Ch. 6.1 #7-49 odd

Ch. 6.1 #7-49 odd
The area is found by looking up z= 0.75 in Table E and subtracting 0.5.
Area = 0.7734- 0.5= 0.2734
The area is found by looking up z= 2.07 in Table E and subtracting from 0.5.
Area = 0.5- 0.0192 = 0.4808
The area is found by looking up z=0.23 in Table E and subtracting it from 1
Area =1-0.5910= 0.4090
The area is found by looking up z= 1.43 in Table E.
Area = 0.0764

15) Between z = 1.05 and z= 1.78
The area is found by looking up the values 1.05 and 1.78 in Table E and subtracting the areas.
Area= 0.9625– 0.8531 = 0.1094.
0.1094
Z=1.78
Z=1.05
58
The area is found by looking up the values z=1.56 and z=1.83 in Table E and subtracting the areas.
Area = 0.0594 - 0.0336 = 0.0258
19) Between z = -1.53 and z= - 2.08
The area is found by looking up the values z=1.53 and z= 2.08 in Table E and subtracting the areas.
Area = 0.0630- 0.0188 = 0.0442
.

The area is found by looking up z= 2.11 in Table E.
Area = 0.9826
23) To the right of z=-0.25
The area is found by looking up z= 0.25 in Table E and subtracting it from 1.
Area= 1- 0.4013 = 0.5987
For z = -0.44, the area is 0.3300. For z = 1.92, the area is 1 - 0.9726 = 0.0274
Area = 0.3300 + 0.0274 = 0.3574
Area = 0.7486 - 0.5 = 0.2486

Area = 0.5 - 0.0582 = 0.4418
Area =1 - 0.9977 = 0.0023
Area = 0.1131

Area= 0.9591 - 0.0069 = 0.9522
Area =0.9985 - 0.9279 = 0.0706
Area = 0.9222
For exercises 40 through 45, find the z value that corresponds to the given area.
41)

Since the z score is on the left side of 0, use the negative z table. Areas in the negative z
table are in the tail, so we will use 0.5 - 0.4175 = 0.0825 as the area. The closest z score corresponding to an area of
0.0825 is z=- 1.39.
43)
Z=- 2.08, found by using the negative z table.
45)
Use the negative z table and 1 - 0.8962 = 0.1038 for the area. The z score is z= -1.26.
47) Find the z values to the left of the mean so that
a) 98.87% of the area under the distribution curve lies to the right of it.
Using the negative z table, area = 1 - 0.9887 = 0.0113. Hence Z=- 2.28.
b) 82.12% of the area under the distribution curve lies to the right of it.
Using the negative z table, area = 1 – 0.8212 = 0.1788. Hence Z=- 0.92.

c) 60.64% of the area under the distribution curve lies to the right of it.
Using the negative z table, area = 1 – 0.6064 = 0.3936. Hence Z=- 0.27.
49) Find two z values, one positive and one negative, so that the areas in the two tails total the
following values.
a) 5%
For total area = 0.05, there will be area = 0.025 in each tail. The z scores are 1.96.
b) 10%
For total area =0.10, there will be area = 0.05 in each tail. The z scores are 1.645.

c) 1%
For total area = 0.01, there will be area = 0.005 in each tail. The z scores are z= 2.58.
Section 6-2 # 1, 2, 9, 11, 15, 21, 22, 23, 26, 28, 30
1. Admission Charges for Movies The average admission charge for a movie is $5.81. If the
distribution of movie admission charges is approximately normal with a standard deviation of $0.81, what
is the probability that a randomly selected admission charge is less than $3.50?
X
$3.50 $5.81
z 2.85
$0.81
then look up z = -2.85 in Table E and you get that the area = 0.0022 or 0.22%
Pz ( 2.85) 0.0022 or 0.22%
-2.85 0
2. Teachers’ Salaries The average annual salary for all U.S. teachers is $47,750. Assume that the
distribution is normal and the standard deviation is $5680. Find the probability that a randomly selected
teacher earns
a. Between $35,000 and $45,000 a year
X
35,000 47,750
z 2.24
5680
45,000 47,750
z
0.48
5680
P( 2.24 z 0.62) 0.3156 0.0125 0.3031 or 30.31%

