Sampling Distribution HW Answer Key

Name _______________________________
Period _____________
Chapter 9 Sampling Distributions Homework
1.
2.
3.
4.
5.
Identify each boldface number as a parameter or a statistic and use the appropriate notation.
A carload lot of ball bearings has a mean diameter of 2.5003 cm. An inspector chooses 100 bearings
from the lot that and finds they have a mean diameter of 2.5009 cm.
𝜇𝜇 = 2.5003
𝑥𝑥̅ = 2.5006
population parameter
sample statistic
𝑝𝑝̂ = .072
sample statistic
The Bureau of Labor Statistics last month interviewed 60,000 members of the U.S. labor force, of
whom 7.2% were unemployed.
A telemarketing firm in Los Angeles uses a device that dials residential telephone numbers in that
city at random. Of the first 100 numbers dialed, 48% are unlisted. This is not surprising because 52%
of all Los Angeles residential phones are unlisted.
𝑝𝑝̂ = .48
𝑝𝑝 = .52
sample statistic
population parameter
𝑥𝑥̅ = 335
𝑥𝑥̅ = 289
sample statistic
sample statistic
A researcher carries out a randomized comparative experiment with young rats. After 8 weeks, the
mean weight gain is 335 grams for the control group and 289 grams for the experimental group.
The figure below shows histograms of four sampling distributions of statistics intended to estimate
the same parameter. Label each distribution relative to the others as having large or small bias and as
having large or small variability.
Large bias; large variability
Small bias; large variability
Small bias; small variability
Large bias; small variability
6.
The IRS plans to examine an SRS of individual federal income tax returns from each state. One
variable of interest is the proportion of returns claiming itemized deductions. The total number of
tax returns in each state varies from almost 14 million in California to fewer than 210,000 in
Wyoming.
(a) Will the sampling variability of the sample proportion change from state to state if an SRS of
2000 tax returns is selected from each state? Explain.
10 ∗ 2000 = 20000 NO, the 10% condition has not been met; both populations are greater
than 20000 people.
(b) Will the sampling variability of the sample proportion change from state to state if an SRS of 1%
of all tax returns is selected in each state? Explain.
7.
. 01 ∗ 14000000 = 140000 𝑎𝑎𝑎𝑎𝑎𝑎 .01 ∗ 210000 = 2100 Yes. A larger sample (140000) will
have less variability.
A USA Today poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl
after they have eaten the cereal. Of the respondents, 67% said that they drink it. Suppose that
70% of U.S. adults actually drink the cereal milk.
(a) Find the mean and the standard deviation of the proportion 𝑝𝑝̂ of the sample that say they drink
the cereal milk.
𝜇𝜇𝑝𝑝� = .7
.7∗.3
𝜎𝜎𝑝𝑝� = �1012
= .014405
(b) Explain why you can use the formula for the standard deviation of 𝑝𝑝̂ in this setting (rule of thumb
#1).
10 ∗ 1012 = 10120 The 10% condition has been met; there are more than 10120 US adults.
(c) Check that you can use the normal approximation for the distribution of 𝑝𝑝̂ (rule of thumb #2).
1012 ∗ .7 ≥ 10, 708.4 ≥ 10
1012 ∗ .3 ≥ 10, 303.6 ≥ 10
(d) Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say the drink the
cereal milk. Do you have any doubts about the result of the poll?
normCdf(-9E999, .67, .7, .014405) = .018643. There is a very small chance (< 2%) that 67%
or fewer drink the milk.
(e) What sample size would be required to reduce the standard deviation of the sample proportion to
one-half the value you found in part (a)?
8.
Four times the sample in (a); 4 ∗ 1012 = 4048 adults in the sample
The Gallop Poll asked a probability sample of 1785 adults whether they attended church or synagogue
during the past week. Suppose that 40% of the adult population did attend. We would like to know the
probability that an SRS of size 1785 would come within plus or minus 3 percentage points of this true
value.
� is the proportion of the sample who did attend church or synagogue, what is the mean of the
(a) If 𝑝𝑝
�? What is the standard deviation?
sampling distribution of 𝑝𝑝
.4∗.6
𝜇𝜇𝑝𝑝� = .4
𝜎𝜎𝑝𝑝� = �1785 = .011595
(b) Explain why you can use the formula for the standard deviation of 𝑝𝑝̂ in this setting (rule of thumb
#1).
10 ∗ 1785 = 17850 The 10% condition has been met; there are more than 17850 adults.
� (rule of thumb #2).
