SPA 3e_ Teachers Edition _ Ch 6

L E S S O N 6.6 • The Central Limit Theorem 443
Lesson 6.6
18/08/16 5:04 PMStarnes_3e_CH06_398-449_Final.indd 443
WhAT DiD y o U LeA rn?
LEARNINg TARgET EXAMPLES EXERCISES
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
Mastering Concepts and Skills
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
x for SRSs of size n 5 100 from the population 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
x for SRSs of size n 5 15 from the population of
homeowners. Justify your answer.
(b) Describe the shape of the sampling distribution of
x for SRSs of size n 5 1000 from the population of
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.
(b) Describe the shape of the sampling distribution of
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
TRM full Solutions to 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
this resource on the TRFD.
Answers to Lesson 6.6 Exercises
1. (a) Because n = 5 < 30, the sampling
distribution of x will also be skewed to the right
but not quite as strongly as the population.
(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.
(b) Describe the shape of the sampling distribution of
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?
Applying the Concepts
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
but not quite as strongly as the population.
(b) Because n = 1000 ≥ 30, the sampling
distribution of x is approximately normal by
the central limit theorem.
3. (a) No; a count only takes on wholenumber
values, so it cannot be normally
distributed.
(b) Because n = 100 ≥ 30, the sampling
distribution of x is approximately normal by
the central limit theorem.
4. (a) No; a count only takes on wholenumber
values, so it cannot be normally
distributed.
(b) Because n = 60 ≥ 30, the sampling
distribution of x is approximately normal
by the central limit theorem.
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
Using technology: Applet/normalcdf
(lower:21000, upper:240, mean:225,
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
normal by the central limit theorem.
300 − 250
z =
158.11 = 0.32;
P( x < 300) = P( Z < 0.32) = 0.6255
Using technology: Applet/normalcdf
(lower:21000, upper:300, mean:250,
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
Shape: Because n = 100 ≥ 30, the sampling
distribution of x is approximately
normal by the central limit theorem.
1.7 − 1.5
z = = 2.67; P( x > 1.7) =
0.075
P( Z > 2.67) = 1− 0.9962 = 0.0038
Using technology: Applet/normalcdf
(lower:1.7, upper:1000, mean:1.5,
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
Shape: Because n = 60 ≥ 30, the sampling
distribution of x is approximately
normal by the central limit theorem.
1.7 − 1.6
z = = 0.65; P( x > 1.7) =
0.155
P( Z > 0.65) = 1− 0.7422 = 0.2578
Using technology: Applet/normalcdf
(lower:1.7, upper:1000, mean:1.6,
SD:0.155) 5 0.2594
Lesson 6.6
L E S S O N 6.6 • The Central Limit Theorem 443
Starnes_3e_ATE_CH06_398-449_v3.indd 443
11/01/17 3:58 PM