. More than $40,000 a year
X
40,000 47,750
z 1.36
5680
Pz ( 1.36) 1 0.0869 0.9131 or 91.31%
c. If you were applying for a teaching position and were offered $31,000 a year, how would you feel
(based on this information)?
Not too happy! It's really at the bottom of the heap!
X
31,000 47,750
z 2.95
5680
Pz ( 2.95) 0.0016
Only 0.16% of salaries are below $31,000.
9. Miles Driven Annually The mean number of miles driven per vehicle annually in the United States is
12,494 miles. Choose a randomly selected vehicle, and assume the annual mileage is normally distributed
with a standard deviation of 1290 miles. What is the probability that the vehicle was driven more than
15,000 miles? Less than 8000 miles? Would you buy a vehicle if you had been told that it had been driven
less than 6000 miles in the past year?
X
15,000 12,494
For x15,000 miles : z 1.94
1290
Pz ( 1.94) 1 0.9738 0.0262
X
8,000 12,494
For x 8,000 miles : z 3.48
1290
Pz ( 3.48) 0.0003
X
6,000 12,494
For x 6,000 miles : z 5.03
1290
Pz ( 5.03) 0.0001
Maybe it would be good to know why it had only been driven less than 6000
miles.
11. Credit Card Debt The average credit card debt for college seniors is $3262. If the debt is normally
distributed with a standard deviation of $1100, find these probabilities.

a) That the senior owes at least $1000
X 1000 3262 2262
z 2.06
1100 1100
Pz ( 2.06) 1 0.0197 0.9803 or98.03%
-2.06 0
b) That the senior owes more than $4000
X
4000 3262 738
z 0.67
1100 1100
Pz ( 0.67) 1 0.7486 0.2514 or 25.14%
0 0.67
c) That the senior owes between $3000 and $4000
X 3000 3262 262
z 0.24
1100 1100
X
4000 3262 738
z 0.67
1100 1100
P( 0.24 z 0.67) 0.7486 0.4052 0.3434 or 34.34%
-0.24 0.67

15. Waiting to Be Seated The average waiting time to be seated for dinner at a popular restaurant is
23.5 minutes, with a standard deviation of 3.6 minutes. Assume the variable is normally distributed.
When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the
following time.
a) Between 15 and 22 minutes
X 15 23.5 8.5
z 2.36
3.6 3.6
X 22 23.5 1.5
z 0.42
3.6 3.6
P( 2.36 z 0.42) 0.3372 0.0091 0.3281
-2.36 -0.42
b) Less than 18 minutes or more than 25 minutes
X 18 23.5 5.5
z 1.53
3.6 3.6
X
25 23.5 1.5
z 0.42
3.6 3.6
P( z 1.53 or z 0.42) 0.0630 (1 0.6628) 0.063 0.3372 0.4002
-1.53 0.42
c) Is it likely that a person will be seated in less than 15 minutes?
X 15 23.5 8.5
z 2.36
3.6 3.6
Pz ( 2.36) 0.0091

Since the probability is small, it is not likely that a person would be seated in less than 15 minutes.
21. Cost of Personal Computers The average price of a personal computer (PC) is $949. If the
computer prices are approximately normally distributed and $100 , what is the probability that a
randomly selected PC costs more than $1200? The least expensive 10% of personal computers cost less
than what amount?
Let X be the amount a least expensive personal computer can cost.
X
1200 949 251
z 2.51
100 100
Pz ( 2.51) 1 0.9940 0.006 or 0.06%
2.51
For the least expensive 10%, the corresponding z-value is z = - 1.28.
x 949
1.28
100
x 1.28(100) 949 $821
Area is .10
Z= -1.28
22. Reading Improvement Program To help students improve their reading, a school district decides
to implement a reading program. It is to be administered to the bottom 5% of the students in the district,
based on the scores on a reading achievement exam. If the average score for the students in the district is
122.6, find the cutoff score that will make a student eligible for the program. The standard deviation is 18.
Assume the variable is normally distributed.
122.6, 18
The bottom 5% (area) is in the left tail of the normal curve. The corresponding z score is found using area = 0.05.
Thus, z= -1.645.
x z
x 1.64518 122.6 92.99 or 93