(c) Check that you can use the normal approximation for the distribution of 𝑝𝑝
1785 ∗ .4 ≥ 10, 714 ≥ 10
1785 ∗ .6 ≥ 10, 1071 ≥ 10
(d) Find the probability that 𝑝𝑝̂ takes a value between 0.37 and 0.43. Will an SRS of size 1785 usually
� within plus or minus 3 percentage points of the true population proportion? Explain.
give a result 𝑝𝑝
9.
normCdf(.37, .43, .4, .011595) = .990327. Yes, over 99% of all samples should give 𝑝𝑝̂ within
±3% of the true population proportion.
Problem #8 asks the probability that 𝑝𝑝̂ from an SRS estimates p = 0.4 within 3 percentage points. Find this
probability for SRSs of size 300, 1200, and 4800. What general fact do your results illustrate?
For n = 300
For n = 1200
For n = 4800
𝜎𝜎𝑝𝑝� = �.4∗.6
= .0283
300
.4∗.6
𝜎𝜎𝑝𝑝� = �1200
= .0141
.4∗.6
𝜎𝜎𝑝𝑝� = �4800
= .00707
normCdf(.37, .43, .4, .0283)
𝑃𝑃(. 37 ≤ 𝑝𝑝̂ ≤ .43) = .7109
normCdf(.37, .43, .4, .0141)
𝑃𝑃(. 37 ≤ 𝑝𝑝̂ ≤ .43) = .9666
normCdf(.37, .43, .4, .00707)
𝑃𝑃(. 37 ≤ 𝑝𝑝̂ ≤ .43) = .99998
For larger sample sizes, the sample proportions are more likely to be close to the true population proportion.
10.
Your mail-order company advertises that it ships 90% of its orders within three working days. You
select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals
that 86 of these orders were shipped on time.
(a) What is the sample proportion of orders shipped on time?
86
𝑝𝑝̂ = 100 = .86
(b) If the company really ships 90% of its orders on time, what is the probability that the proportion
in an SRS of 100 orders is as small as the proportion in your sample or smaller?
10 ∗ 100 = 1000 The 10% condition has been met; there are more than 1000 orders.
100 ∗ .9 ≥ 10, 90 ≥ 10 ; 100 ∗ .1 ≥ 10, 10 ≥ 10; 𝜇𝜇𝑝𝑝� = .9
11.
𝜎𝜎𝑝𝑝� = �.9∗.1
= .03
100
normCdf(-9E999, .86, .9, .03) = .091211.
𝑃𝑃(𝑝𝑝̂ ≤ .86) = .091211
It happens about 9% of the time, a small probability, but not unlikely.
Explain why you can not use the methods you learned in 9.2 to find the following probabilities:
(a) A factory employs 3000 unionized workers, of whom 30% are Hispanic. The 15-member union
executive committee contains 3 Hispanics. What would be the probability of 3 or fewer Hispanics
if the executive committee were chosen at random from all the workers?
15 ∗ .3 ≥ 10, 4.5 ≥ 10
Not Normal, try binomial probability
(b) A university is concerned about the academic standing of its intercollegiate athletes. A study
committee chooses an SRS of 50 of the 316 athletes to interview in detail. Suppose that in fact
40% of the athletes have been told by coaches to neglect their studies on at least one occasion.
What is the probability that at least 15 in the sample are among this group?
10 ∗ 50 = 500 The 10% condition has not been met; there are only 316 athletes.
(c) How could you find the probability for (a)? Find it.
12.
𝑃𝑃(𝑥𝑥 ≤ 3) binomialCdf(15, .3, 0, 3) = .296868
Taxi fares are normally distributed with mean fare $22.27 and standard deviation $2.20.
a)
Which should have the greater probability of falling between $21 and $24 – the mean of 10
random taxi fares or the amount of a single random taxi fare? Why?
b)
The mean of 10 random taxi fares. Larger sample size has less variability; more likely to be closer
to the true mean
Which should have a greater probability of being greater than $24 – the mean of 10 random
taxi fares or the amount of a single taxi fare? Why?
A single taxi fare. It is more likely that 1 value will be farther away from the mean as opposed to
the mean of a sample of 10 taxi fares..
13.
Suppose a sample of n = 50 items is drawn from a population of manufactured products and the
weight, x, of each item is recorded. Prior experience has shown that the weight has a mean of 6
ounces and standard deviation of 2.5 ounces.
a)
What is the shape of the sampling distribution of x ?
The distribution will be approximately Normal due to the large sample size.
b)
c)
d)
What is the mean and standard deviation of the sampling distribution?
