known facts 

about the population. About 12% of American adults 

identify themselves as black. Suppose we take an SRS 

of 1500 American adults and let X be the number of 

blacks in the sample. 

1. Calculate the mean and standard deviation of the 

sampling distribution of X. Interpret the standard 

deviation. 

2. Justify that the sampling distribution of X is 


Exercises 

Lesson 6.3 



Calculate the mean and the standard deviation of the sampling 

distribution of a sample count and interpret the standard deviation. 

Determine if the sampling distribution of a sample count is 


If appropriate, use the normal approximation to the binomial 

distribution to calculate probabilities involving a sample count. 


1. Lefties Eleven percent of students at a large high 

school are left-handed. A statistics teacher selects a 

random sample of 100 students and records X 5 

the number of left-handed students in the sample. 

(a) Calculate the mean and standard deviation of the 

sampling distribution of X. 


2. Hip dysplasia Dysplasia is a malformation of the 

hip socket that is very common in certain dog 

breeds and causes arthritis as a dog gets older. 

According to the Orthopedic Foundation for 

Animals, 11.6% of all Labrador retrievers have 

pg 418 

Lesson 6.3 

3. Calculate the probability that an SRS of 1500 American 

adults will contain between 155 and 205 blacks. 

4. Explain how a polling organization could use the 

results from the previous question to check for 

undercoverage and other sources of bias. 

p. 418 1–4 

p. 420 5–8 

p. 421 9–12 

hip dysplasia. 6 A veterinarian tests a random 

sample of 50 Labrador retrievers and records 

Y 5 the number of Labs with dysplasia in the 

sample. 


sampling distribution of Y. 


3. NASCAR cards and cereal boxes In an attempt to 

increase sales, a breakfast cereal company decides to 

offer a promotion. Each box of cereal will contain 

a collectible card featuring one NASCAR driver: 

Kyle Busch; Dale Earnhardt, Jr.; Kasey Kahne; 

Danica Patrick; or Jimmie Johnson. The company 

says that each of the 5 cards is equally likely to 

Rawpixel Ltd/Getty Images 


Exercises 


lesson by clicking on the link in the TEbook, 

logging into the Teacher’s Resource 

site, or accessing this resource on the TRFD. 

Answers to Lesson 6.3 

Exercises 

1. (a) m X = np = 100(0.11) = 11; 

s X = "np(1−p) = "100(0.11)(1− 0.11) 

= 3.13 

(b) If many samples of size 100 were 

taken, the number of students who are 

left-handed would typically vary by 

about 3.13 from the mean of 11. 


2. (a) m Y = np = 50(0.116) = 5.8; 

s Y = "np(1− p) = "50(0.116)(1 − 0.116) 

= 2.26 

(b) If many samples of size 50 were taken, the 

number of Labs that have dysplasia would 

typically vary by about 2.26 from the mean 

of 5.8. 

3. (a) m X = np = 12a 1 5 b = 2.4; 

s X = "np(1 − p) 

= Å 

12a 1 5 b a1−1 5 b = 1.39 


number of Kyle Busch cards would typically 

vary by about 1.39 from the mean of 2.4. 

4. (a) m Y = np = 50(0.20) = 10; 

s Y = "np(1− p) = "50(0.2)(1− 0.2) = 2.83 


number of individuals who have never married 

would typically vary by about 2.83 from the 

mean of 10. 

5. Yes; because np = 100(0.11) = 11$ 10 

and n(1−p) = 100(1−0.11) = 89 ≥ 10, the 

sampling distribution of X is approximately 

normal. 

6. No; because np = 50(0.116)= 5.8 < 10, 

the sampling distribution of Y is not 


7. No; because np = 12a 1 b = 2.4 < 10, 

5 

the sampling distribution of X is not 



422 



11/01/17 3:55 PM


423 


appear in the 100,000 boxes of cereal that are part 

of this promotion. You buy 12 boxes and let X 5 

the number of Kyle Busch cards in the sample. 




4. What, me marry? In the United States, 20% of 

adults ages 25 and older have never been married, 

more than double the figure recorded for 1960. 7 

Select a random sample of 50 U.S. adults ages 25 

and older and let Y 5 the number of individuals in 

the sample who have never married. 


sampling distribution of Y. 


5. Are lefties normal? Refer to Exercise 1. Would it be 

pg 420 appropriate to use a normal distribution to model 

the sampling distribution of X 5 the number of 

left-handed students in the sample? Justify your 

answer. 

6. Is hip dysplasia normal? Refer to Exercise 2. Would it 

be appropriate to use a normal distribution to model 

the sampling distribution of Y 5 the number of Labs 

with dysplasia in the sample? Justify your answer. 

7. Is NASCAR normal? Refer to Exercise 3. Would 

it be appropriate to use a normal distribution to 

model the sampling distribution of X 5 the number 

of Kyle Busch cards in the sample? Justify your 

answer. 

8. A normal marriage? Refer to Exercise 4. Would 


model the sampling distribution of Y 5 the number 

of individuals in the sample who have never married? 

Justify your answer. 

9. Lefties are all right Refer to Exercises 1 and 5. 

pg 421 Calculate the probability that at least 15 of the 

members of the sample are left-handed. 

10. Never been married Refer to Exercises 4 and 8. 

Calculate the probability that at most 5 of the individuals 

in the sample have never been married. 

11. Public transportation In a large city, 34% of residents 

use public transportation at least once per 

week. If the mayor selects a random sample of 200 

residents, calculate the probability that at most 60 

residents in the sample use public transportation at 

least once per week. 

12. U.S. quarters According to www.usmint.gov, 54% 

of the quarters minted in 2014 were produced by 

the U.S. Mint in Denver, Colorado (the rest were 

produced in Philadelphia). In a random sample of 

200 quarters, what is the probability that at least 

115 of them were minted in Denver? 

8. Yes; because np = 50(0.20) = 10 $ 10 

and n(1− p) = 50(1− 0.20)= 40 ≥ 10, the 

sampling distribution of Y is approximately 

normal. 

9. From Exercises 1 and 5, X is approximately 

normal with a mean of 11 and standard 

deviation of 3.13. 

15 − 11 

z = ≈ 1.28; P(X ≥ 15)≈ P(Z ≥ 1.28) 

3.13 

= 1− 0.8997 = 0.1003 

Using technology: Applet/normalcdf(lower:15, 

upper:1000, mean:11, SD:3.13) 5 0.1006 

10. From Exercises 4 and 8, Y is 

approximately normal with a mean of 10 and 

standard deviation of 2.83. 

z = 5 − 10 

2.83 ≈ −1.77; 

P(Y ≤ 5) ≈ P(Z ≤ −1.77) = 0.0384 


13. Tasty chips For a statistics project, Zenon decided 

to investigate if students at his school prefer namebrand 

potato chips to store-brand potato chips. He 

prepared two identical bowls of chips, filling one 

with name-brand chips and the other with storebrand 

chips. Then, he selected a random sample 

of 30 students, had each student try both types of 

chips in random order, and recorded which type of 

chip each student preferred. Assume that 50% of 

students at Zenon’s school prefer the name-brand 

chips. Let X 5 the number of students in the sample 

that prefer the name-brand chips. 8 



deviation. 

(b) Justify that the distribution of X is approximately 

normal. 

(c) Calculate the probability that 19 or more of the 

students will prefer the name-brand chips. 

14. Blood types About 10% of people in the United 

States have type B blood. Suppose we take a random 

sample of 120 U.S. residents, and let X 5 the number 

of residents in the sample who have type B blood. 



deviation. 

(b) Justify that the distribution of X is approximately 

normal. 

(c) Calculate the probability that 16 or more individuals 

in the sample have type B blood. 

15. More chips! Refer to Exercise 13. In Zenon’s study, 

19 of the 30 students chose the name-brand chips. 

Based on your answer to Exercise 13(c), does this 

provide convincing evidence that more than half of 

the students at Zenon’s school prefer name-brand 

potato chips? Explain. 

16. More on blood type Refer to Exercise 14. Some people 

believe that one’s blood type has an impact on personality. 

For example, people with type B blood are 

supposed to be more creative, active, and passionate. 

To test this hypothesis, Jason selects a random sample 

of 120 art, music, and drama majors at his college 

and finds that 16 of them have type B blood. Based on 

your answer to Exercise 14(c), does this provide convincing 

evidence that art, music, and drama majors at 

Jason’s college are more likely than the general population 

to have type B blood? Explain. 


17. Binomial transportation Refer to Exercise 11. Use 

a binomial distribution to calculate the probability 

that at most 60 residents in the sample use public 

transportation at least once per week. Hint: See 

Lesson 5.4. 

18/08/16 5:01 PM 


−1000, upper:5, mean:10, SD:2.83) 5 0.0386 

11. X 5 the number of residents who use 

public transportation at least once per week. 

Mean: m X = np = 200(0.34) = 68; 

SD: s X = "np(1 − p) 

= "200(0.34)(1 − 0.34) = 6.70 

Shape: Approximately normal because 

np = 200(0.34) = 68 ≥ 10 and 

n(1−p) = 200(1− 0.34) = 132 ≥ 10. 

60 − 68 

z = ≈ −1.19; 

6.70 

P(X ≤ 60) ≈ P(Z ≤ −1.19) = 0.1170 


(lower:−1000, upper:60, mean:68, 

SD:6.70) 5 0.1162 

12. X 5 the number of quarters minted 

in Denver. 

Mean: m X = np = 200(0.54) = 108; 

SD: s X = "np(1 − p) 

= "200(0.54)(1 − 0.54) = 7.05 


np = 200(0.54) = 108 ≥ 10 and 

n(1− p) = 200(1− 0.54) = 92 ≥ 10. 

115 − 108 

z = ≈ 0.99; P(X ≥ 115) 

7.05 

≈ P(Z ≥ 0.99) = 1− 0.8389 = 0.1611 



SD:7.05) 5 0.1604 

13. (a) m X = np = 30(0.5) = 15 

students; s X = "np(1− p) 

= "30(0.5)(1 − 0.5) = 2.74 students. 

If many samples of size 30 were taken, the 

number of students who prefer namebrand 

chips would typically vary by about 

2.74 from the mean of 15. 

(b) Because np = 30(0.5) = 15 ≥ 10 

and n(1 − p) = 30(1 − 0.5) = 15 ≥ 10, 

the sampling distribution of X is 


19 − 15 

(c) z = ≈ 1.46; P(X ≥ 19) ≈ 

2.74 

P(Z ≥ 1.46) = 1− 0.9279 = 0.0721 



SD:2.74) 5 0.0722 

14. (a) m X = np = 120(0.1) = 12 

residents; s X = "np(1− p) 

= "120(0.1)(1 − 0.1) = 3.29 residents. 

If many samples of size 120 were 

taken, the number of residents who have 

type B blood would typically vary by 

about 3.29 from the mean of 12. 

(b) Because np = 120(0.1) = 12 $ 10 

and n(1− p) = 120(1 − 0.1) = 108 $ 10, 

the sampling distribution of X is 


16 − 12 

(c) z = ≈ 1.22; P(X ≥ 16) ≈ 

3.29 

P(Z ≥ 1.22) = 1 − 0.8888 = 0.1112 



SD:3.29) 5 0.112 

15. No; assuming that 50% of students 

prefer name-brand chips, there’s about a 

7% chance that the number of students 

who prefer name-brand chips (out of 

an SRS of 30) is 19 or more. The results 

from Zenon’s study could have happened 

purely by chance, so we do not have 

convincing evidence that more than half 

of the students prefer name-brand chips. 



Lesson 6.3 


11/01/17 3:55 PM


16. No; assuming that 10% of a group 

have type B blood, there’s about an 11% 

chance that the number who have type 

B blood (out of an SRS of 120) is 16 or 

more. The results from Jason’s study 

could have happened purely by chance, 

so we do not have convincing evidence 

that art, music, and drama majors at 

Jason’s college are more likely than the 

general population to have type B blood. 

17. X 5 the number of residents who 

use public transportation at least once 

per week. Using the applet (Probability, 

Binomial distribution, n 5 200, 

p 5 0.34, “at most 60 successes”) or 

binomcdf(n 5 200, p 5 0.34, X 5 60) 

gives P(X ≤ 60) = 0.131. (Compare to 

the answer of 0.1162 from Exercise 11.) 

18. (a) To provide a baseline for 

comparing the effects of the treatment. 

Otherwise, we wouldn’t be able to tell 

if the books or something else (e.g., 

students maturing) caused an increase in 

reading ability. 

(b) The difference in the reading scores 

for the third-grade girls group was too 

large to be due only to chance variation 

in the random assignment to treatments. 

19. 

(a) 

Sister’s height (in.) 

70 

69 

68 

67 

66 

65 

64 

63 

62 

61 

60 

59 

58 

d 

65 

d 

d 

d d d 

66 67 68 69 70 71 72 73 

d 

d 

Brother’s height (in.) 

Direction: Positive. Form: Linear. Strength: 

Moderate. Outliers: No obvious outliers. 

(b) y^ = 27.635 + 0.527x, where x 5 

brother’s height and y 5 sister’s height. 

The slope of the line is 0.527, which tells 

us the predicted sister’s height increases 

by 0.527 inch for each additional increase 

of 1 inch in the brother’s height. 

(c) y^ = 27.635 + 0.527(70) = 64.525 

inches 

(d) The actual height of a sister is typically 

about 2.247 inches away from her 

predicted height using the least-squares 

regression line. 

d 

d 

424 



18. Summer reading (3.6, 3.8) A group of educational 

researchers studied the impact of summer reading 

with a randomized experiment involving secondand 

third-graders in North Carolina. Students 

were randomly assigned to either a group that was 

mailed one book a week for 10 weeks or a control 

group that was not mailed any books. Both groups 

were given a reading comprehension test at the 

start and end of the summer. Third-grade girls who 

were mailed books showed a statistically significant 

increase in reading ability, but third-grade boys and 

second-graders of both genders did not. 9 

(a) Explain the purpose of including a control group in 

this experiment. 

(b) Explain what is meant by “statistically significant 

increase” in the last sentence. 


PD LESSONS 6.4–6.6 Overview 

Watch the Lessons 6.4–6.6 overview video for 

guidance on teaching the content in these 

lessons. Find it in the Teacher’s Resource 

Materials by clicking on the link in the TEbook, 

logging into the Teacher’s Resource site, 

or accessing it on the TRFD. 


19. Sisters and brothers (2.2, 2.5, 2.7) How strongly 

do physical characteristics of sisters and brothers 

correlate? Here are data on the heights (in inches) 

of 11 adult pairs: 10 

Brother 71 68 66 67 70 71 70 73 72 65 66 

Sister 69 64 65 63 65 62 65 64 66 59 62 

(a) Construct a scatterplot using brother’s height as the 

explanatory variable. Describe what you see. 

(b) Use technology to compute the least-squares 

regression line for predicting sister’s height from 

brother’s height. Interpret the slope in context. 

(c) Damien is 70 inches tall. Predict the height of his 

sister Tonya. 

(d) The standard deviation of residuals for this model 

is s 5 2.247. Interpret this value in context. 

Lesson 6.4 


of a Sample Proportion 

d Calculate the mean and standard deviation of the sampling distribution of a 

sample proportion p^ and interpret the standard deviation. 

d Determine if the sampling distribution of p^ is approximately normal. 

d If appropriate, use a normal distribution to calculate probabilities involving p^ . 

What proportion of U.S. teens know that 1492 was the year in which Columbus 

“discovered” America? A Gallup poll found that 210 out of a random sample of 501 

American teens aged 13 to 17 knew this historically important date. 11 The sample 

proportion p^ 5 210/501 5 0.42 is the statistic that we use to estimate the unknown 

population proportion p. Because a random sample of 501 teens is unlikely to perfectly 

represent all teens, we can only say that “about” 42% of U.S. teenagers know 

that Columbus voyaged to America in 1492. 

To understand how much p^ varies from p and what values of p^ are likely to 

happen by chance, we want an understanding of the sampling distribution of the 

sample proportion p^ . 


The problems in the test bank are keyed 

to the learning targets using these 

numbers: 

d 6.4.1 

d 6.4.2 

d 6.4.3 


Teaching Tip 

Be picky with students about the correct 

symbols! The sample proportion is denoted p^ 

and the population proportion is denoted p. 

BELL RINGER 

Mrs. Gallas loves blue M&M’S® milk chocolate 

candies. According to the manufacturer, the 

true proportion of blue M&M’S chocolate 

candies is 0.24. If Mrs. Gallas takes a random 

sample of 50 candies, what proportion of blue 

candies should she expect to have? Is she 

certain to get this proportion? Why or why not? 

