SPA 3e_ Teachers Edition _ Ch 6

438
C H A P T E R 6 • Sampling Distributions
9. (a) m x = m = 48 months;
s x = s !n = 8.2 = 2.90 months
!8
(b) In SRSs of size n 5 8, the sample mean
battery life will typically vary by about 2.90
months from the true mean of 48 months.
(c) The sampling distribution is normal
because the population distribution is
normal.
42.2 − 48
z = = −2.00; P( x < 42.2) =
2.90
P( Z < −2.00) = 0.0228
Using technology: Applet/normalcdf
(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
(b) In SRSs of size n 5 15, the sample mean
birth weight will typically vary by about
0.129 kg from the true mean of 3.4 kg.
(c) The sampling distribution is normal
because the population distribution is
normal.
3.55 − 3.4
z = = 1.16; P( x > 3.55) =
0.129
P( Z > 1.16) = 1− 0.8770 = 0.1230
Using technology: Applet/normalcdf
(lower:3.55, upper:1000, mean:3.4,
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
Using technology: Applet/normalcdf
(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).
(a) Calculate the mean and standard deviation of the
sampling distribution of x for SRSs of size 8.
(b) Interpret the standard deviation from part (a).
(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.
(b) Interpret the standard deviation from part (a).
(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?
Starnes_3e_CH06_398-449_Final.indd 438
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
Using technology: Applet/normalcdf(lower:297,
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?
Extending the Concepts
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.
Recycle and Review
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
Shape: Normal because the population
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
Using technology: Applet/normalcdf(lower:
3.25, upper:1000, mean:3, SD:0.11) 5 0.0115
Answers 18–19 are on page 439
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