𝜇𝜇𝑥𝑥̅ = 6 ; 𝜎𝜎𝑥𝑥̅ =
2.5
√50
= .353553; 10(50) = 500; There are more than 500 manufactured products
What is the probability that the manufacturer’s sample has a mean weight of less than 5
ounces? If you did get a sample mean of 5 or less, does this indicate that the manufacturing
process may be faulty? Explain.
Use CLT n = 50 > 30. normCdf(-9E999, 5, 6, .353553); 𝑃𝑃(𝑥𝑥̅ ≤ 5) = .002339. This will
happen less than 1% of the time, the manufacturing process is most likely faulty.
How would the sampling distribution of x change if the sample size, n, were increased from 50
to 100?
2.5
𝜎𝜎𝑥𝑥̅ =
√100
= .25; It would have less variability; the sampling distribution would be
closer to the mean.
14.
A soft-drink bottle vendor claims that its production process yields bottles with a mean internal
strength of 157 psi (pounds per square inch) and a standard deviation of 3 psi and is normally
distributed. As part of its vendor surveillance, a bottler strikes an agreement with the vendor that
permits the bottler to sample from the vendor’s production to verify the vendor’s claim.
N(157, 3)
a)
b)
c)
Suppose the bottler randomly selects a single bottle to sample. What is the mean and
3
standard deviation?
𝜇𝜇𝑥𝑥̅ = 157
; 𝜎𝜎𝑥𝑥̅ =
√1
=3
What is the probability that the psi of the single bottle is 1.3 psi or more below the process
mean?
normCdf(-9E999, 155.7, 157, 3); 𝑃𝑃(𝑥𝑥 ≤ 155.7) = .332386.
Suppose the bottler randomly selected 40 bottles from the last 10,000 produced. Describe
the sampling distribution of the sample means?
10(40) = 400 and we are looking at the last 10,000.
N(157, .474342)
d)
e)
𝜇𝜇𝑥𝑥̅ = 157
; 𝜎𝜎𝑥𝑥̅ =
3
√40
= .474342
What is the probability that the sample mean of the 40 bottles is 1.3 psi or more below the
process mean?
normCdf(-9E999, 155.7, 157, .474342); 𝑃𝑃(𝑥𝑥̅ ≤ 155.7) = .003066.
In order to reduce the standard deviation 50% (half), how large would the sample size need to
be?
4 times as large; 4(40) = 160. 𝜎𝜎𝑥𝑥̅ =
3
√160
= .237171
15.
The distribution of weights of chips in these bags is normally distributed with mean 10 ounces and
standard deviation 0.12 ounces.
N(10, .12)
a)
b)
If a bag of chips is selected at random, what is the probability that it weighs less than 9.95
ounces?
normCdf(-9E999, 9.95, 10, .12); 𝑃𝑃(𝑥𝑥 ≤ 9.95) = .338461.
Describe the sampling distribution of sample means for samples of size 12 (n = 12).
10(12) = 120; there are more than 120 bags of chips. 𝜇𝜇𝑥𝑥̅ = 10; 𝜎𝜎𝑥𝑥̅ =
N(10, .034641)
c)
d)
normCdf(-9E999, 9.95, 10, .034641); 𝑃𝑃(𝑥𝑥̅ ≤ 9.95) = .074457.
For random samples of 5 bags, what mean weight is at the 60th percentile?
invNorm(.6, 10, .053666)
𝑥𝑥̅ = 10.0136
.12
√5
= .053666
You randomly select 20 mortgage institutions and determine the current mortgage rate at each. The
current mean mortgage interest rate is known to be 6.93% with a standard deviation of 0.42% and is
known to be approximately normal.
a)
b)
c)
17.
= .034641
What’s the probability that the mean weight of a random sample of 12 bags weighs less than
9.95 ounces?
10(5) = 50; there are more than 50 bags of chips. 𝜇𝜇𝑥𝑥̅ = 10; 𝜎𝜎𝑥𝑥̅ =
16.
.12
√12
Construct the sampling distribution for the sample of 20 mortgage institutions.
10(20) = 200; there are more than 200 mortgage institutions. 𝜇𝜇𝑥𝑥̅ = 6.93;
.42
𝜎𝜎𝑥𝑥̅ =
= .093915
N(6.93, .093915) or N(.0693, .00093915)
√20
What is the probability that a sample of mortgage rates is above 7%?
normCdf(7, 9E999, 6.93, .09315); 𝑃𝑃(𝑥𝑥̅ ≥ 7) = .228029.
What is the probability that a sample of mortgage rates is between 6% and 6.5%?
normCdf(6, 6.5, 6.93, .09315); 𝑃𝑃(6 < 𝑥𝑥̅ < 6.5) = .000002.