424 



11/01/17 3:55 PM

L E S S O N 6.4 • The Sampling Distribution of a Sample Proportion 425 

DEFINITION Sampling distribution of the sample proportion p^ 

The sampling distribution of the sample proportion p^ describes the distribution of 

values taken by the sample proportion p^ in all possible samples of the same size from the 

same population. 

When Mr. Ramirez’s class did the Penny for Your Thoughts activity at the beginning 

of the chapter, his students produced the “dotplot” in Figure 6.8 showing the 

simulated sampling distribution of p^ 5 the sample proportion of pennies from the 

2000s in 50 samples of size n 5 20. 

p 

p 

p p 

p p 

p p p 

p p p p 

p p p p p p 

p p p p p p 

p p p p p p p 

p p p p p p p p 

p p p p p p p p p 

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 

Sample proportion of pennies from 2000s (n = 20) 

This distribution is roughly symmetric, with a mean of about 0.65 and a standard 

deviation of about 0.10. By the end of this lesson, you should be able to anticipate the 

shape, center, and variability of distributions like this one without getting your hands 

dirty in a jar of pennies. 

p 


sampling distribution of 

the sample proportion of 

pennies in 50 samples of 

size n 5 20 from a population 

of pennies. 

Teaching Tip 

In Figure 6.8, ask students what the 

“dot” at 0.35 represents. It is the sample 

proportion of pennies from the 2000s 

for one sample of 20 pennies. That is, 

there were 7 pennies from the 2000s in 

one sample of 20 pennies, resulting in a 

sample proportion of 0.35. 

Teaching Tip 

This figure is a good opportunity to refer 

to the dotplots made by your students in 

the “A penny for your thoughts?” activity 

from Lesson 6.1. Compare the results 

from your class with those from 

Mr. Ramirez’s class. 

Lesson 6.4 


When we select random samples of size n from a population with proportion of 

successes p, the value of p^ will vary from sample to sample. As with the sampling 

distribution of the sample count X, there are formulas that describe the center and 

variability of the sampling distribution of p^ . 

How to Calculate μ p^ and σ p^ 

Suppose that p^ is the proportion of successes in an SRS of size n drawn from a large population 

with proportion of successes p. Then: 

Teaching Tip 

Point out that any description of a 

sampling distribution must include 

information about shape, center, and 

variability. Center and variability will be 

examined in more detail in this lesson. 

• The mean of the sampling distribution of p^ is m p^ = p. 

p(1 − p) 

• The standard deviation of the sampling distribution of p^ is s p^ = . Å n 

Here are some important facts about the mean and standard deviation of the sampling 

distribution of the sample proportion p^ : 

• The sample proportion p^ is an unbiased estimator of the population proportion 

p. This is because the mean of the sampling distribution m p^ is equal to the 

population proportion p. 

• The standard deviation of the sampling distribution of p^ describes the typical 

distance between p^ and the population proportion p. 

FYI 

The formulas for m p^ and s p^ are true for 

the sampling distribution of p^ no matter 

what shape it has. 

Teaching Tip 

This is consistent with previous 

definitions of standard deviation as the 

typical distance a value falls from the 

mean of a distribution. 


FYI 

The exact formula for the standard deviation 

of p^ when sampling without replacement is 

p(1 − p) 

s p^ = # N − n 

. When n is small 

Å n Å N − 1 

relative to N (less than 10%), the value of 

N − n 

is approximately 1 and therefore 

Å N − 1 

p(1 − p) 

doesn’t change the value of Å n 

N − n 

much. The factor is sometimes 

Å N − 1 

called the finite population correction factor. 

18/08/16 5:01 PM 



11/01/17 3:55 PM

426 


• The sampling distribution of p^ is less variable for larger samples. This is indicated 

by the !n in the denominator of the standard deviation formula. 

• The formula for the standard deviation of the distribution of p^ requires that the 

observations are independent. In practice, we are safe assuming independence 

when sampling without replacement as long as the sample size is less than 10% 

of the population size. 


Expensive ride? 

Mean and SD of the sampling 

distribution of p^ 

PROBLEM: Suppose that 26% of all 

high school students at a large school 

spend about half or more of their 

earnings on a car. A random sample 

of 150 students from this school is 

surveyed. Let p^ 5 the proportion of 

students in the sample who spend about 

half or more of their earnings on a car. 


deviation of the sampling distribution of p^ . 



SOLUTION: 

(a) m p^ = 0.26 and 

0.26(1− 0.26) 

s p^ = 

= 0.036 

Å 150 

(b) In SRSs of size n 5 150, the sample 

proportion of students who spend about 

half or more of their earnings on a car 

will typically vary by about 0.036 from 

the true proportion of p 5 0.26. 

a 

e XAMPLe 

What proportion of students have a smartphone? 

Mean and SD of the sampling distribution of p^ 

PROBLEM: Suppose that 43% of students at a large high school own a smartphone. As part of a 

schoolwide technology study, the principal surveys an SRS of n 5 100 students. Let p^ 5 the proportion 

of students in the sample who own a smartphone. 

(a) Calculate the mean and the standard deviation of the sampling distribution of p^ . 


SOLUTION: 

0.43(1 − 0.43) 

(a) m p^ = 0.43 and s p^ = = 0.050 

Å 100 

(b) In SRSs of size n 5 100, the sample proportion of students 

who own a smartphone will typically vary by about 0.050 from 


d 

ThinK ABoUT iT Is the sampling distribution of p^ (the sample proportion of 

successes) related to the sampling distribution of X (the sample count of successes)? 

Yes! 

number of successes in sample 

p^ = = X sample size 

n 

For example, here are dotplots showing the simulated sampling distribution of X 5 the 

number of pennies from the 2000s and p^ 5 the proportion of pennies from the 2000s 

for samples of size 20 in Mr. Ramirez’s class. The distributions are exactly the same, 

other than the scale on the axis. 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

6 8 10 12 14 16 18 20 

Number of pennies from 2000s 

d 

d 

d 

d 

d 

d 

d 

d 

d d 

d d 

d d d 

d 

In this context, n 5 100 and p 5 0.43. The mean is m p^ 5 p 

p(1 − p) 

and the standard deviation is s p^ 5 . Å n 


d 

d d 

d d 

d d d 

d d d d 

d d d d d d 

d d d d d d 

d d d d d d d 



0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 

Proportion of pennies from 2000s 


Common Error 

Students may refer to the sampling 

distribution of the sample proportion of 

pennies as a binomial distribution. Tell them 

this is not technically correct because a 

binomial random variable is a count, not a 

proportion. However, these two distributions 

are mathematical transformations of one 

another. 


426 



11/01/17 3:56 PM



Also, the formulas for the mean and standard deviation of the sampling distribution 

of p^ are derived from the formulas we learned in the previous lesson: 

Shape 

s p^ 5 s X 

n 

m p^ 5 m X 

n = np n = p 

"np(1 − p) np(1 − p) p(1 − p) 

= = = 

n Å n 2 Å n 

Both the sample size and the proportion of successes in the population affect the shape 

of the sampling distribution of the sample proportion p^ . The following activity helps 

you explore the effect of these two factors. 

AcT iviT y 

Sampling from the candy machine 

Imagine a very large candy machine filled with 

orange, brown, and yellow candies. When you insert 

money, the machine dispenses a sample of candies. 

In this activity, you will use an applet to investigate 

the sample-to-sample variability in the proportion of 

orange candies dispensed by the machine. 

1. Launch the Reese’s Pieces® applet at 

www.rossmanchance.com/applets. Click the 

button for “Proportion of orange” to have the 

applet calculate and record the value of p^ 5 the 

sample proportion of orange candies. 

2. Click on the “Draw Samples” button. An animated 

simple random sample of n 5 25 candies should 

be dispensed. The screen shot shows the results 

of one such sample. Was your sample proportion 

of orange candies close to the actual population 

proportion, p 5 0.50? 

Teaching Tip: 

Differentiate 

Students who have difficulty with algebra can 

ignore the mathematical derivations of m p^ 

and s p^ at the top of this page. They are not 

important for understanding the big ideas in 

this and other lessons. 

3. Click “Draw Samples” 9 more times, so that you 

have a total of 10 sample proportions. Look at the 

dotplot of your p^ values. Does the distribution 

have a recognizable shape? 

4. To take many more samples quickly, enter 990 

in the “number of samples” box. Click on the 

“ Animate” box to turn off the animation. Then 

click “Draw Samples.” You have now taken a 

total of 1000 samples of n 5 25 candies from 

the machine. Describe the shape of the simulated 

sampling distribution of p^ shown in the 

dotplot. 

5. How does the shape of the sampling distribution 

of p^ change if the proportion of orange 

candies in the machine is p 5 0.10 instead of 

p 5 0.50? Set the probability of orange 

candies to p 5 0.10 and draw 1000 samples of 

size n 5 25. What if p 5 0.90? Describe how the 

value of p affects the shape of the sampling 

distribution of p^ . 

6. How does the shape of the sampling distribution 

of p^ change if the sample size increases? 

Set the probability of orange to p 5 0.90 and 

the number of candies to n 5 25 and draw 

1000 samples. Then, repeat with sample sizes of 

n 5 100 and n 5 500. Describe how the value of 

n affects the shape of the sampling distribution 

of p^ . 



18/08/16 5:01 PM 

Materials: An Internet-connected device for 

each student or group of students 

Teaching Advice: This activity helps students 

understand the shape of the sampling 

distribution of p^ . Because the sampling 

distribution of p^ is closely related to the 

sampling distribution of the sample count X, 

this activity can be skipped if your students 

understood Lesson 6.3 well. 

If you don’t have enough devices, students 

can work in groups or you can demonstrate 

the applet to the entire class. Showing the 

applet as a demonstration also saves time, 

even if it doesn’t engage students as much. 

This applet gives a visual of the 

population distribution (the candy 

machine), the distribution of one 

sample (the candies in the dish), and the 

sampling distribution (the dotplot). Point 

out these three distributions to your 

students. 

Beginning at Step 4, emphasize the 

difference between “number of candies” 

(sample size) and “number of samples.” 

Students will often be confused about 

the meaning of these terms. Make sure 

students are using the correct values. 

There is nothing special about 1000 

samples. This activity would work just 

as well if students generated 10,000 

samples. What matters is to repeat the 

sampling process often enough to see 

the pattern in the sampling distribution. 

The shape of the distribution of p^ 

follows the same rules as the shape of the 

distribution of the corresponding binomial 

random variable X. This should be evident 

from comparing the results of this activity 

to the results of the activity in Lesson 6.3. 

Answers: 

1. Students should launch the applet 

and click “Proportion of orange.” 

2. Student results will vary. Most students 

should get a sample proportion close to 

0.50, but some students may not. 

3. With only 10 dots on the dotplot, 

the distribution shouldn’t have a 

recognizable shape. 

4. The distribution should look moundshaped 

and roughly symmetric. 

5. When p 5 0.1, the shape of the 

distribution is skewed to the right. 

When p 5 0.9, it is skewed to the left. 

Values of p that are lower than 0.5 

result in right-skewed distributions 

and values higher than 0.5 result in 

left-skewed distributions. 

6. As n increases, the sampling 

distribution of p^ gets more 

approximately normal. This result 

holds true for all values of p. 

Teaching Tip 

The preceding activity helps students 

visualize sampling. This is a good time 

to remind them that the word “sample” 

does not refer to a single individual, 

but to a group of many individuals. 

Make sure your students don’t refer to 

individuals as “samples.” 

Lesson 6.4 



11/01/17 3:56 PM

428 


As you learned in the activity, the shape of the sampling distribution of p^ will be 

closer to normal when the value of p is closer to 0.5 and the sample size is larger. 

These relationships are the same as those you discovered in the previous lesson. And 

the Large Counts condition is the same as well. 

DEFINITION the Large counts condition 

Suppose p^ is the proportion of successes in a random sample of size n from a population 

with proportion of successes p. The Large Counts condition says that the distribution of 

p^ will be approximately normal when 

np ≥ 10 and n(1 2 p) ≥ 10 


Cooking the books? 

Shape of the sampling distribution of p^ 

PROBLEM: To audit one department 

of a large corporation, an accountant 

selects a random sample of n 5 80 

transactions. Historically, the true 

proportion of transactions that are more 

than $1000 is p 5 0.42. Would it be 

appropriate to use a normal distribution 

to model the sampling distribution of p^ 

for samples of size n 5 80? Justify your 

answer. 

SOLUTION: 

Yes; because np 5 80(0.42) 5 33.6 ≥ 10 

and n(1 2 p) 5 80(1 2 0.42) 5 46.4 ≥ 10, 

the sampling distribution of p^ is approximately 

normal. 

a 

a 

e XAMPLe 

A penny for your thoughts? 

Shape of the sampling distribution of p^ 

PROBLEM: Mr. Ramirez’s class did the Penny for Your 

Thoughts Activity from the beginning of this chapter. 

In his population of pennies, the proportion of pennies 

from the 2000s is p 5 0.627. Would it be appropriate 

to use a normal distribution to model the sampling 

distribution of p^ for samples of size n 5 16? Justify 

your answer. 

SOLUTION: 

No. Because n (1 2 p ) 5 16(1 2 0.627) 5 5.968 < 10, 

the sampling distribution of p^ is not approximately normal. 

e XAMPLe 

Finding Probabilities Involving p^ 

When the Large Counts condition is met, we can use a normal distribution to calculate 

probabilities involving p^ 5 the proportion of successes in a random sample 

of size n. 

How far from home do you attend college? 

Normal calculations involving p^ 

Check the Large Counts condition to determine if p^ will 

have an approximately normal distribution. 

PROBLEM: A polling organization asks an SRS of 1500 first-year college students how far away 

their home is. Suppose that 35% of all first-year students attend college within 50 miles of home. 

Find the probability that the random sample of 1500 students will give a result within 2 percentage 

points of this true value. 


Andrew Unangst/Getty Images 


College bound? 

Normal calculations involving p^ 

PROBLEM: In a recent year, a fact sheet 

for the state of Pennsylvania stated that 

72.6% of all public high school seniors 

were planning to attend college the next 

year. If an SRS of 400 public high school 

seniors in Pennsylvania were surveyed, 

what is the probability that the sample 

proportion who are going to college next 

year will be within 3 percentage points 

of the true value? 

• Mean: m p^ 5 0.726 


0.726(1 − 0.726) 

• SD: s p^ = 

= 0.022 

Å 400 


np 5 400(0.726) 5 290.6 ≥ 10 and 

n(1 2 p) 5 400(1 2 0.726) 5 109.6 ≥ 10 

0.696 

0.756 

0.696 − 0.726 


0.022 

0.756 − 0.726 

and z = = 1.36 

0.022 

P(21.36 ≤ Z ≤ 1.36) 5 0.9131 2 0.0869 

5 0.8262 


(lower:0.696, upper:0.756, mean:0.726, 

SD:0.022) 5 0.8273 


SOLUTION: 

Let p^ 5 the proportion in the sample 

who are going to college next year, 

where p 5 0.726 and n 5 400. 

0.660 0.682 0.704 0.726 0.748 0.770 0.792 

Sample proportion who 

are going to college 

428 



11/01/17 3:56 PM


SOLUTION: 

• Mean: m P^ 5 0.35 

0.35(1 − 0.35) 

• SD: s p^ = = 0.0123 

Å 1500 

• Shape: Approximately normal because np 5 1500(0.35) 

5 525 ≥ 10 and n (12 p) 5 1500(12 0.35) 5 975 ≥ 10 

Let p^ 5 the proportion in the sample who attend college 

within 50 miles of home, where p 5 0.35 and n 5 1500. 

To use a normal distribution to calculate probabilities 

involving p^ , we have to know the mean, standard 

deviation, and shape of the sampling distribution of p^ . 

Recall that the mean is m p^ = p and the standard 

p(1 − p) 

deviation is s p^ = . Å n 

Teaching Tip 

In this example, if students wonder what 

to do if the shape isn’t approximately 

normal, the answer is that the 

approximate probability would have 

to be found by using a binomial 

distribution. 

Lesson 6.4 

1. Draw a normal distribution. 

0.3131 0.3254 0.3377 0.35 0.3623 0.3746 0.3869 

0.33 0.37 

Sample proportion who live within 50 miles 

Using Table A: 

0.33 − 0.35 

0.37 − 0.35 

z = = −1.63 and z = = 1.63 

0.0123 

0.0123 

P (21.63 ≤ Z ≤ 1.63) 5 0.94842 0.05165 0.8968 

Using technology: Applet/normalcdf (lower:0.33, upper:0.37, 

mean:0.35, SD: 0.0123) 5 0.8961 







Lesson App 

Answers 


What’s that spot on my potato chip? 