Therefore, a cutoff score of 93 will make a student eligible for the program.
23. Used Car Prices an automobile dealer finds that the average price of a previously owned vehicle is
$8256. He decides to sell cars that will appeal to the middle 60% of the market in terms of price. Find the
maximum and minimum prices of the cars the dealer will sell. The standard deviation is $1150, and the
variable is normally distributed.
$8256, $1150 . The middle 60% means that 30% of the area will be on either side of the mean.
The corresponding z score is found using area = 0.20.
The z scores are 0.84.
Min z * 0.84*1150 8256 $7290
Max z * 0.84*1150 8256 $9222
26. High School Competency Test a mandatory competency test for high school sophomores has a
normal distribution with a mean of 400 and a standard deviation of 100.
a) The top 3% of students receive $500. What is the minimum score you would need to receive this
award?
400, 100
10.03 0.97
z 1.88
Min z * 1.88*100 400 588
588 is minimum score to receive the award.
b) The bottom 1.5% of students must go to summer school. What is the minimum score you would need
to stay out of this group?
400, 100
For the bottom 1.5%, the area is 0.015.
z 2.17
Min z * 2.17*100 400 183
183 is the minimum score to avoid summer school.

28. Bottled Drinking Water Americans drank an average of 23.2 gallons of bottled water per capita in
2004. If the standard deviation is 2.7 gallons and the variable is normally distributed, find the probability
that a randomly selected American drank more than 25 gallons of bottled water. What is the probability
that the selected person drank between 18 and 26 gallons?
23.2, 2.7
For x 25 gallons:
25 23.2
z 0.67; the corresponding area=0.7486
2.7
Pz ( 0.67) 1 0.7486 0.2514
0.67
For 18 x 26gallons:
18 23.2
z 1.93; the corresponding area=0.0268
2.7
26 23.2
z 1.04; the corresponding area=0.8508
2.7
P(18 x1.04) 0.8508 0.0268 = 0.824
-1.93 1.04
30. Security Officer Stress Tolerance to qualify for security officers’ training, recruits is tested for
stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8. If
only the top 15% of recruits are selected, find the cutoff score.
62, 8
10.15 0.85
The corresponding z score is 1.04
Cutoff z
1.048 62 70.32
6.3 #9, 13, 15, 17, 21, 23
9. College Costs The mean undergraduate cost for tuition, fees, room, and board for four-year
institutions was $26,489 for the 2004–2005 academic year. Suppose that σ =$3204 and that 36 four-year
institutions are randomly selected. Find the probability that the sample mean cost for these 36 schools is
a. Less than $25,000
µ=26,489, n=36, σ= 3204, P( x < 25.000)

x x 25,000 26,489
Z
2.79
3204
n 36
P( z < -2.79)=0.0026
-2.97
b. Greater than $26,000
µ=26,489, n=36, σ= 3204, P( x >26.000)
Z
x
x 26,000 26,489
0.92
3204
n 36
-0.92
P( z > -0.92)=1-0.1788 = 0.8212
c. Between $24,000 and $26,000.
µ=26,489, n=36, σ= 3204, P($24,000 < x

x x
Z
x
x 20 21
1.72
2.9
n 25
P(z

Pz ( 0.90) 1 0.8159 0.1841 18.41%
c) Why is the probability for part a greater than that for part b?
Because the individual values are more variable than means.
17. Water Use The Old Farmer’s Almanac reports that the average person uses 123 gallons of water
daily. If the standard deviation is 21 gallons, find the probability that the mean of a randomly selected
sample of 15 people will be between 120 and 126 gallons. Assume the variable is normally distributed.
µ=123, n=15, σ= 21, normally distributed
P(120< x