A potato-chip producer and its main supplier agree 

that each shipment of potatoes must meet certain 

quality standards. If the producer finds convincing 

evidence that more than 8% of the potatoes in the 

entire shipment have “blemishes,” the truck will be 

sent away to get another load of potatoes from the 

supplier. Otherwise, the entire truckload will be used 

to make potato chips. To make the decision, a supervisor 

will inspect a random sample of potatoes from the 

shipment. Suppose that the proportion of blemished 

potatoes in the entire shipment is p 5 0.08 and that 

the supervisor randomly selects n 5 500 potatoes for 

inspection. 

Teaching Tip 

The supervisor has made a mistake about 

the proportion of all blemished potatoes 

in the truckload, because getting a sample 

proportion of 0.11 or higher if the true 

proportion in the truckload is 0.08 would 

be highly unlikely to happen just by chance. 

In Chapter 8, we’ll learn that this is called a 

Type I error, or a false positive. Not rejecting 

a truckload with a proportion of blemished 

potatoes higher than 0.08 would be a Type II 

error, or false negative. 

1. Calculate the 

mean and 

standard 

deviation of the sampling distribution of p^ . 

Interpret the standard deviation. 

2. Justify that the sampling distribution of p^ is 


3. Calculate the probability that at least 11% of the 

potatoes in the sample are blemished. 

4. Based on your answer to Question 3, what should 

the supervisor conclude if he selects an SRS of size 

n 5 500 and finds p^ 5 0.11? Explain. 

Africa Studio/Shutterstock 

18/08/16 5:02 PM 

p(1− p) 

1. m p^ = p = 0.08; s p^ = 

Å n 

0.08(1− 0.08) 

= 

= 0.012 

Å 500 

In SRSs of size n 5 500, the sample 

proportion of blemished potatoes will 

typically vary by about 0.012 from the 

true proportion of p 5 0.08. 

2. Because np = (500)(0.08) = 40 ≥ 10 

and n(1 − p) = (500)(1 − 0.08) = 460 

$ 10, the sampling distribution of p^ is 


0.11 − 0.08 

3. z = = 2.5; P( p^ ≥ 0.11) 

0.012 

= P(Z ≥ 2.5) = 1 − 0.9938 = 0.0062 


(lower:0.11, upper:1000, mean:0.08, 

SD:0.012) 5 0.0062 

4. Send the shipment back. Assuming 

the true proportion of blemished 

potatoes is 0.08, there is only a 0.62% 

chance of getting a sample proportion of 

0.11 or higher purely by chance. Because 

this result is unlikely (less than 5%), we 

have convincing evidence that more 

than 8% of the potatoes in this shipment 

have blemishes. 



11/01/17 3:56 PM

430 


TRM chapter 6 Activity: 

Sampling Movies 

This activity reviews the sampling 

distribution of p^ by sampling from a 

population of movies. Access this resource 

by clicking on the link in the TE-book, 

logging into the Teacher’s Resource site, or 

accessing this resource on the TRFD. 


Exercises 


lesson by clicking on the link in the TEbook, 

logging into the Teacher’s Resource 

site, or accessing this resource on the TRFD. 


p(1− p) 

1. (a) m p^ = p = 0.20; s p^ = 

Å n 

0.20(1− 0.20) 

= 

= 0.089 

Å 20 


proportion of orange Skittles® will 

typically vary by about 0.089 from the 

true proportion of p 5 0.20. 

p(1 − p) 

2. (a) m p^ = p = 0.55; s p^ = 

Å n 

0.55(1 − 0.55) 

= = 0.022 

Å 500 


proportion of Democrats will typically 

vary by about 0.022 from the true 

proportion of p 5 0.55. 

p(1 − p) 

3. (a) m p^ = p = 0.90; s p^ = 

Å n 

0.90(1 − 0.90) 

= = 0.03 

Å 100 


proportion of orders shipped within three 

working days will typically vary by about 

0.03 from the true proportion of p 5 0.90. 

p(1 − p) 

4. (a) m p^ = p = 0.59; s p^ = 

Å n 

0.59(1 − 0.59) 

= = 0.07 

Å 50 


proportion of couples in which both 

parents work outside the home will 

typically vary by about 0.07 from the true 

proportion of p 5 0.59. 

5. No; because np = (10)a 16 

120 b =1.33 

< 10, the sampling distribution of p^ is 

not approximately normal. 

6. No; because np = (7)a 42 

100 b = 2.94 

< 10, the sampling distribution of p^ is 

not approximately normal. 

430 

Exercises 


Lesson 6.4 



Calculate the mean and standard deviation of the sampling 

distribution of a sample proportion p^ and interpret the standard 

deviation. 


7. Yes; because np = (100)(0.90) = 90 ≥ 10 

and n(1 − p) = (100)(1 − 0.90) = 10 ≥ 10, 


normal. 

8. Yes; because np = (50)(0.59) = 29.5 ≥ 10 

and n(1− p) = (50)(1− 0.59) = 20.5 ≥ 10, 


normal. 

p(1 − p) 

9. Mean: m p^ = p = 0.70; SD: s p^ = 

Å n 

0.70(1 − 0.70) 

= = 0.028 

Å 267 


np = 267(0.70) = 186.9 ≥ 10 and 

n(1 − p) = 267(1 − 0.70) = 80.1 ≥ 10 

0.75 − 0.70 

z = = 1.79; P( p^ ≥ 0.75) 

0.028 

= P(Z ≥ 1.79) = 1 − 0.9633 = 0.0367 

p. 426 1–4 

Determine if the sampling distribution of p^ is approximately normal. p. 428 5–8 

If appropriate, use a normal distribution to calculate probabilities 

involving p^ . 


1. Orange Skittles ® The makers of Skittles claim that 

20% of Skittles candies are orange. You select a 

random sample of 20 Skittles from a large bag. Let 

p^ 5 the proportion of orange Skittles in the sample. 

(a) Calculate the mean and the standard deviation of 

the sampling distribution of p^ . 


2. Registered voters In a congressional district, 55% 

of registered voters are Democrats. A polling 

organization selects a random sample of 500 registered 

voters from this district. Let p^ 5 the proportion 

of Democrats in the sample. 




3. On-time shipping A large mail-order company 

advertises that it ships 90% of its orders within 

3 working days. You select an SRS of 100 of the 

orders received in the past week for an audit. Let 

p^ 5 the proportion of orders in the last week that 

were shipped within 3 working days. 




4. Married with children According to a recent U.S. 

Bureau of Labor Statistics report, the proportion of 

married couples with children in which both parents 

work outside the home is 59%. 12 You select 

an SRS of 50 married couples with children and 

let p^ 5 the sample proportion of couples in which 

both parents work outside the home. 




pg 426 

Lesson 6.4 

p. 428 9–12 

5. Airport safety The Transportation Security Administration 

(TSA) is responsible for airport safety. On 

pg 428 

some flights, TSA officers randomly select passengers 

for an extra security check before boarding. One 

such flight had 120 passengers—16 in first class and 

104 in coach class. TSA officers selected an SRS of 

10 passengers for screening. Would it be appropriate 

to use a normal distribution to model the sampling 

distribution of p^ 5 the proportion of first-class passengers 

in the sample? Justify your answer. 

6. Only vowels? In the game of Scrabble, each player 

begins by drawing 7 tiles from a bag containing 

100 tiles. There are 42 vowels, 56 consonants, and 

2 blank tiles in the bag. Cait chooses an SRS of 

7 tiles. Would it be appropriate to use a normal 

distribution to model the sampling distribution of 

p^ 5 the proportion of vowels in her sample? Justify 

your answer. 

7. Model shipping? Refer to Exercise 3. Would it be 

appropriate to use a normal distribution to model 

the sampling distribution of p^ 5 the proportion of 

orders in the last week that were shipped within 3 

working days? Justify your answer. 

8. A model marriage? Refer to Exercise 4. Would 


model the sampling distribution of p^ 5 the sample 

proportion of couples in which both parents work 

outside the home? Justify your answer. 

9. Women on diets Suppose that 70% of all college 

women have been on a diet within the past 12 

pg 428 

months. A sample survey interviews an SRS of 267 

college women. Find the probability that 75% or 

more of the women in the sample have been on a diet. 

10. Percentage of Harleys Harley-Davidson motorcycles 

make up 14% of all the motorcycles registered 

in the United States. You plan to interview an SRS 

Using technology: Applet/normalcdf(lower:0.75, 

upper:1000, mean:0.70, SD:0.028) 5 0.0371 

p(1 − p) 

10. Mean: m p^ = p = 0.14; SD: s p^ = 

Å n 

0.14(1 − 0.14) 

= = 0.0155 

Å 500 

Shape: Approximately normal because np = 

(500)(0.14) = 70 ≥ 10 and n(1− p) = 500 

(1− 0.14) = 430 ≥ 10 

0.20 − 0.14 

z = = 3.87; 

0.0155 

P( p^ ≥ 0.20) = P( Z ≥ 3.87) ≈ 0 

Using technology: Applet/normalcdf(lower:0.20, 

upper:1000, mean:0.14, SD:0.0155) 5 0.0001 



11/01/17 3:56 PM



of 500 motorcycle owners. Find the probability 

that 20% or more of the motorcycle owners in the 

sample own Harleys. 

11. Success on Kickstarter The fundraising site Kickstarter 

regularly tracks the success rate of projects 

that seek funding on its site. Recently, the percentage 

of projects that were successfully funded was 

38.7%. 13 You select a random sample of 50 Kickstarter 

projects. What is the probability that less 

than 30% of them were successfully funded? 

12. Parlez-vous français? Quebec is the only province in 

Canada where the one official language is French. 

According to a recent census, 79.7% of Quebec 

residents identify French as their mother tongue. 

You select an SRS of 165 Quebec residents. What is 

the probability that less than 80% of them identify 

French as their mother tongue? 


13. Drinking the cereal milk? A USA Today poll asked 

a random sample of 1012 U.S. adults what they do 

with the milk in their cereal bowl after they have 

eaten. Let p^ be the proportion of people in the sample 

who drink the cereal milk. A spokesman for the 

dairy industry claims that 70% of all U.S. adults 

drink the cereal milk. Suppose this claim is true. 


sampling distribution of p^ . Interpret the standard 

deviation. 

(b) Justify that the sampling distribution of p^ is 


(c) Calculate the probability that at most 67% of the 

people in the sample drink the cereal milk. 

14. Who goes to church? A Gallup poll asked a random 

sample of 1785 adults if they attended church during 

the past week. Let p^ be the proportion of people 

in the sample who attended church. A newspaper 

report claims that 40% of all U.S. adults went to 

church last week. Suppose this claim is true. 


sampling distribution of p^ . Interpret the standard 

deviation. 

(b) Justify that the sampling distribution of p^ is 


(c) Calculate the probability that at least 44% of the 

people in the sample attended church. 

15. Who drinks the cereal milk? Refer to Exercise 13. Of 

the poll respondents, 67% said that they drink the 

cereal milk. Based on your answer to part (c), does 

this poll give convincing evidence that less than 70% 

of all U.S. adults drink the cereal milk? Explain. 

16. Do you go to church? Refer to Exercise 14. Of the 

poll respondents, 44% said they attended church 

last week. Based on your answer to part (c), does 

p(1− p) 

11. Mean: m p^ = p = 0.387; SD: s p^ = 

Å n 

0.387(1 − 0.387) 

= = 0.069 

Å 50 


np = (50)(0.387) = 19.35 ≥ 10 and 

n(1 − p) = 50(1 − 0.387) = 30.65 ≥ 10 

0.30 − 0.387 

z = = −1.26; P( p^ < 0.30) 

0.069 

= P(Z < −1.26) = 0.1038 


−1000, upper:0.30, mean:0.387, SD:0.069) 

5 0.1037 

p(1 − p) 

12. Mean: m p^ = p = 0.797; SD: s p^ = 

Å n 

0.797(1 − 0.797) 

= = 0.031 

Å 165 

this poll give convincing evidence that more than 

40% of all U.S. adults attended church last week? 

Explain. 


17. More milk drinkers Refer to Exercise 13. What 

sample size would be required to reduce the standard 

deviation of the sampling distribution to onehalf 

the value you found in part (a)? Justify your 

answer. 

18. Off to college The example on page 428 used the 

sampling distribution of the sample proportion p^ 

to find the probability that the random sample of 

1500 students from a population where p 5 0.35 

will give a p^ between 0.33 and 0.37. You can also 

find this probability using the sampling distribution 

of the sample count X, where X 5 the number 

of students in the sample who attend college within 

50 miles of their home. 

(a) Find the mean and standard deviation of the sampling 

distribution of the sample count X. 

(b) Justify that X has an approximately normal 

distribution. 

(c) Find the values of X that would result in sample 

proportions of p^ 5 0.33 and p^ 5 0.37. 

(d) Calculate the probability that X is between the two 

values from part (c). 


19. Waiting with intent (1.8, 3.9) Do drivers take longer 

to leave their parking spaces when another car 

is waiting? Researchers hung out in a parking lot 

and collected some data. The graphs and numerical 

summaries display information about how long it 

took drivers to exit their spaces. 

Someone waiting? 

Yes 

No 

30 40 50 60 70 80 90 

Time (sec) 

Descriptive Statistics: Time 

* 

* 

Waiting N Mean StDev Min Q 1 Median Q 3 Max 

No 20 44.42 14.10 33.76 35.61 39.56 48.48 84.92 

Yes 20 54.11 14.39 41.61 43.41 47.14 66.44 85.97 


np = (165)(0.797) = 131.5 ≥ 10 and 

n(1 − p) = 165(1 − 0.797) = 33.5 ≥ 10 

0.80 − 0.797 

z = = 0.10; 

0.031 

P(p^ < 0.80) = P(Z < 0.10) = 0.5398 


−1000, upper:0.80, mean:0.797, SD:0.031) 

5 0.5385 

p(1 − p) 

13. (a) m p^ = p = 0.70; s p^ = 

Å n 

0.70(1 − 0.70) 

= = 0.014 

Å 1012 

In SRSs of size n 5 1012, the sample 

proportion of people who drink the cereal 

milk will typically vary by about 0.014 from 


18/08/16 5:02 PM 

(b) Because np = (1012)(0.70) = 708.4 

≥ 10 and n(1 − p) = (1012)(1 − 0.70) 

= 303.6 ≥ 10, the sampling distribution 

of p^ is approximately normal. 

0.67 − 0.70 

(c) z = =−2.14; P( p^ ≤ 0.67) 

0.014 

= P( Z ≤ −2.14) = 0.0162 


(lower:−1000, upper:0.67, mean:0.70, SD: 

0.014) 5 0.0161 

p(1 − p) 

14. (a) m p^ = p = 0.40; s p^ = 

Å n 

0.40(1 − 0.40) 

= 

= 0.012 

Å 1785 

In SRSs of size n 5 1785, the sample proportion 

of people who attended church 

will typically vary by about 0.012 from 


(b) Because np = (1785)(0.40) = 714 $ 

10 and n(1− p) = (1785)(1 − 0.40) = 

1071 ≥ 10, the sampling distribution of 

p^ is approximately normal. 

0.44 − 0.40 

(c) z = = 3.33; P( p^ ≥ 0.44) 

0.012 

= P( Z ≥ 3.33) = 1 − 0.9996 = 0.0004 



SD:0.012) 5 0.0004 

15. Yes; assuming the true proportion 

of people who drink the cereal milk is 

0.70, there is only about a 2% chance 

of getting a sample proportion of 0.67 

or lower purely by chance. Because this 

result is unlikely (less than 5%), we have 

convincing evidence that less than 70% of 

all U.S. adults drink the cereal milk. 

16. Yes; assuming the true proportion 

of people who attended church is 0.40, 

there is about a 0.0004% chance of getting 

a sample proportion of 0.44 or higher 

purely by chance. Because this result is 

unlikely (less than 5%), we have convincing 

evidence that more than 40% of all U.S. 

adults attended church last week. 

17. Because the standard deviation is 

found by dividing by !n, using 4n for the 

sample size halves the standard deviation 

¢"4n = 2"n≤ ; we would have to 

sample 1012(4) = 4048 adults. 

18. (a) m X = np = 1500(0.35) = 525; 

s X = "np(1−p) ="1500(0.35)(1−0.35) 

= 18.47 

(b) Because np = 1500(0.35) = 525 ≥ 10 

and n(1− p) = 1500(1− 0.35)= 975 $ 10, 

the sampling distribution of X is approximately 

normal. 

Answers 18(c,d)–19 are on page 432 

Lesson 6.4 



11/01/17 3:56 PM


X 

18. (c) p^ = 

n , so X = p^ n; (0.33)(1500) = 

495 and (0.37)(1500) = 555 

495 − 525 

(d) z = = −1.62; 

18.47 

555 − 525 

z = = 1.62 

18.47 

P(495 # X # 555)= P(−1.62 ≤ Z # 1.62) 

= 0.9474 − 0.0526 = 0.8948 


495, upper:555, mean:525, SD:18.47) 

5 0.8957. This answer matches the answer 

from the example that uses the sampling 

distribution of p^ , except for rounding. 

19. (a) Shape: Both distributions are 

skewed to the right. Center: Drivers 

generally take longer to leave when 

someone is waiting for the space. Spread: 

There is more variability for the drivers 

with someone waiting. Outliers: There 

were no outliers for those with someone 

waiting, but there were two high outliers 

for those with no one waiting. 

(b) Not necessarily; the researchers 

merely observed what was happening, 

and they did not randomly assign the 

treatments of either having a person 

waiting or not to the drivers of the cars 

leaving the lot. 

20. (a) 

Relative frequency 

1.05 

1 

0.95 

0.9 

0.85 

0.8 

0.75 

0.7 

0.65 

0.6 

0.55 

0.5 

0.45 

0.4 

0.35 

0.3 

0.25 

0.2 

0.15 

0.1 

0.05 

0 

Key 

Other 

Hazel 

Green 

Brown 

Blue 

432 


(a) Write a few sentences comparing these distributions. 

(b) Can we conclude that a waiting car causes drivers to 

leave their spaces more slowly? Why or why not? 

20. Those baby blues (2.1, 4.4) The two-way table 

summarizes information about eye color and gender 

in a random sample of 200 high school students. 

Gender 

Male Female Total 

Blue 21 29 50 

Brown 35 40 75 

Eye color green 14 21 35 

Hazel 12 23 35 

Other 2 3 5 

Total 84 116 200 

(a) Is there an association between eye color and gender 

in this group of students? Support your answer 

with an appropriate graphical summary of the data. 

(b) Select a student at random. Are the events “Student 

is male” and “Student has blue eyes” independent? 


Lesson 6.5 


of a Sample Mean 


d 

d 

Find the mean and standard deviation of the sampling distribution of a sample 

mean x and interpret the standard deviation. 

Use a normal distribution to calculate probabilities involving x when sampling 

from a normal population. 

When sample data are categorical, we often use the count or proportion of successes 

in the sample to make an inference about a population. When sample data are quantitative, 

we often use the sample mean x to estimate the mean m of a population. When 

we select random samples from a population, the value of x will vary from sample 

to sample. To understand how much x varies from m and what values of x are likely 

to happen by chance, we want to understand the sampling distribution of the sample 

mean x. 

Male 

Gender 

Female 

Yes, because the percentages for a 

given eye color are not the same for 

each gender. In other words, knowing 

a person’s gender helps us predict eye 

color. Males are more likely than females 

to have brown eyes, while females are 

more likely than males to have hazel or 

green eyes. 

(b) P(blue eyes | male) 5 21/84 5 0.25; 

P(blue eyes | female) 5 29/116 5 0.25. 

Because the probabilities are equal, 

the events “male” and “blue eyes” are 

independent. Knowing that a student 

is male does not change the probability 

that he has blue eyes. 






d 6.5.1 

d 6.5.2 

BELL RINGER 

Thinking back to the “A penny for your 

thoughts?” activity for samples of 

5 pennies, did every sample produce the 

same sample mean year? What is the 

name for this phenomenon? 

Teaching Tip 

Be picky with students about using the correct 

symbols! The sample mean is denoted x and 

the population mean is denoted m. 

Common Error 

This lesson and the next are about means, not 

proportions. Make sure students don’t use the 

symbols p and p^ when working with means. 


432 



11/01/17 3:56 PM

L E S S O N 6.5 • The Sampling Distribution of a Sample Mean 433 

DEFINITION Sampling distribution of the sample mean x 

The sampling distribution of the sample mean x describes the distribution of values taken 

by the sample mean x in all possible samples of the same size from the same population. 

When Mr. Ramirez’s class did the Penny for Your Thoughts activity at the beginning 

of the chapter, his students produced the “dotplot” in Figure 6.9 showing the simulated 

sampling distribution of x 5 the sample mean year of pennies in 50 samples of size n 5 5. 

x 

xx 

xxxxxx 

x 

x 

x 

1990 1995 2000 2005 2010 2015 

Sample mean year (n = 5) 

This distribution is slightly skewed to the left, with a mean of about 2002 and a 

standard deviation of about 5 years. By the end of Lesson 6.6, you should be able to 

anticipate the shape, center, and variability of distributions like this one without having 

to do a simulation. 


When we select random samples of size n from a population with mean m and standard 

deviation s, the value of x will vary from sample to sample. As with the sampling 

distribution of p^ , there are formulas that describe the center and variability of the 

sampling distribution of x. 

How to Calculate μ x and σ x 

Suppose that x is the mean of an SRS of size n drawn from a large population with mean m 

and standard deviation s. Then: 

• The mean of the sampling distribution of x is m x = m. 

• The standard deviation of the sampling distribution of x is s x = s "n . 

The behavior of x in repeated samples is much like that of the sample proportion p^ : 

• The sample mean x is an unbiased estimator of the population mean m. This is 

because the mean of the sampling distribution m x is equal to the mean of the 

population m. 

• The standard deviation of the sampling distribution of x describes the typical 

distance between the sample mean x and the population mean m. 

• The distribution of x is less variable for larger samples. This is indicated by the 

!n in the denominator of the standard deviation formula. 

• The formula for the standard deviation of the distribution of x requires that the 

observations be independent. In practice, we are safe assuming independence 

when we are sampling without replacement as long as the sample size is less 

than 10% of the population size. 

These facts about the mean and standard deviation of x are true no matter what shape 

the population distribution has. 


sampling distribution of 

the sample mean year 

x in 50 samples of size 


of pennies. 

Teaching Tip 

In Figure 6.9, ask students what the 

“dot” at 1997 represents. It is the sample 

mean/average year for one sample of 

5 pennies. 

Teaching Tip 

This figure is a good opportunity to refer 

to the dotplots made by your students in 


from Lesson 6.1. Compare the results 

from your class with those from 

Mr. Ramirez’s class. 

FYI 

The formulas for m x and s x are true for 

the sampling distribution of x no matter 

what shape it has. 

Teaching Tip: 

Differentiate 

There are many symbols in this section, 

which may be difficult for some students. 

These students may find it easier to 

read the bullet points by substituting 

the words “sample mean” for x and 

“population mean” for m. 

Teaching Tip 

This is consistent with previous 

definitions of standard deviation as the 

typical distance a value falls from the 

mean of a distribution. 

Lesson 6.5 


18/08/16 5:02 PM 



11/01/17 3:56 PM

434 



Thick hair? 

Mean and standard deviation of the 

sampling distribution of x 

PROBLEM: Suppose that the true mean 

number of hair follicles on a human head 

is 100,000 with a standard deviation of 

40,000 follicles. The mean number of hair 

follicles on the heads of 20 randomly 

selected humans will be computed. 


deviation of the sampling distribution of x. 



SOLUTION: 

(a) m x = 100,000 follicles and 

s x = 40,000 = 8944 follicles 

"20 


mean number of hair follicles will typically 

vary by about 8944 follicles from the 

population mean of 100,000 follicles. 





Teaching Advice: This activity helps 

students understand the shape of the 

sampling distribution of x. Although 

the applet doesn’t have high-resolution 

graphics, it is an excellent visual display 

of key concepts in this lesson. 

If you don’t have enough devices, 

students can work in groups or you 

can demonstrate the applet to the 

entire class. Showing the applet as a 

demonstration also saves time, although 

it doesn’t engage students as much. 

Even if students are doing the activity 

individually, it is helpful to show them 

the layout of the applet and demonstrate 

taking a few samples. Point out that 

sample size is denoted with a capital N, 

instead of the usual lowercase n. 


population distribution (the top/first 

number line), the distribution of one 

sample (the second number line), and 

the sampling distribution (the third and 

fourth number lines). Point out these three 

distributions. Note that this activity doesn’t 

make use of the fourth number line. 

There are two mysterious values 

reported by the applet: skew and kurtosis. 

Neither is important for this course. 

a 

e XAMPLe 

Seen any good movies lately? 

Mean and standard deviation of the sampling distribution of x 

PROBLEM: The number of movies viewed in the last year by students at a large high school has 

a mean of 19.3 movies with a standard deviation of 15.8 movies. Suppose we take an SRS of 100 

students from this school and calculate the mean number of movies viewed by the members of 

the sample. 

(a) Calculate the mean and standard deviation of the sampling distribution of x. 


SOLUTION: 

(a) m x = 19.3 movies and s x = 15.8 = 1.58 movies 

"100 

(b) In SRSs of size n 5 100, the sample mean number of 

movies will typically vary by about 1.58 movies from 

the population mean of 19.3 movies. 

AcT iviT y 

Shape 

Sampling from a normal population 

Professor David Lane of Rice University has 

developed a wonderful applet for investigating 

the sampling distribution of x. In this activity, you’ll 

use Professor Lane’s applet to explore the shape of 

the sampling distribution when the population is 

normally distributed. 

1. go to http://onlinestatbook.com/stat_sim/ 

sampling_dist/ or search for “online statbook 

sampling distributions applet” and go to the website. 

When the BEgIN button appears on the left 

side of the screen, click on it. You will then see a 

yellow page entitled “Sampling Distributions” like 

the one in the screen shot. 

2. There are choices for the population distribution: 

normal, uniform, skewed, and custom. The 


Skewness is a measure of the skewness of the 

distribution; kurtosis measures how light or 

heavy the tails of the distribution are relative to a 

normal distribution. 

Answers: 

1. Students should launch the applet. 

2. The black boxes represent individuals 

being randomly selected from the 

population. The blue square represents the 

sample mean of the sample on the second 

number line. 

3. 

• The simulated sampling distribution has 

an approximately normal shape. 

Recall that m x = m and s x = s "n . 


The shape of the sampling distribution of the sample mean x depends on the shape of 

the population distribution. In the following activity, you will explore what happens 

when sampling from a normal population. 

default is normal. Click the “Animated” button. 

What happens? Click the button several more 

times. What do the black boxes represent? What 

is the blue square that drops down onto the 

plot below? 

• The mean and median of the sampling 

distribution are 16, just like the population. 

(It is possible that some students will get 

values slightly different from 16.) 

• The standard deviation of the sampling 

distribution is smaller than the standard 

deviation of the population. 

4. The sampling distribution of x for n 5 20 

has the same shape and center, but the 

variability is even less than the sampling 

distribution for n 5 5. 

5. The shape of the sampling distribution is 

normal when the population distribution 

has a normal shape. 


434 



11/01/17 3:57 PM


3. Click on “Clear lower 3” to start clean. Then click 

on the “100,000” button under “Sample:” to simulate 

taking 100,000 SRSs of size n 5 5 from the 

population. Answer these questions: 

• Does the simulated sampling distribution 

(blue bars) have a recognizable shape? Click 

the box next to “Fit normal.” 

• To the left of each distribution is a set of 

summary statistics. Compare the mean of 

the simulated sampling distribution with the 

mean of the population. 

Describing a Sampling Distribution of a Sample Mean 

When Sampling from a Normal Population 

• How is the standard deviation of the simulated 

sampling distribution related to the 

standard deviation of the population? 

4. Click “Clear lower 3.” Use the drop-down menus 

to set up the bottom graph to display the mean 

for samples of size n 5 20. Then sample 100,000 

times. How do the two distributions of x compare: 

shape, center, and variability? 

5. What have you learned about the shape of the 

sampling distribution of x when the population 

has a normal shape? 

As the activity demonstrates, if the population distribution is normal, so is the 

sampling distribution of x. This is true no matter what the sample size is. 

Suppose that a population is normally distributed with mean m and standard deviation s. Then 

the sampling distribution of x for SRSs of size n has a normal distribution with mean m and 

standard deviation s "n. 

FYI 

The exact formula for the standard 

deviation of x when sampling without 

replacement is s x = s # N − n 

"n Å N − 1 . 

When n is small relative to N (less than 

N − n 

10%), the value of Å N − 1 is 

approximately 1 and therefore doesn’t 

s 

change the value of much. The 

"n 

N − n 

factor is sometimes called the 

Å N − 1 

finite population correction factor. 

Lesson 6.5 


In the next lesson, you will learn what happens when sampling from a non-normal 

population. 

Finding Probabilities Involving x 

Now we have enough information to calculate probabilities involving x when the 

population distribution is normal. 

Are those peanuts underweight? 

Probabilities involving x 

PROBLEM: At the P. Nutty Peanut Company, dry-roasted, shelled peanuts 

are placed in jars labeled “16 ounces” by a machine. The distribution of 

weights in the jars is approximately normal with a mean of 16.1 ounces and 

a standard deviation of 0.15 ounces. Find the probability that the mean 

weight of 10 randomly selected jars is less than the advertised weight of 

16 ounces. 

e XAMPLe 

SOLUTION: 

Let x 5 the sample mean weight of 10 randomly 

• Mean: m x - 5 16.1 ounces 

selected jars. To find P(x ≤ 16), we have to know the 

• SD: s x - = 0.15 

mean, standard deviation, and shape of the sampling 

"10 = 0.047 ounces distribution of x. Recall that m x = m and s x = s "n . 

Charles Nesbit/Getty 

Images 

18/08/16 5:03 PM 


How hungry are the hounds? 


PROBLEM: The local SPCA (Society for the 

Prevention of Cruelty to Animals) feeds the 

animals it shelters, but the amount of food 

needed per day varies. The distribution of 

the weight of dry dog food used per day 

is approximately normal, with a mean of 

30 pounds and standard deviation of 5.1 

pounds. Find the probability that the mean 

weight of dry dog food for 6 randomly 

selected days is greater than 33 pounds. 

SOLUTION: 

Let x 5 the sample mean weight of dog 

food on 6 randomly selected days. 

• Mean: m x = 30 pounds 

• SD: s x = 5.1 = 2.08 pounds 

!6 

• Shape: Approximately normal 

because the population distribution is 

approximately normal 

33 

23.76 25.84 27.92 30.00 32.08 34.16 36.24 

Sample mean weight of dry dog food 

33 − 30 

Using Table A: z = = 1.44 and 

2.08 

P(Z ≥ 1.44) 5 1 2 0.9251 5 0.0749 


33, upper: 10000, mean: 30, SD: 2.08) 5 

0.0746 



11/01/17 3:57 PM

436 


• Shape: Approximately normal because the population 

distribution is approximately normal 

1. Draw a normal curve. 

Teaching Tip 

After the “Are those peanuts 

underweight?” example, you can have 

students calculate the probability that 

the weight of a single peanut X is less 

than 16 grams. Using technology, 

P(X < 16) 5 0.2525. This should make 

intuitive sense to students because the 

weight of a single peanut is more likely 

to be far from the true mean than the 

average weight of 10 peanuts (which 

is likely to include a mix of light and 

heavy peanuts). These probabilities are 

illustrated in Figure 6.10. 

Lesson App 

Answers 

1. m x = m = 64.5 inches; 

s x = s !n = 2.5 = 0.645 inches 

!15 

2. In SRSs of size n 5 15, the sample 

mean heights of young women will 

typically vary by about 0.645 inch from 

the true mean of 64.5 inches. 

66.5 − 64.5 

3. z = = 3.10; 

0.645 

P( x > 66.5) = P(Z > 3.10) = 1 − 0.9990 

= 0.0010 



SD:0.645) 5 0.001 

4. Assuming the true mean height of 

women at this college is 64.5 inches, 

there is about a 0.1% chance of selecting 

an SRS of 15 women and getting a 

sample mean of 66.5 or higher purely 

by chance. Because this result is unlikely 

(less than 5%), we have convincing 

evidence that the mean height for all 

young women at this college is greater 

than 64.5 inches. 

TRM sAmpling Distribution Summary 

Chart 

15.959 16.006 16.053 16.1 16.147 16.194 16.241 

16 

Sample mean weight x 

– 

(oz) 


16 − 16.1 

z = =−2.13 

0.047 

P (Z ≤ 22.13) 5 0.0166 

Using technology: Applet/normalcdf(lower: 21000, upper: 16, 

mean: 16.1, SD: 0.047) 5 0.0167 

FigUre 6.10 Normal 

curves showing the 

distribution of weights 

for individual jars of peanuts 

(purple curve) and 

distribution of sample 

mean weights for SRSs of 

10 jars (blue curve). 


Are college women taller? 

A helpful summary chart is available in the 

Teacher’s Resource Materials. This chart helps 

students organize the sampling distributions 

for sample counts (Lesson 6.3), sample 

proportions (Lesson 6.4), and sample means 

(Lesson 6.5). Access this resource by clicking 

on the link in the TE-book, logging into the 

Teacher’s Resource site, or accessing it on the 

TRFD. 