43
46.2 3.2
3.2
z 2.83
8 8 1.13
50 7.07
23. Cholesterol Content The average cholesterol content of a certain brand of eggs is 215 milligrams,
and the standard deviation is 15 milligrams. Assume the variable is normally distributed.
a. If a single egg is selected, find the probability that the cholesterol content will be greater than 220
milligrams.
µ=215, σ= 15, P( x >220), normally distributed
Z
220 215
15
5
0.33
15
Pz ( 0.33) 1 0.6293 0.3707 or 37.07%
b. If a sample of 25 eggs is selected, find the probability that the mean of the sample will be larger than
220 milligrams.
µ=215, n=25, σ= 15, P( x >220)
220 215
z
15
25
5
15
5
5
1.67
3
Pz ( 1.67) 1 0.9525 0.0475 or 4.75%
Ch 6.4 p.344 #3a, b, c, d, 7, 8, 11
3. Check each binomial distribution to see whether it can be approximated by a normal distribution (i.e.,
are np ≥ 5 and nq ≥ 5?).
a. n=20, p=0.5
q = 1–q = 1 – 0.5 = 0.5
np = (20)(0.5) =10
nq = (20)(0.5) = 10
Yes, because np ≥ 5 and nq ≥ 5.
b. n=10, p=0.6

q = 1–q = 1 – 0.6 = 0.4
np = (10)(0.6) =6
nq = (10)(0.4) = 4
No because np ≥ 5 and nq ≤ 5.
c. n=40, p=0.9
q = 1–q = 1 – 0.9 = 0.1
np = (40)(0.9) =36
nq = (40)(0.1) = 4
No because np ≥ 5 and nq ≥ 5.
d. n=50, p=0.2
q = 1–q = 1 – 0.2 = 0.8
np = (50)(0.2) =10
nq = (50)(0.8) = 40
Yes, because np ≥ 5 and nq ≥ 5..
7. Percentage of Americans Who Have Some College Education The percentage of Americans 25
years or older who have at least some college education is 53.1%. In a random sample of 300 Americans
25 years old or older, what is the probability that more than 175 have at least some college education?
n=300, p=0.531
q = 1–q = 1 – 0.531 = 0.469
Since np = (300)(0.531) = 159.3 and nq = (300)(0.469) = 140.7, the normal distribution can be used to
approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np = (300)( 0.531) = 159.3
σ = npq = (300)(0.531)(0.469) 8.64
For binomial distribution, P(X >175)
For normal distribution, P(X >175 + 0.5) = P(X > 175.5)

153.9 175.5
The z value is z = 175.5 159.3 1.88
8.64
From table E (Normal Distribution ), z=1.88 → 0.9699
1 – 0.9699 = 0.0301 or 3.01%
8. Household Computers According to recent surveys, 60% of households have personal computers. If a
random sample of 180 households is selected, what is the probability that more than 60 but fewer than
100 have a personal computer?
n=180, p=0.60
q = 1–q = 1 – 0.60 = 0.40
Since np = (180)(0.60) = 108 and nq = (180)(0.40) = 72, the normal distribution can be used to
approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np = (180)(0.60) = 108
σ = npq = (180)(0.60)(0.40) 6.57
For binomial distribution, P(60 x100)
For normal distribution, P(60.5 < X < 99.5)
= P(60.5 < X < 99.5)
z = 99.5 108 1.29 area=0.0985
6.57
z = 60.5 108 7.23 area=0.0001
6.57
P(60.5 < X < 99.5)= P(-7.23 < z< -1.29)=0.0985-0.0001=0.0984 or 9.84%
0.4955 + 0.4945 = 0.99 or 0.99%

11. Elementary School Teachers Women comprise 80.3% of all elementary school teachers. In a
random sample of 300 elementary teachers, what is the probability that more than three-fourths are
women?
n=300, p=0.803
q = 1–q = 1 – 0.803= 0.197
Since np = (300)(0.803) = 240.9 and nq = (300)(0.197) = 59.1, the normal distribution can be used to
approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np= (300)(0.803) = 240.9
σ = npq = (300)(0.803)(0.197) 6.89
For binomial distribution, P(X >225) ¾ (300)=225
For normal distribution, P(X >255 + 0.5) = P(X > 255.5)
240.9 255.5
The z value is z = 225.5 240.9 2.24 area=0.0125
6.89
P(z > 225.5)=P(X > -2.24)=1- 0.0125 =0.9875 or 98.75%