Individual values vary more than averages, so randomly selecting a single jar that 

is under the advertised weight is more likely than getting a sample mean for 10 jars 

that is less than the advertised weight. This is illustrated in Figure 6.10. 

Population distribution 






Sampling distribution of x – 

16 16.1 

The fact that averages of several observations are less variable than individual observations 

is important in many settings. For example, it is common practice in science and 

medicine to repeat a measurement several times and report the average of the results. 

The heights of young women follow a normal distribution with mean m 5 64.5 

inches and standard deviation s 5 2.5 inches. 

1. Calculate the mean and standard deviation of the sampling distribution of x 

for SRSs of size 15. 

2. Interpret the standard deviation from Question 1. 

3. Find the probability that the mean height of an SRS of 15 young women 

exceeds 66.5 inches. 

4. Suppose that the mean height in a sample of n 5 15 young women from a local college is x 5 66.5. Based on your 

answer to Question 3, what would you conclude about the mean height for all young women at this college? 


© Dave & Les Jacobs/Blend Images/Corbis 


436 



11/01/17 3:57 PM


Lesson 6.5 




Find the mean and standard deviation of the sampling distribution of 

a sample mean x and interpret the standard deviation. 

Use a normal distribution to calculate probabilities involving x when 

sampling from a normal population. 

Exercises 


1. Short songs? David’s iPod has about 10,000 songs. 

The distribution of the play times for these songs is 

heavily skewed to the right with a mean of 225 seconds 

and a standard deviation of 60 seconds. Suppose 

we choose an SRS of 10 songs from this population 

and calculate the mean play time x of these songs. 




2. Grinding auto parts A grinding machine in an autoparts 

factory prepares axles with a target diameter 

of m 5 40.125 millimeters (mm). The machine has 

some variability, so the standard deviation of the 

diameters is s 5 0.002 millimeter. The machine 

operator inspects a random sample of 4 axles each 

hour for quality-control purposes and records the 

sample mean diameter x. 




pg 434 

Lesson 6.5 

3. Fresh tomatoes A local garden center says that a 

certain variety of tomato plant produces tomatoes 

with a mean weight of 250 grams and a standard 

deviation of 42 grams. You take a random sample 

of 20 tomatoes produced by these plants and calculate 

their mean weight x. 




4. Fuel efficiency Driving styles differ, so there is variability 

in the fuel efficiency of the same model automobile 

driven by different people. For a certain car 

model, the mean fuel efficiency is 23.6 miles per 

gallon with a standard deviation of 2.5 miles per 

gallon. 14 Take a simple random sample of 25 owners 

of this model and calculate the sample mean 

fuel efficiency x. 


Exercises 

You can find the full solutions for this lesson 

by clicking on the link in the TE-book, logging 

into the Teacher’s Resource site, or accessing 

this resource on the TRFD. 


1. (a) m x = m = 225 seconds; 

s x = s !n = 60 = 18.97 seconds 

!10 

(b) In SRSs of size n 5 10, the sample mean play 

time of songs will typically vary by about 18.97 

seconds from the true mean of 225 seconds. 

p. 434 1–4 

p. 435 5–8 




5. How much cereal? A company’s cereal boxes advertise 

pg 435 that 9.65 ounces of cereal are contained in each box. 

In fact, the amount of cereal in a randomly selected 

box follows a normal distribution with mean m 5 9.70 

ounces and standard deviation s 5 0.03 ounce. What 

is the probability that the mean amount of cereal x in 

5 randomly selected boxes is at most 9.65? 

6. Finch beaks One dimension of bird beaks is 

“depth”—the height of the beak where it arises 

from the bird’s head. During a research study on 

one island in the Galapagos archipelago, the beak 

depth of all Medium Ground Finches on the island 

was found to be normally distributed with mean 

m 5 9.5 millimeters and standard deviation s 5 1.0 

millimeter. 15 What is the probability that the mean 

depth x in 10 randomly selected Medium Ground 

Finches is at least 10 millimeters? 

7. Estimating cholesterol Suppose that the blood cholesterol 

level of all men aged 20 to 34 follows a 

normal distribution with mean m 5 188 milligrams 

per deciliter (mg/dl) and standard deviation s 5 41 

mg/dl. In an SRS of size 100, find the probability 

that x estimates m to within ±3 mg/dl. 

8. Bottlers at work A company uses a machine to fill 

plastic bottles with cola. The contents of the bottles 

vary according to a normal distribution with mean 

m 5 298 milliliters and standard deviation s 5 3 

milliliters. In an SRS of size 16, find the probability 

that x estimates m to within ±1 milliliter. 


9. Why won’t the car start? An automaker has found 

that the lifetime of its batteries varies from car to 

car according to a normal distribution with mean 

m 5 48 months and standard deviation s 5 8.2 

months. The company installs a new battery on an 

SRS of 8 cars. 

2. (a) m x = m = 40.125 mm; 

s x = s !n = 0.002 = 0.001 mm 

!4 

18/08/16 5:03 PM 

(b) In SRSs of size n 5 4, the sample mean 

axle diameter will typically vary by about 

0.001 mm from the true mean of 40.125 mm. 

3. (a) m x = m = 250 grams; 

s x = s !n = 42 = 9.39 grams 

!20 


weight of tomatoes will typically vary by 

about 9.39 grams from the true mean of 

250 grams. 

4. (a) m x = m = 23.6 miles per gallon; 

s x = s !n = 2.5 = 0.5 mile per gallon 

!25 


mean fuel efficiency will typically vary by 

about 0.5 mile per gallon from the true 

mean of 23.6 miles per gallon. 

5. Mean: m x = m = 9.70; 

SD: s x = s !n = 0.03 

!5 = 0.013 

Shape: Normal because the population 

distribution is normal 

9.65 − 9.70 

z = = −3.85; 

0.013 

P( x ≤ 9.65) = P(Z ≤ −3.85) ≈ 0 


(lower:−1000, upper:9.65, mean:9.70, 

SD:0.013) 5 0.0001 

6. Mean: m x = m = 9.5; 

SD: s x = s !n = 1.0 

!10 = 0.316 



10 − 9.5 

z = = 1.58; P(x ≥ 10) = 

0.316 

P(Z ≥ 1.58) = 1−0.9429 = 0.0571 


(lower:10, upper:1000, mean:9.5, 

SD:0.316) 5 0.0568 

7. Mean: m x = m = 188; 

SD: s x = s !n = 41 

!100 = 4.1 



185 − 188 

z = = −0.73; 

4.1 

191 − 188 

z = = 0.73 

4.1 

P(185 ≤ x ≤ 191) = P(−0.73 ≤ 

Z ≤ 0.73) = 0.7673 − 0.2327 = 0.5346 


(lower:185, upper:191, mean:188, SD:4.1) 

5 0.5357 

8. Mean: m x = m = 298; 

SD: s x = s !n = 3 

!16 = 0.75 



297 − 298 

z = = −1.33; 

0.75 

299 − 298 

z = = 1.33 

0.75 

P(297 ≤ x ≤ 299)= P(−1.33 ≤ Z ≤ 

1.33) = 0.9082 − 0.0918 = 0.8164 



SD:0.75) 5 0.8176 

Lesson 6.5 



11/01/17 3:57 PM

438 


9. (a) m x = m = 48 months; 

s x = s !n = 8.2 = 2.90 months 

!8 


battery life will typically vary by about 2.90 

months from the true mean of 48 months. 

(c) The sampling distribution is normal 


normal. 

42.2 − 48 

z = = −2.00; P( x < 42.2) = 

2.90 

P( Z < −2.00) = 0.0228 


(lower:−1000, upper:42.2, mean:48, 

SD:2.90) 5 0.0228 

10. (a) m x = m = 3.4 kg; 

s x = s !n = 0.5 = 0.129 kg 

!15 


birth weight will typically vary by about 

0.129 kg from the true mean of 3.4 kg. 

(c) The sampling distribution is normal 


normal. 

3.55 − 3.4 

z = = 1.16; P( x > 3.55) = 

0.129 

P( Z > 1.16) = 1− 0.8770 = 0.1230 



SD:0.129) 5 0.1225 

11. Yes; assuming the true mean battery 

life is 48 months, there is only about a 2% 

chance of getting a sample mean of 42.2 

or lower purely by chance. Because this 

result is unlikely (less than 5%), we have 

convincing evidence that the population 

mean battery life is less than 48 months. 

12. No; assuming the true mean birth 

weight is 3.4 kg, there is about a 12% 

chance of getting a sample mean of 3.55 

or higher purely by chance. Because this 

result is plausible (greater than 5%), we 

do not have convincing evidence that 

the population mean is more than 3.4 kg. 

13. (a) Less likely; individual values 

vary more than averages, so getting an 

individual value that is close to the true 

mean is less likely. 

185 − 188 

(b) z = = −0.07; 

41 

z = 191−188 = 0.07 

41 

P(185 ≤ X ≤ 191)= P(−0.07 ≤ Z 

≤ 0.07) = 0.5279 − 0.4721 = 0.0558 


(lower:185, upper:191, mean:188, SD:41) 

5 0.0583. This probability of 0.0583 is 

much less than probability calculated in 

Exercise 7 (0.5357). 


sampling distribution of x for SRSs of size 8. 


(c) Find the probability that the sample mean life is 

less than 42.2 months. 

10. Birth weights The birth weights of males born full term 

are normally distributed with mean m 5 3.4 kilograms 

and standard deviation s 5 0.5 kilogram. 16 A large 

city hospital selects a random sample of 15 full-term 

males born in the last six months. 

(a) Calculate the mean and standard deviation of 

the sampling distribution of x for SRSs of size 15. 


(c) Find the probability that the sample mean weight is 

greater than 3.55 kilograms. 

11. Could it be the battery? Refer to Exercise 9. Suppose 

that the average life of the batteries on these 

8 cars turns out to be x 5 42.2 months. Based on 

your answer to Exercise 9, is there convincing evidence 

that the population mean m is really less than 

48 months? Explain. 

12. More birth weights Refer to Exercise 10. Suppose 

that the average birth weight of the 15 babies turns 

out to be 3.55 kilograms. Based on your answer to 

Exercise 10, is there convincing evidence that the 

population mean m is really more than 3.4 kilograms? 

Explain. 

13. One man’s cholesterol In Exercise 7, you calculated 

the probability that x would estimate the 

true mean cholesterol level within 63 mg/dl of m in 

samples of size 100. 

(a) If you randomly selected one 20- to 34-year-old 

male instead of 100, would he be more likely, less 

likely, or equally likely to have a cholesterol level 

within 63 mg/dl of m? Explain this without doing 

any calculations. 

(b) Calculate the probability of the event described in 

part (a) to confirm your answer. 

14. One bottle of cola In Exercise 8, you calculated the 

probability that x would estimate the true mean 

amount of cola within 61 milliliter of m in samples 

of size 16. 

(a) If you randomly selected one bottle instead of 16, 

would it be more likely, less likely, or equally likely 

to contain an amount of cola within 61 milliliter of 

m? Explain this without doing any calculations. 

(b) Calculate the probability of the event described in 

part (a) to confirm your answer. 

15. Sampling music Refer to Exercise 1. How many 

songs would you have to sample if you wanted the 

standard deviation of the sampling distribution of 

x to be 30 seconds? 


14. (a) Less likely; individual values vary more 

than averages, so getting an individual value 

that is close to the true mean is less likely. 

297 − 298 

(b) z = = −0.33; 

3 

299 − 298 

z = = 0.33 

3 

P(297 ≤ X ≤ 299)= P(−0.33 ≤ Z ≤ 0.33) 

= 0.6293 − 0.3707= 0.2586 


upper:299, mean:298, SD:3) 5 0.2611. This 

probability of 0.2611 is much less than the 

probability calculated in Exercise 8 (0.8176). 

15. s x = s 60 

→ 30 = 

!n "n → n 5 4; 

we would have to sample 4 songs. 

16. Sampling auto parts Refer to Exercise 2. How many 

axles would you have to sample if you wanted the 

standard deviation of the sampling distribution of 

x to be 0.0005 millimeter? 


17. Orange overage Mandarin oranges from a certain 

grove have weights that follow a normal distribution 

with mean m 5 3 ounces and standard deviation 

s 5 0.5 ounce. Bags are filled with an SRS of 

20 mandarin oranges. What is the probability that 

the total weight of oranges in a bag is greater than 

65 ounces? Hint: Re-express the total weight of 20 

oranges in terms of the average weight x. 


18. Let’s text (1.2) We used Census at School’s “Random 

Data Selector” to choose a sample of 50 Canadian 

students who completed the survey in a recent year. 

The bar graph displays data on students’ responses 

to the question “Which of these methods do you 

most often use to communicate with your friends?” 

25 

Frequency 

20 

15 

10 

5 

0 

Text 

In 

person 

Social 

media 

Phone 

Other 

Method of communication 

(a) Would it be appropriate to make a pie chart for 

these data? Why or why not? 

(b) Jerry says that he would describe this bar graph as 

skewed to the right. Explain why Jerry is wrong. 

19. Shut it down and go to sleep! (5.1, 5.2) A National 

Sleep Foundation survey of 1103 parents asked, 

among other questions, how many electronic 

devices (TVs, video games, smartphones, computers, 

MP3 players, and so on) children had in their 

bedrooms. 17 Let X 5 the number of devices in a 

randomly chosen child’s bedroom. Here is the 

probability distribution of X. 

number of 

0 1 2 3 4 5 

devices 

Probability 0.28 0.27 0.18 0.16 0.07 0.04 

(a) Show that this is a legitimate probability distribution. 

(b) What is the probability that a randomly chosen child 

has at least 1 electronic device in her bedroom? 

(c) Calculate the expected value and standard deviation 

of X. 

16. s x = s 0.002 

→ 0.0005 = 

!n !n → n 5 16; 

we would have to sample 16 axles. 

17. A bag of 20 oranges that weighs 65 

ounces would give an average orange weight 

of x = 3.25 ounces. So we want to find 

P( x > 3.25). 

Mean: m x = m = 3; SD: s x = s !n = 0.5 

!20 = 0.11 


distribution is normal z = 3.25 − 3 = 2.27; 

0.11 

P(total weight > 65) = P( x > 3.25) = 

P( Z > 2.27) = 1− 0.9884 = 0.0116 


3.25, upper:1000, mean:3, SD:0.11) 5 0.0115 



438 



11/01/17 3:57 PM


Lesson 6.6 

The central Limit Theorem 


d Determine if the sampling distribution of x is approximately normal when 

sampling from a non-normal population. 

d If appropriate, use a normal distribution to calculate probabilities involving x. 

In Lesson 6.5, you learned about the sampling distribution of the sample mean x 

when sampling from a normally distributed population. The following activity will 

help you explore what happens when you sample from non-normal populations. 

AcT iviT y 

Sampling from a non-normal population 

In this activity, we will use an applet to investigate the 

sampling distribution of the sample mean x when 

sampling from a non-normal population. 

1. go to http://onlinestatbook.com/stat_sim/sampling_dist/ 

or search for “online statbook sampling 

distributions applet” and go to the website. Launch 

the applet and select the “Skewed” population. Set 

the bottom two graphs to display the mean—one 


18. (a) Yes, a pie chart is appropriate 

here because the categories (method of 

communication) form parts of a whole. 

(b) The graph should not be described as 

skewed to the right because this is a distribution 

of categorical data not quantitative data. 

The categories could be graphed in any order. 

19. (a) The probabilities are all between 0 

and 1. Also, the sum of the probabilities is 

0.28 1 0.27 1 0.18 1 0.16 1 0.07 1 0.04 5 1 

(b) Using the complement rule, 

P(X ≥ 1) = 1− P(X = 0) = 1− 0.28 = 0.72 

(c) E(X) = m X = 0(0.28) + 1(0.27) + 2(0.18) + 

+ 3(0.16) + 4(0.07) + 5(0.04) = 1.59 devices 

s X = 1.422 devices 

for samples of size 2 and the other for samples of 

size 5. Click the “Animated” button a few times to be 

sure you see what’s happening. Then “Clear lower 3” 

and take 100,000 SRSs. Describe what you see. 

2. Change the sample sizes to n 5 10 and n 516 and 

repeat Step 1. What do you notice? 

3. Now change the sample sizes to n 5 20 and 

n 5 25 and take 100,000 samples. Did this confirm 

what you saw in Step 2? 

4. Clear the page, and select “Custom” distribution 

from the drop-down menu at the top of the page. 

Click on a point on the horizontal axis, and drag 

up to create a bar. Make a distribution that looks 

as strange as you can. (Note: You can shorten a bar 

or get rid of it completely by clicking on the top 

of the bar and dragging down to the axis.) Then 

repeat Steps 1 to 3 for your custom distribution. 

Cool, huh? 

5. Summarize what you learned about the shape of 

the sampling distribution of x. 

439 





d 6.6.1 

d 6.6.2 

18/08/16 5:03 PM 

BELL RINGER 

Thinking back to the “A penny for your 

thoughts?” activity, does the sampling 

distribution of the sample mean x always 

have the same shape? Does it have the 

same shape as the population distribution? 





Teaching Advice: This activity helps 

students understand the shape of the 

sampling distribution of x from nonnormal 

populations. Contrast this with 

the activity from the previous lesson 

where the population was normal. As 

noted in previous activities, it is best to 

have students work individually or in 

pairs, but the applet work can be done 

in larger groups or as an entire class. If 

your class didn’t do the activity in Lesson 

6.5, show the layout of the applet and 

demonstrate taking a few samples. In 

particular, show the animations for the 

second and third number line. 


population distribution (the top/first 

number line), the distribution of one 

sample (the second number line), and 

the sampling distribution (the third and 

fourth number lines). Point out these 

three distributions to your students. 

Make sure students click “Animated” 

in Step 1! Don’t let them miss the 

visual reminder of the process of 

random sampling. Also in Step 1, make 

sure students select “Mean” from the 

dropdown box next to the fourth 

number line in the applet. 

There are two mysterious statistics 

reported by the applet: skew and kurtosis. 

Neither is important for this course. The 

skewness statistic is a measure of the 

skewness of the distribution; the kurtosis 

statistic measures how light or heavy the 

tails of the distribution are relative to a 

normal distribution. 

Answers: 

1. The sampling distribution of the sample 

mean x for n 5 2 and n 5 5 have a 

mean near 8, which is the mean of the 

population. The standard deviation of 

the sampling distribution for n 5 5 is 

less than the variability of the sampling 

distribution for n 5 2, which is less than 

the variability of the population. The 

shape of both sampling distributions 

is skewed right, but the sampling 

distribution for n 5 5 seems to be a 

little less skewed. 

2. The sampling distributions have the 

same mean as the population. The 

variability in the sampling distribution 

decreases as n increases. The shape of 

the sampling distribution is less skewed 

as n increases. 

Activity answers continue on page 440 

Lesson 6.6 

L E S S O N 6.6 • The Central Limit Theorem 439 


11/01/17 3:57 PM

Activity answers continued 

3. Yes, this confirms our observations in 

Step 2. 

4. Students should paint their own 

population distribution. Yes, this is 

cool. The results from Steps 1 to 3 

hold true for this new population (no 

matter what the population looks like)! 

5. The sampling distribution of x has 

the same mean as the population 

distribution. As n increases, the 

variability in the sampling distribution 

decreases. The shape of the 

sampling distribution becomes more 

approximately normal as n increases. 

440 


The Central Limit Theorem 

It is a remarkable fact that as the sample size increases, the sampling distribution 

of x changes shape: It looks less like that of the population and more like a normal 

distribution. This is true no matter what shape the population distribution has. This 

famous fact of probability theory is called the central limit theorem (sometimes abbreviated 

as CLT). 

DEFINITION central limit theorem (cLt) 

Draw an SRS of size n from any population with mean m and finite standard deviation s. 

The central limit theorem (CLT) says that when n is large, the sampling distribution of 

the sample mean x is approximately normal. 

How large a sample size n is needed for the sampling distribution of x to be close 

to normal depends on the population distribution. A larger sample size is required 

if the shape of the population distribution is far from normal. In that case, the sampling 

distribution of x will also be far from normal if the sample size is small. To use 

a normal distribution to calculate probabilities involving x, check the Normal/Large 

Sample condition. 

Teaching Tip 

It is hard to overstate the importance 

of the central limit theorem. Students 

should know this theorem by name. 

Make sure students understand that the 

CLT applies to sample means and refers 

to the shape (and only the shape) of the 


Teaching Tip 

There is nothing magical about 30 as 

the boundary value for a “large” sample 

size. Some statisticians and authors 

recommend n 5 40 as the boundary. 

In truth, some populations need 

sample sizes much larger than 40 to 

have sampling distributions that are 

approximately normal. For our purposes, 

30 is a reasonable value that works 

relatively well for most populations. 

Make sure that students understand that 

n ≥ 30 is just a guideline (like the Large 

Counts condition for proportions). 

a 

e XAMPLe 

A few more pennies for your thoughts? 


DEFINITION normal/Large Sample condition 

The Normal/Large Sample condition says that the distribution of x will be approximately 

normal when either of the following is true: 

• The population distribution is approximately normal. This is true no matter what the 

sample size n is. 

• The sample size is large. If the population distribution is not normal, the sampling 

distribution of x will be approximately normal in most cases if n ≥ 30. 

PROBLEM: Mr. Ramirez’s class did the Penny for Your 

Thoughts Activity from the beginning of this chapter. 

The histogram shows the distribution of ages for the 

2341 pennies in their collection. 

(a) Describe the shape of the sampling distribution 

of x for SRSs of size n 5 2 from the population of 

pennies. Justify your answer. 

(b) Describe the shape of the sampling distribution 

of x for SRSs of size n 5 50 from the population of 

pennies. Justify your answer. 

Frequency 

140 

120 

100 

80 

60 

40 

20 

0 

1950 1960 1970 1980 1990 2000 2010 2020 

Year 


Will you show me the money? 


PROBLEM: Early in 2016, the 

opening weekend earnings from 

the top-earning movies of all time 

were reported by the website Box 

Office Mojo. The histogram shows the 

distribution of earnings, rounded to the 

nearest million dollars, for the top 1799 

movies based on opening weekend 

gross income. 


Frequency 

900 

800 

700 

600 

500 

400 

300 

200 

100 

0 

50 100 150 200 250 300 

Opening weekend gross 

earnings ($ millions) 

(a) Describe the shape of the sampling 

distribution of x for SRSs of size n 5 60 from 

this population of movies. Justify your answer. 

(b) Describe the shape of the sampling 

distribution of x for SRSs of size n 5 10 from 

this population of movies. Justify your answer. 

SOLUTION: 

(a) Because n 5 60 ≥ 30, the sampling 

distribution of x will be approximately normal 

by the central limit theorem. 

(b) Because n 5 10 < 30, the sampling 

distribution of x will also be skewed to the right, 

but not quite as strongly as the population. 


440 



11/01/17 3:58 PM


SOLUTION: 

(a) Because n 5 2 < 30, the sampling distribution of x will be skewed to the left, but not quite as strongly as 

the population. 

(b) Because n 5 50 ≥ 30, the sampling distribution of x will be approximately normal by the central limit 

theorem. 


Lesson 6.6 

The dotplots in Figure 6.11 show the simulated sampling distributions of the sample 

mean for (a) 500 SRSs of size n 5 2 and (b) 500 SRSs of size n 5 50. 

1970 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

1980 1990 2000 2010 

Sample mean (n = 2) 

2020 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d 

d d 

d ddddddddddddddddddddddddddddddddddddddddddddd d ddddddddddddddddddddddddddddddddddddddddddddddddd d ddddddddddddddddddddddddddddddddddddddddddddddddddddddd d 

1998 2000 2002 2004 2006 

Sample mean (n = 50) 

As expected, the simulated sampling distribution of x for SRSs of size n 5 2 is 

skewed to the left and the simulated sampling distribution of x for SRSs of size n 5 50 

is approximately normal—thanks to the central limit theorem. 

Probabilities Involving x 

Using the central limit theorem, we can do probability calculations involving x even 

when the population is non-normal. 

Mean texts? 


PROBLEM: Suppose that the number of texts sent during a typical day by the population of students 

at a large high school follows a right-skewed distribution with a mean of 45 and a standard 

deviation of 35. How likely is it that a random sample of 100 students will average at least 50 texts 

per day? 

SOLUTION: 

• Mean: m- x 5 45 texts 

• SD: s x - = 35 5 3.5 texts 

"100 

• Shape: Approximately normal by the CLT because n 5 100 ≥ 30 

d 

d 

d 

d 

d 

d 

d 



the sample mean age for 

(a) 500 SRSs of size n 5 2 

and (b) 500 SRSs of size 


of pennies. 

e XAMPLe 

Let x = the sample mean number of texts. To find 

P(x ≥ 50), we have to know the mean, standard 

deviation, and shape of the sampling distribution of x. 

Recall that m x -= m and s– x = s "n . 

Common Error 

Remind your students that the CLT only 

addresses the shape of the sampling 

distribution of x. It doesn’t tell us about 

the center or variability of the sampling 

distribution. 


Will you show me more money? 


PROBLEM: The opening weekend 

earnings for the top 1799 movies of all time 

have a distribution that is strongly rightskewed 

with a mean of $26.2 million and 

standard deviation of $22.9 million. What is 

the probability that a random sample of 50 

of these movies will have average opening 

weekend earnings of less than $19 million? 

SOLUTION: 

• Mean: m x = $26.2 million 

• SD: s x = $22.9 = $3.24 million 

"50 

• Shape: Approximately normal by the 

CLT because n 5 50 ≥ 30 


18/08/16 5:04 PM 

19 

16.48 19.72 

22.96 26.20 

29.44 32.68 35.92 

Sample mean opening weekend gross 

earnings ($ millions) 

19 − 26.2 


3.24 

and P(Z < 22.22) 5 0.0132 

Using technology: Applet/normalcdf (lower: 

2100000, upper: 19, mean: 26.2, SD: 3.24) 

5 0.0131 



11/01/17 3:58 PM

442 


Teaching Tip: 

Differentiate 

By this point in the chapter, advanced 

math students may notice similarities 

between the sampling distribution of p^ 

and the sampling distribution of x. This 

is not a coincidence. Thinking back to 


in Lesson 6.1, we can let the random 

variable X be 0 if the penny is not from 

the 2000s, and let it be 1 if it is from the 

2000s. That is, X is 0 for a “failure” and 1 

for a “success.” Then in a sample of n 5 5 

pennies, the sample proportion is just 

the sum of the values of X, divided by 5. 

In other words, the sample proportion 

can be thought of as a special case of the 

sample mean. 

34.5 38 41.5 45 48.5 52 55.5 

50 

Sample mean number of texts 


50 − 45 

z = = 1.43 

3.5 

P (Z ≥ 1.43) 5 1 2 0.9236 5 0.0764 

Using technology: 

Applet/normalcdf (lower: 50, upper: 100000, mean: 45, 

SD: 3.5) 5 0.0766 


Keeping things cool with statistics? 

1. Draw a normal curve. 






Lesson App 

Answers 

1. Because n = 70 ≥ 30, the sampling 

distribution of x is approximately normal 


2. Mean: m x = m = 1 hour; 

SD: s x = s !n = 1.5 = 0.179 hour 

!70 

1.1 − 1.0 

z = = 0.56; P( x > 1.1) 

0.179 

= P(Z > 0.56) = 1 − 0.7123 = 0.2877 


(lower:1.1, upper:1000, mean:1, SD:0.179) 

5 0.2882 

3. No; there is a 29% chance that the 

time allotted will not be enough. 

Your company has a contract to perform preventive maintenance 

on thousands of air-conditioning units in a large city. Based on 

service records from the past year, the time (in hours) that a technician 

requires to complete the work follows a strongly right-skewed 

distribution with m 5 1 hour and s 5 1.5 hours. As a promotion, 

your company will provide service to a random sample of 70 airconditioning 

units free of charge. You plan to budget an average of 

1.1 hours per unit for a technician to complete the work. Will this be 

enough time? 

1. What is the shape of the sampling distribution of x for samples of size 

n 5 70 from this population? Justify. 

2. Calculate the probability that the average maintenance time x for 70 

units exceeds 1.1 hours. 

3. Based on your answer to the previous problem, did the company 

budget enough time? Explain. 

simazoran/iStock/Getty Images 


TRM Quiz 6B: Lessons 6.4–6.6 

You can find a prepared quiz for Lessons 

6.4–6.6 by clicking on the link in the TE-book, 

logging into the Teacher’s Resource site, or 

accessing this resource on the TRFD. 

TRM chapter 6 Activity: Sampling 

Movies (The Sequel) 

This activity reviews the sampling distribution 

of x by sampling from a population of movies. 

Access this resource by clicking on the link 

in the TE-book, logging into the Teacher’s 

Resource site, or accessing this resource on 

the TRFD. 


442 



11/01/17 3:58 PM


Lesson 6.6 




Determine if the sampling distribution of x is approximately normal 

when sampling from a non-normal population. 

If appropriate, use a normal distribution to calculate probabilities 

involving x. 

Exercises 


1. Songs on an iPod David’s iPod has about 10,000 

songs. The distribution of the play times for these 

songs is heavily skewed to the right with a mean 

of 225 seconds and a standard deviation of 60 

seconds. 

(a) Describe the shape of the sampling distribution of 

x for SRSs of size n 5 5 from the population of 

songs on David’s iPod. Justify your answer. 

(b) Describe the shape of the sampling distribution of 


songs on David’s iPod. Justify your answer. 

2. Insurance claims An insurance company claims 

that in the entire population of homeowners, the 

mean annual loss from fire is m 5 $250 with a standard 

deviation of s 5 $5000. The distribution of 

losses is strongly right-skewed: Many policies have 

$0 loss, but a few have large losses. 

(a) Describe the shape of the sampling distribution of 


homeowners. Justify your answer. 



homeowners. Justify your answer. 

3. How many in a car? A study of rush-hour traffic in 

San Francisco counts the number of people in each 

car entering a freeway at a suburban interchange. 

Suppose that the number of people per car in the 

population of all cars that enter at this interchange 

during rush hours has a mean of m 5 1.5 and a 

standard deviation of s 5 0.75. 

(a) Could the distribution of the number of people 

per car be normal for the population of all cars 

entering the interchange during rush hours? 

Explain. 


x for SRSs of size n 5 100 from the population 

of all cars that enter this interchange during rush 

hours. Justify your answer. 

pg 440 

Lesson 6.6 


Exercises 

You can find the full solutions for this lesson 

by clicking on the link in the TE-book, logging 

into the Teacher’s Resource site,or accessing 



1. (a) Because n = 5 < 30, the sampling 

distribution of x will also be skewed to the right 


(b) Because n = 100 ≥ 30, the sampling 

distribution of x is approximately normal by 

the central limit theorem. 

p. 440 1–4 

p. 441 5–8 

4. Flawed carpets A supervisor at a carpet factory 

randomly selects 1-square-yard pieces of carpet 

and counts the number of flaws in each piece. The 

number of flaws per square yard in the population 

of 1-square-yard pieces varies with mean m 5 1.6 

and standard deviation s 5 1.2. 

(a) Could the distribution of the number of flaws be 

normal for the population of all 1-square-yard 

pieces of carpet? Explain. 


x for SRSs of size n 5 60 from the population of all 

1-square-yard pieces of carpet. Justify your answer. 

5. More songs on an iPod Refer to Exercise 1. What 

pg 441 is the probability that the mean length in a random 

sample of 100 songs is less than 4 minutes (240 

seconds)? 

6. More insurance claims Refer to Exercise 2. Suppose 

that the insurance company charges $300 

for each policy. What is the probability that the 

insurance company will make money on a random 

sample of 1000 homeowners? That is, what is the 

probability that the mean loss for a random sample 

of homeowners is less than $300? 

7. More people in a car Refer to Exercise 3. What 

is the probability that the mean number of people 

in a random sample of 100 cars that enter at this 

interchange during rush hours is at least 1.7? 

8. More flawed carpets Refer to Exercise 4. What is 

the probability that the mean number of flaws in 

a random sample of sixty 1-square-yard pieces of 

carpet is at least 1.7? 


9. Where does lightning strike? The number of lightning 

strikes on a square kilometer of open ground 

in a year has a mean of 6 and standard deviation 

of 2.4. The National Lightning Detection Network 

(NLDN) uses automatic sensors to watch for lightning 

in a random sample of fifty 1-square-kilometer 

18/08/16 5:04 PM 

2. (a) Because n = 15 < 30, the sampling 

distribution of x will also be skewed to the right 





3. (a) No; a count only takes on wholenumber 

values, so it cannot be normally 

distributed. 




4. (a) No; a count only takes on wholenumber 

values, so it cannot be normally 

distributed. 




5. Mean: m x = m = 225 seconds; 

SD: s x = s !n = 60 

!100 = 6 seconds 

Shape: Because n = 100 ≥ 30, the sampling 

distribution of x is approximately 

normal by the central limit theorem. 

240 − 225 

z = = 2.5; 

6 

P( x < 240) = P(Z < 2.5) = 0.9938 



SD:6) 5 0.9938 

6. Mean: m x = m = $250; 

SD: s x = s !n = 5000 

!1000 = $158.11 

Shape: Because n = 1000 ≥ 30, the 

sampling distribution of x is approximately 


300 − 250 

z = 

158.11 = 0.32; 

P( x < 300) = P( Z < 0.32) = 0.6255 



SD:158.11) 5 0.6241 

7. Mean: m x = m = 1.5 people; 

SD: s x = s !n = 0.75 = 0.075 people 

!100 




1.7 − 1.5 

z = = 2.67; P( x > 1.7) = 

0.075 

P( Z > 2.67) = 1− 0.9962 = 0.0038 



SD:0.075) 5 0.0038 

8. Mean: m x = m = 1.6 flaws; 

SD: s x = s !n = 1.2 = 0.155 flaws 

!60 




1.7 − 1.6 

z = = 0.65; P( x > 1.7) = 

0.155 

P( Z > 0.65) = 1− 0.7422 = 0.2578 



SD:0.155) 5 0.2594 

Lesson 6.6 



11/01/17 3:58 PM

9. (a) Because n = 50 ≥ 30, the 



(b) Mean: m x = m = 6 lightning strikes; 

SD: s x = s !n = 2.4 = 0.339 lightning 

!50 

strikes; z = 5 − 6 

0.339 = −2.95; 

P( x < 5) = P( Z < −2.95) = 0.0016 



SD:0.339) 5 0.0016 

10. (a) Because n = 45 ≥ 30, the 



(b) Mean: m x = m = 12.8 minutes; 

SD: s x = s !n = 7.2 = 1.073 minutes 

!45 

15 − 12.8 

z = = 2.05; P( x > 15) = 

1.073 

P( Z > 2.05) = 1 − 0.9798 = 0.0202 


(lower:15, upper:1000, mean:12.8, 

SD:1.073) 5 0.0202 

11. (a) Because n = 10 < 30, we can’t 

be sure that the sampling distribution of 

x will be approximately normal. 

(b) Greater than; the variability of the 

sampling distribution of x will be greater 

with the smaller sample size of 10, making 

it more likely to get a sample mean 

that is far away from the true mean of 6 

lightning strikes (such as x 5 5 or lower). 

12. (a) Because n = 5 < 30, we can’t 

be sure that the sampling distribution of 

x will be approximately normal. 

(b) Greater than; the variability of the 

sampling distribution of x will be greater 

with the smaller sample size of 5, making 

it more likely to get a sample mean that 

is far away from the true mean of 12.8 

minutes (such as x 5 15 or higher). 

13. From Exercise 7, we know that when 

the true mean number of people in the 

car is 1.5, there is almost a 0% chance 

that the mean number of people in the 

car will be at least 1.7 in a random sample 

of 100 cars. Because the observed result 

is unlikely to happen purely by chance 

(less than 5%), the researcher has good 

evidence to conclude that people are 

more likely to drive with other people in 

the car on Sundays. 

14. From Exercise 8, we know that when 

the true mean number of flaws is 1.6, 

there is about a 26% chance that the 

mean number of flaws will be at least 

1.7 in a random sample of 60 pieces of 

carpet. Because this is a large probability, 

it is plausible that the supervisor’s result 

occurred purely by chance and that the 

machine is still working properly. 

444 


plots of land. Let x be the average number of lightning 

strikes in the sample. 

(a) What is the shape of the sampling distribution of x for 

samples of size n 5 50 from this population? Justify. 

(b) Calculate the probability that the average number of 

lightning strikes per square kilometer x is less than 5. 

10. Please hold The customer care manager at a cell 

phone company keeps track of how long each 

help-line caller spends on hold before speaking to 

a customer service representative. He finds that the 

distribution of wait times for all callers has a mean 

of 12.8 minutes with a standard deviation of 7.2 

minutes. The distribution is moderately skewed to the 

right. Suppose the manager takes a random sample of 

45 callers and calculates their mean wait time x. 

(a) What is the shape of the sampling distribution of 

x for samples of size n 5 45 from this population? 

Justify. 

(b) Calculate the probability that the mean wait time x 

is more than 15 minutes. 

11. Lightning strikes twice? Refer to Exercise 9. 

(a) Explain why you cannot calculate the probability that 

the average number of lightning strikes per square 

kilometer x is less than 5 for samples of size n 5 10. 

(b) Will the probability referred to in part (a) be less 

than, greater than, or about the same as the probability 

in Exercise 9(b)? Explain. 

12. Please continue to hold Refer to Exercise 10. 

(a) Explain why you cannot calculate the probability 

that the average wait time for customer service x is 

more than 15 minutes for samples of size n 5 5. 

(b) Will the probability referred to in part (a) be less 

than, greater than, or about the same as the probability 

in Exercise 10(b)? Explain. 

13. Even more people in a car Refer to Exercise 7. A 

researcher selects a random sample of 100 cars 

that enter this interchange on a Sunday and finds 

x 5 1.7 people per car. Because the sample mean 

is greater than 1.5, the researcher concludes that 

people are more likely to drive with other people 

in the car on Sundays. Based on your answer to 

Exercise 7, what would you say to the researcher? 

14. Even more flaws in carpets Refer to Exercise 8. 

A supervisor selects a random sample of sixty 

1-square-yard pieces of carpet and finds that 

x 5 1.7 flaws. Because the sample mean is more 

than the expected mean of 1.6 flaws, the supervisor 

is thinking about shutting down the machine 

for inspection. Based on your answer to Exercise 8, 

what would you say to the supervisor? 


15. No; the graph of the sample will resemble 

the shape of the population distribution, 

regardless of the sample size. The student 

should say that the graph of the sampling 

distribution of the sample mean (x) looks more 

and more normal as you take larger and larger 

samples from a population. 

16. No; the central limit theorem is only 

about the shape of the sampling distribution 

of the sample mean. The statement is 

otherwise correct. 

17. A total of $45,000 for 50 students yields 

an average cost per student of $900. We want 

to find P(x > 900). 

Mean: m x = m = $836; 

SD: s x = s !n = 388 

!50 = $54.87 

15. What does the CLT say? Asked what the central 

limit theorem says, a student replies, “As you take 

larger and larger samples from a population, the 

graph of the sample values looks more and more 

normal.” Is the student right? Explain your answer. 

16. Is this what the CLT says? Asked what the central 

limit theorem says, a student replies, “As you take 

larger and larger samples from a population, the 

variability of the sampling distribution of the sample 

mean decreases.” Is the student right? Explain your 

answer. 


17. Cost of textbooks The cost of textbooks for students 

at a particular college has a mean of m 5 $836 

per year with a standard deviation of s 5 $388. 

What is the probability that a random sample of 

50 students spends a total of more than $45,000 

on books this year? Hint: Re-express the total cost 

in terms of the average cost per student x. 


18. Are rich people mean? (3.1, 3.6) Psychologist 

Paul Piff from the University of California, 

Berkeley, studies the relationship between 

wealth and lawful behavior. In one such study, 

he had assistants cross a road at a crosswalk and 

recorded if drivers obeyed the law and stopped 

to let the person cross or kept driving and cut 

off the pedestrian. He compared the response of 

people driving expensive cars and inexpensive 

cars. Here are his results. 18 type of car 

Driver 

behavior 

Yielded to 

pedestrian 

Cut off 

pedestrian 

Expensive car 

Inexpensive car 

32 67 

26 27 

(a) The report on this study stated that the researcher 

who determined if the cars could be classified as 

expensive or inexpensive was “blind to the hypothesis 

of the study.” Explain what this means. 

(b) Is this an observational study or an experiment? 


19. How do rich people drive? (4.3, 4.4) Suppose we 

choose a driver at random from the results of the study 

in Exercise 18. Show that the events “Yielded to pedestrians” 

and “Expensive car” are not independent. 




900 − 836 

z = = 1.17; P(total cost > 45,000) 

54.87 

= P( x > 900) = P(Z > 1.17) = 1 − 0.8790 

= 0.1210 


upper:10000, mean:836, SD:54.87)5 0.1217 

18. (a) This means that the researcher 

who determined whether the cars could be 

classified as expensive or inexpensive didn’t 

know the other variable being measured 

(driver behavior). If the researcher knew the 

hypothesis of the study, it could have affected 

the classification of the cars. 

Answers 18(b)–19 are on page 445 


444 



11/01/17 3:58 PM

Chapter 6 Main Points 

445 


© Kirsty Pargeter/Alamy Stock Photo 

Main Points 

The Idea of a Sampling Distribution 

j 

j 

Chapter 6 


18. (b) Observational study. There were no 

treatments imposed on the drivers. In other 

words, drivers weren’t assigned to drive an 

expensive car or an inexpensive car. 

19. P(yielded to pedestrians | expensive car) 

32 

5 

32 + 26 = 32 

58 = 0.55 

P(yielded to pedestrians | inexpensive car) 

67 

5 

67 + 27 = 67 

94 = 0.71 

Because the probabilities are not equal, the 

events “Yielded to pedestrians” and 

“Expensive car” are not independent. 

Knowing that a randomly selected car is 

expensive decreases the probability that the 

car yielded to pedestrians. 


How can we build “greener” batteries? 

Refer to the STATS applied! on page 399. When the manufacturing 

process is working properly, the distribution of battery lifetimes 

has mean m 5 17 hours with standard deviation s 5 0.8 hour, and 

73% last at least 16.5 hours. 

1. Assume that the manufacturing process is working properly, and let p^ 5 the sample 

proportion of batteries that last at least 16.5 hours. Calculate the mean and standard 

deviation of the sampling distribution of p^ for random samples of 50 batteries. 

2. Describe the shape of the sampling distribution of p^ for random samples of 50 

batteries. Justify your answer. 

3. In the sample of 50 batteries, only 68% lasted at least 16.5 hours. Find the probability 

of obtaining a random sample of 50 batteries where p^ is 0.68 or less if the 

manufacturing process is working properly. 

4. Assume that the process is working properly, and let x 5 the sample mean lifetime 

(in hours). Calculate the mean and standard deviation of the sampling distribution 

of x for random samples of 50 batteries. 

5. Describe the shape of the sampling distribution of x for random samples of 50 

batteries. Justify your answer. 

6. In the sample of 50 batteries, the mean lifetime was only 16.718 hours. Find the 

probability of obtaining a random sample of 50 batteries with a mean lifetime of 

16.718 hours or less if the manufacturing process is working properly. 

7. Based on your answers to Questions 3 and 6, should the company be worried that 

the manufacturing process isn’t working properly? Explain. 


A parameter is a number that describes some characteristic 

of the population. A statistic is a number 

that describes some characteristic of a sample. We 

use statistics to estimate parameters. 

The sampling distribution of a statistic is the distribution 

of values taken by the statistic in all possible 

samples of the same size from the same population. 

j 

To determine a sampling distribution, list all possible 

samples of a particular size, calculate the value 

of the statistic for each sample, and graph the 

distribution of the statistic. If there are many possible 

samples, use simulation to approximate the 

sampling distribution: Repeatedly select random 

samples of a particular size, calculate the value of 

the statistic for each sample, and graph the distribution 

of the statistic. 

18/08/16 5:04 PM 

TRM chapter 6 Learning Targets Grid 

You can find a grid with all of the learning 

targets for this chapter by clicking on the link 


Resource site, or accessing this resource on 

the TRFD. An extra column has been added for 

students to track their progress. 

Teaching Tip: 


This STATS applied! reviews the skills and 

ideas from this chapter. Many students 

will have difficulty switching between 

sample proportions and sample means. 

They may also struggle with using the 

correct symbols. Monitor their progress 

and provide help and practice as needed. 

Answers to STATS applied! 

p(1− p) 

1. m p^ = p = 0.73; s p^ = Å n 

0.73(1 − 0.73) 

= = 0.063 

Å 50 

2. Because np = (50)(0.73) = 36.5 $ 10 

and n(1− p) = (50)(1− 0.73)=13.5 $ 10, 

the sampling distribution of p^ is 


0.68 − 0.73 

3. z = = −0.79; 

0.063 

P( p^ ≤ 0.68) = P( Z ≤ −0.79) = 0.2148 


(lower:21000, upper:0.68, mean:0.73, 

SD:0.063) 5 0.2137 

4. m x = m = 17 hours; 

s x = s !n = 0.8 = 0.113 hour 

!50 

5. Because n = 50 ≥ 30, the sampling 



6. z = 16.718−17 = −2.50; 

0.113 

P( x ≤ 16.718) = P( Z ≤ −2.5) = 0.0062 


(lower:21000, upper:16.718, mean:17, 

SD:0.113) 5 0.0063 

7. From Question 3, we know that if 

the manufacturing process is working 

properly (p = 0.73), there is about a 

21% chance that the sample proportion 

of batteries that last less than 16.5 

hours would be less than 0.68. Because 

this outcome is plausible, the sample 

proportion of 0.68 does not provide 

strong evidence that the true proportion 

is less than 0.73, and the company 

should not be worried. 

However, from Question 6, we know 

that if the manufacturing process is 

working properly (m = 17 hours), there 

is only about a 1% chance that the mean 

battery life will be 16.718 hours or less. 

Because this outcome is unlikely (less 

than 5%), the sample mean of 16.718 

provides strong evidence that the true 

mean battery life is less than 17 hours, 

and the company should be worried. 

Main Points 

C H A P T E R 6 • Main Points 445 


11/01/17 3:58 PM

446 


Answers to Chapter 6 Review 

Exercises 

1. Population: All eggs shipped in one 

day. Sample: The 200 eggs examined. 

Parameter: The proportion p of eggs 

shipped that day that had salmonella. 

Statistic: The proportion of eggs in 

the sample that had salmonella, 

p^ = 9 

200 = 0.045. 

2. (a) 

Sample #1: 64, 66, 71 Median 5 66 

Sample #2: 64, 66, 73 Median 5 66 

Sample #3: 64, 66, 76 Median 5 66 

Sample #4: 64, 71, 73 Median 5 71 

Sample #5: 64, 71, 76 Median 5 71 

Sample #6: 64, 73, 76 Median 5 73 

Sample #7: 66, 71, 73 Median 5 71 

Sample #8: 66, 71, 76 Median 5 71 

Sample #9: 66, 73, 76 Median 5 73 

Sample #10: 71, 73, 76 Median 5 73 

d 

d 

d 

d 

d 

d 

66 67 68 69 70 71 72 73 

Sample median book length 

66 + 66 + 66 + 71 + 71 + 

71 + 71 + 73 + 73 + 73 

(b) m median = 

10 

= 701 

10 = 70.1 

The sample median is a biased estimator 

of the population median. The mean 

of the sampling distribution is equal to 

70.1, which is less than the value of the 

population median of 71. 

(c) The sampling distribution of the 

sample median will be less variable 

because the sample size is larger. The 

estimated median book length will 

typically be closer to the true median 

book length. In other words, the estimate 

will be more precise. 

3. (a) m X = np = 500(0.24) = 120 

people; s X = "np(1 − p) 

= "500(0.24)(1 − 0.24) = 9.55 people 

(b) If many samples of size 500 were 

taken, the number of people who are 

under 18 years old would typically vary 

by about 9.55 from the mean of 120. 

(c) The sampling distribution of X 

is approximately normal because 

np = 500(0.24) = 120 ≥ 10 and 

n(1 − p) = 500(1 − 0.24) = 380 ≥ 10 

j 

j 

j 

We can use sampling distributions to determine 

what values of a statistic are likely to happen by 

chance alone and how much a statistic typically 

varies from the parameter it is trying to estimate. 

A statistic used to estimate a parameter is an 

unbiased estimator if the mean of its sampling 

distribution is equal to the value of the parameter 

being estimated. That is, the statistic doesn’t 

consistently overestimate or consistently underestimate 

the value of the parameter when many 

random samples are selected. 

The sampling distribution of any statistic will 

have less variability when the sample size is larger. 

That is, the statistic will be a more precise estimator 

of the parameter with larger sample sizes. 

Sample Counts and Sample Proportions 

j 

j 

Let X 5 the number of successes in a random sample 

of size n from a large population with proportion 

of successes p. The sampling distribution of a 

sample count X describes the distribution of values 

taken by the sample count X in all possible samples 

of the same size from the same population. 

j The mean of the sampling distribution of X 

is m X = np. The mean describes the average 

value of X in repeated random samples. 

j The standard deviation of the sampling distribution 

of X is s X = !np(1 − p). The standard 

deviation describes how far the values of X typically 

vary from m X in repeated random samples. 

j The shape of the sampling distribution of X 

will be approximately normal when the Large 

Counts condition is met: np ≥ 10 and n(1 2 p) 

≥10. 

Let p^ 5 the proportion of successes in a random 

sample of size n from a large population with proportion 

of successes p. The sampling distribution of a 

sample proportion p^ describes the distribution of values 

taken by the sample proportion p^ in all possible 

samples of the same size from the same population. 

j The mean of the sampling distribution of p^ is 

m p^ = p. The mean describes the average value 

of p^ in repeated random samples. 

j The standard deviation of the sampling 

p(1 − p) 

distribution of p^ is s p^ 5 . The 

Å n 


100 − 120 

(d) z = = −2.09; 

9.55 

110 − 120 

z = = −1.05 

9.55 

P(100 ≤ X ≤ 110) ≈ P(−2.09 ≤ 

Z ≤ − 1.05) = 0.1469 − 0.0183 = 0.1286 



5 0.1294 

j 

standard deviation describes how far the values 

of p^ typically vary from p in repeated random 

samples. 

The shape of the sampling distribution of 

p^ will be approximately normal when the 

Large Counts condition is met: np ≥ 10 and 

n(1 2 p) ≥ 10. 

Sample Means 

j 

j 

Let x 5 the mean of a random sample of size n 

from a large population with mean m and standard 

deviation s. The sampling distribution of 

a sample mean x describes the distribution of 

values taken by the sample mean x in all possible 

samples of the same size from the same 

population. 

j The mean of the sampling distribution of x is 

m x = m. The mean describes the average value 

of x in repeated random samples. 

j The standard deviation of the sampling distribution 

of x is s− x = s . The standard deviation 

describes how far the values of x typically 

"n 

vary from m in repeated random samples. 

The shape of the sampling distribution of 

x will be approximately normal when the 

Normal/Large Sample condition is met: The 

population is normal or the sample size is large 

(n ≥ 30). The fact that the sampling distribution 

of x becomes approximately normal— 

even when the population is non-normal—as 

the sample size increases is called the central 

limit theorem. 

Probability Calculations 

j 

j 

When the sampling distribution of a statistic is 

approximately normal, you can use z-scores and 

Table A or technology to do probability calculations 

involving the statistic. 

To determine which sampling distribution to use, 

consider whether the variable of interest is categorical 

or quantitative. If it is categorical, use the 

sampling distribution of a sample count X or the 

sampling distribution of a sample proportion p^ . If 

it is quantitative, use the sampling distribution of 

a sample mean x. 

TRM chapter 6 Review Exercise Videos 

Video solutions to the Chapter 6 Review 

Exercises are available to teachers and 

students. Access them by clicking on the link 



on the TRFD. 


446 



11/01/17 3:58 PM

Chapter 6 Review Exercises 447 

Chapter 6 Review Exercises 

1. Bad eggs (6.1) People who eat eggs that are contaminated 

with salmonella can get food poisoning. A large 

egg producer takes an SRS of 200 eggs from all the eggs 

shipped in one day. The laboratory reports that 9 of 

these eggs had salmonella contamination. Identify the 

population, the parameter, the sample, and the statistic. 

2. Five books (6.1, 6.2) An author has written 5 children’s 

books. The number of pages in these books are 

64, 66, 71, 73, and 76. 

(a) List all 10 possible SRSs of size n 5 3, calculate the 

median number of pages for each sample, and display 

the sampling distribution of the sample median on a 

dotplot. 

(b) Show that the sample median is a biased estimator of 

the population median for this population. 

(c) Describe how the variability of the sampling distribution 

of the sample median would change if the sample 

size was increased to n 5 4. 

3. Kids these days (6.3) According to the 2010 U.S. 

Census, 24% of U.S. residents are under 18 years old. 

Suppose we take a random sample of 500 U.S. residents. 

Let X 5 the number of people in the sample 

who are under 18 years old. 




(c) Justify that it is appropriate to use a normal distribution 

to model the sampling distribution of X. 

(d) Using a normal distribution, calculate the probability 

that the number of people under 18 years old in a random 

sample of size 500 is between 100 and 110. 

4. Five-second rule (6.1, 6.4) A report claimed that 20% 

of respondents subscribe to the “5-second rule.” That 

is, they would eat a piece of food that fell onto the 

kitchen floor if it was picked up within 5 seconds. Assume 

this figure is accurate for the population of U.S. 

adults. Let p^ = the proportion of people who subscribe 

to the 5-second rule in an SRS of size 80 from 

this population. 

(a) Calculate the mean and the standard deviation of the 

sampling distribution of p^ . 


(c) Justify that it is appropriate to use a normal distribution 

to model the sampling distribution of p^ . 

(d) In an SRS of size 80, only 10% subscribed to the 

5-second rule. Does this result provide convincing evidence 

that the proportion of all U.S. adults who subscribe 

to the 5-second rule is less than 0.20? Calculate 

P(p^ ≤ 0.10) and use this result to support your answer. 

5. Normal IQ scores? (6.5, 6.6) The Wechsler Adult 

Intelligence Scale (WAIS) is a common “IQ test” for 

adults. The distribution of WAIS scores for persons 

over 16 years of age is approximately normal with 

mean 100 and standard deviation 15. Let x 5 the 

mean WAIS score in a random sample of 10 people 

over 16 years of age. 




(c) What is the probability that the average WAIS score is 

105 or greater for a random sample of 10 people over 

16 years of age? Show your work. 

(d) Would your answers to any of parts (a), (b), or (c) be 

affected if the distribution of WAIS scores in the adult 

population were distinctly non-normal? Explain. 

6. Watching for gypsy moths (6.1, 6.6) The gypsy moth 

is a serious threat to oak and aspen trees. A state agriculture 

department places traps throughout the state 

to detect the moths. Each month, an SRS of 50 traps 

is inspected, the number of moths in each trap is recorded, 

and the mean number of moths is calculated. 

Based on years of data, the distribution of moth 

counts is strongly skewed with mean 0.5 and standard 

deviation 0.7. 

(a) Explain why it is reasonable to use a normal distribution 

to approximate the sampling distribution of x for 

SRSs of size 50. 

(b) Calculate the probability that the mean number of 

moths in a sample of size 50 is at least 0.6 moths. 

(c) In a recent month, the mean number of moths in an SRS 

of size 50 was 0.6. Based on this result, should the state 

agricultural department be worried that the moth population 

is getting larger in their state? Explain. 

(c) The sampling distribution of x is 

approximately normal because the 

population distribution is approximately 

normal. 

105 − 100 

z = = 1.05; P ( x ≥ 105) 

4.74 

= P(Z ≥ 1.05) = 1− 0.8531 = 0.1469 



SD:4.74) 5 0.1457 

(d) The answer to parts (a) and (b) 

would be the same because the 

mean and standard deviation do not 

depend on the shape of the population 

distribution. We could not answer part 

(c) because we could not be sure that the 

sampling distribution is approximately 

normal. 

6. (a) Because n = 50 ≥ 30, the 



(b) Mean: m x = m = 0.5 moth; 

SD: s x = s !n = 0.7 = 0.099 moth 

z = 

!50 

0.6 − 0.5 

0.099 = 1.01; 

P( x ≥ 0.6) = P(Z ≥ 1.01) = 1 − 0.8438 

= 0.1562 



SD:0.099) 5 0.1562 

(c) Assuming the true mean number of 

moths is 0.5, there is about a 16% chance 

that the mean number of moths will 

be at least 0.6 in a sample of 50 traps. 

Because this result is plausible (more 

than 5%), we do not have convincing 

evidence that the moth population is 

getting larger. 

Review 



4. (a) m p^ = p = 0.20; 

p(1− p) 

s p^ = Å n 

= 0.045 

= Å 

0.20(1− 0.20) 

80 


proportion of people who subscribe to the 

5-second rule will typically vary by about 

0.045 from the true proportion of p 5 0.20. 

(c) Because np = (80)(0.2) = 16 $ 10 and 

n(1− p) = (80)(1− 0.20) = 64 ≥ 10, the 

sampling distribution of p^ is approximately 

normal. 

0.10 − 0.20 

(d) z = = −2.22; 

0.045 

P( p^ ≤ 0.10) = P( Z ≤ −2.22) = 0.0132 

18/08/16 5:04 PM 


(lower:21000, upper:0.10, mean:0.20, SD:0.045) 

5 0.0131. 

Assuming the true proportion of people who 

subscribe to the 5-second rule is 0.20, there is 

only about a 1% chance of getting a sample 

proportion of 0.10 or less purely by chance. 

Because this result is unlikely (less than 5%), 

we have convincing evidence that the 

proportion of all U.S. adults who subscribe to 

the 5-second rule is less than 0.20. 

5. (a) m x = m = 100; 

s x = s !n = 15 

!10 = 4.74 


WAIS score will typically vary by about 4.74 

from the true mean of 100. 

TRM full Solutions to Chapter 

Review Exercises and Test 

You can find the full solutions by clicking 


the Teacher’s Resource Site, or accessing 


C H A P T E R 6 • Review Exercises 447 


11/01/17 3:58 PM

448 


Answers to Chapter 6 Practice 

Test 


Practice Test 

1. b 

2. c 

3. c 

4. b 

5. a 

6. a 

7. b 

Section I: Multiple choice Select the best answer for each question. 

1. A study of voting chose 663 registered voters at random 

shortly after an election. Of these, 72% said they 

had voted in the election. Election records show that 

only 56% of registered voters voted in the election. 

Which of the following statements is true about these 

percentages? 

(a) 72% and 56% are both statistics. 

(b) 72% is a statistic and 56% is a parameter. 

(c) 72% is a parameter and 56% is a statistic. 

(d) 72% and 56% are both parameters. 

2. Vermont is particularly beautiful in early October 

when the leaves begin to change color. At that time of 

year, a large proportion of cars on Interstate 91 near 

Brattleboro have out-of-state license plates. Suppose a 

Vermont state trooper randomly selects 50 cars driving 

past Exit 2 on I-91, records the state identified 

on the license plate, and calculates the proportion of 

cars with out-of-state plates. Which of the following 

describes the sampling distribution of the sample proportion 

in this context? 

(a) The distribution of state for all cars in the trooper’s 

sample of cars passing this exit 

(b) The distribution of state for all cars passing this exit 

(c) The distribution of the proportion of cars with 

out-of-state plates in all possible samples of 50 cars 

passing this exit 

(d) The distribution of the proportion of cars with 

out-of-state plates in the trooper’s sample of 50 cars 

passing this exit 

3. A polling organization wants to estimate the proportion 

of voters who favor a new law banning smoking 

in public buildings. The organization decides to 

increase the size of its random sample of voters from 

about 1500 people to about 4000 people right before 

an election. The effect of this increase is to 

(a) reduce the bias of the estimate. 

(b) increase the bias of the estimate. 

(c) reduce the variability of the estimate. 

(d) increase the variability of the estimate. 

4. A machine is designed to fill 16-ounce bottles of 

shampoo. When the machine is working properly, the 

amount poured into the bottles follows a normal distribution 

with mean 16.05 ounces and standard deviation 

0.1 ounce. Assume that the machine is working 

properly. If 4 bottles are randomly selected and the 

number of ounces in each bottle is measured, then 

there is about a 95% chance that the sample mean 

will fall in which of the following intervals? 

(a) 16.00 to 16.10 ounces 

(b) 15.95 to 16.15 ounces 

(c) 15.90 to 16.20 ounces 

(d) 15.85 to 16.25 ounces 

5. The central limit theorem is important in statistics 

because it allows us to use the normal distribution to 

find probabilities involving the sample mean if the 

(a) sample size is reasonably large for any population 

shape. 

(b) sample size is reasonably large and the population is 

normally distributed. 

(c) population size is reasonably large for any population 

shape. 

(d) population size is reasonably large and the population 

is normally distributed. 

6. At a high school, 85% of students are right-handed. 

Let X 5 the number of students who are right-handed 

in a random sample of 10 students from the school. 

Which one of the following statements about the 

mean and standard deviation of the sampling distribution 

of X is true? 

(a) m x 5 8.5; s x ≈ 1.129 

(b) m x 5 8.5; s x ≈ 0.113 

(c) m x 5 8.5; s x ≈ cannot be determined from the information 

given. 

(d) Neither the mean nor the standard deviation can be 

determined from the information given. 

7. The student newspaper at a large university asks an 

SRS of 250 undergraduates, “Do you favor eliminating 

the carnival from the end-of-term celebration?” 

In the sample, 150 of the 250 undergraduates are 

in favor. Suppose that 55% of all undergraduates 

favor eliminating the carnival. If you took a very 

large number of SRSs of size n 5 250 from this 

population, the sampling distribution of the sample 

proportion p^ would have which of the following 

characteri stics? 

(a) Mean 0.55, standard deviation 0.03, shape unknown 

(b) Mean 0.55, standard deviation 0.03, approximately 

normal 

(c) Mean 0.60, standard deviation 0.03, shape unknown 

(d) Mean 0.60, standard deviation 0.03, approximately 

normal 



448 



11/01/17 3:58 PM

Chapter 6 Practice Test 

449 

8. Scores on the mathematics part of the SAT exam in a 

recent year followed a normal distribution with mean 

515 and standard deviation 114. You choose an SRS 

of 100 students and calculate x5 mean SAT Math 

score. Which of the following are the mean and standard 

deviation of the sampling distribution of x? 

(a) Mean 5 515, SD 5 114 

114 

(b) Mean 5 515, SD 5 

"100 

(c) Mean 5 515 114 

, SD 5 

100 100 

(d) Mean 5 515 

100 , SD 5 114 

"100 

9. In a congressional district, 55% of the registered voters 

are Democrats. Which of the following is closest to 

the probability of getting less than 50% Democrats in 

a random sample of size 100? 

(a) 0.157 (b) 0.496 (c) 0.504 (d) 0.843 

10. A statistic is an unbiased estimator of a parameter 

when 

(a) the statistic is calculated from a random sample. 

(b) in all possible samples of a specific size, the distribution 

of the statistic has a shape that is approximately 

normal. 

(c) in all possible samples of a specific size, the values of 

the statistic are very close to the value of the parameter. 

(d) in all possible samples of a specific size, the values of 

the statistic are centered at the value of the parameter. 

Section II: Free Response 

11. Here are histograms of the values taken by three 

sample statistics in several hundred samples from the 

same population. The true value of the population 

parameter is marked with an arrow on each histogram. 

Which statistic would provide the best estimate 

of the parameter? Explain. 

A B C 

12. The amount that households pay service providers for 

access to the Internet varies quite a bit, but the mean 

monthly fee is $48 and the standard deviation is $20. 

The distribution is not normal: Many households pay 

a base rate for low-speed access, but some pay much 

more for faster connections. A sample survey asks an 

SRS of 500 households with Internet access how much 

they pay per month. Let x be the mean amount paid 

by the members of the sample. 

(a) Calculate the mean and standard deviation of the sampling 

distribution of x. Interpret the standard deviation. 

(b) What is the shape of the sampling distribution of x? 

Justify. 

(c) Find the probability that the average amount paid by 

the sample of households exceeds $50. 

13. According to government data, 22% of American children 

under the age of 6 live in households with incomes 

less than the official poverty level. A study of learning in 

early childhood chooses an SRS of 300 children. 

(a) Let X 5 the count of children in this sample who live in 

households with incomes less than the official poverty 

level. What is the shape of the sampling distribution of 

X? Justify your answer. 

(b) Find the probability that more than 20% of the sample 

are from poverty-level households. 

8. b 

9. a 

10. d 

11. Statistic A. Both statistics A and B 

appear to be unbiased, with the center of 

their sampling distributions equal to the 

value of the parameter, but statistic A has 

less variability than statistic B. 

12. (a) m x = m = $48; 

s x = s !n = 20 

!500 = $0.89 

In SRSs of size n 5 500, the sample mean 

amount paid for Internet will typically 

vary by about $0.89 from the true mean 

of $48. 




50 − 48 

(c) z = = 2.25; P( x > 50) 

0.89 

= P(Z > 2.25) = 1 − 0.9878 = 0.0122 



5 0.0123 

13. (a) The sampling distribution of 

X is approximately normal because 

np = 300(0.22) = 66 ≥ 10 and 

n(1 − p) = 300(1 − 0.22) = 234 ≥ 10. 

(b) 20% of 300 is 60, so we want to find 

P( X > 60). 

m X = np = 300(0.22) = 66; 

s X = "np(1 − p) 

= "300(0.22)(1 − 0.22) = 7.17 

60 − 66 

z = = −0.84; 

7.17 

P(X > 60) = P(Z > −0.84) = 0.7995 



5 0.7987 

Practice Test 


18/08/16 5:04 PM 

C H A P T E R 6 • Practice Test 449 


11/01/17 3:58 PM

SPA 3e_ Teachers Edition _ Ch 6

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