You randomly select 50 newly constructed houses in a particular area of Oregon. The mean
construction cost in this area is $181,000 with a standard deviation of $28,000.
a)
b)
c)
Construct a sampling distribution for a sample of size 50.
10(50) = 500; there are more than 500 newly constructed home in a particular area of
28000
Oregon. 𝜇𝜇𝑥𝑥̅ = 181000; 𝜎𝜎𝑥𝑥̅ =
= 3959.8
N(181000, 3959.8)
√50
What is the probability that a sample of new houses averages less than $170,000?
normCdf(-9E999, 170000, 181000, 3959.8); 𝑃𝑃(𝑥𝑥̅ ≤ 170000) = .002735.
What is the minimum cost of a house in the 80th percentile?
Can’t do the problem. n = 1 and it is not stated that the population is
approximately normally distributed.
18.
You randomly select 18 adult male athletes and measure the resting heart rate of each. The mean
heart rate is 64 beats per minute with a standard deviation of 2.5 beats per minute. It is safe to
assume the resting heart rates are approximately normal.
a)
b)
c)
What range of resting heart beats mark the middle 95% of all the heart rates?
Within 2𝜎𝜎;
64 ± 2(2.5)
(59, 69)
What is the probability that an adult male athlete has resting heart rate below 60 beats per
minute?
normCdf(-9E999, 60, 64, 2.5); 𝑃𝑃(𝑥𝑥 < 60) = .054799.
Would you consider it unusual if your sample of athletes had a resting heart rate above 65.4
beats per minute?
10(18) = 180; there are more than 180 adult male athletes. 𝜇𝜇𝑥𝑥̅ = 64;
2.5
𝜎𝜎𝑥𝑥̅ =
= .589256
N(64, .589256)
√18
19.
normCdf(65.4, 9E999, 64, .589256); 𝑃𝑃(𝑥𝑥̅ ≥ 65.4) = .008754.
Yes because this happens less than 1% of the time.
A football filled with helium was kicked on a windless day at Katy High School. The distances (in
yards) are listed below. The average distance kicked of helium footballs is known to be 27.5 yards
with a standard deviation of 6.3 yards. (Note: you aren’t told that the population distribution is
normal, and the n < 30, you can infer normality from sample data. Check graphically!).
11
29
a)
b)
12
30
14
30
22
30
23
30
24
31
26
31
26
31
26
32
27
32
28
33
28
34
29
35
29
39
29
What is the probability that a sample this size will have a mean distance of more than 30
yards?
What is the probability that a sample this size will have a mean distance between 25 and 30
yards?
The histogram does not appear to be Normally distributed. The boxplot shows
outliers on both sides. We can’t use Normal calculations for this distribution.
20.
Fourteen randomly selected people with a bachelor’s degree in economics were asked their monthly
salary. According to the U.S. Bureau of Labor Statistics, the average monthly salary is approximately
$4418 with a standard deviation of $389. The monthly salaries are listed below:
4450.66
4527.64
4596.73
4407.34
4366.66
5036.64
Check normality graphically!
a)
b)
4455.40
5083.73
4151.70
3946.47
3727.08
4023.41
4283.76
4806.22
The histogram, boxplot and NPP do not have unusual features;
we can use Normal calculations for this distribution.
What is the probability that a sample of 14 economic graduates will have an average monthly
salary of over $4600?
10(14) = 140; there are more than 140 economic graduates. 𝜇𝜇𝑥𝑥̅ = 4418;
389
𝜎𝜎𝑥𝑥̅ =
= 103.965
N(4418, 103.965)
√14
normCdf(4600, 9E999, 4418, 103.965); 𝑃𝑃(𝑥𝑥̅ > 4600) = .040008.
How would the sampling distribution’s standard deviation change if the sample size is
quadrupled? Be specific.
𝑊𝑊ℎ𝑒𝑒𝑒𝑒 𝑛𝑛 = 14, 𝜎𝜎𝑥𝑥̅ =
389
√14
√4 = 2, 𝑠𝑠𝑠𝑠 𝑖𝑖𝑖𝑖 𝑤𝑤𝑤𝑤𝑤𝑤𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 ℎ𝑎𝑎𝑎𝑎𝑎𝑎.
389
= 103.965, 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑛𝑛 = 56, 𝜎𝜎𝑥𝑥̅ =
= 51.9823
√56
21.
According to the EPA, the mean waste generated per person per day was 4.5 pounds with a standard
deviation of 1.2 pounds.
a)
What is the probability that a sample of 20 people would have a mean waste of more than 5.3
pounds per day?
n = 20 we don’t have the data and we were not told anything about the population so
we can’t verify Normality.
b)
What is the probability that a sample of 45 people would have a mean waste of more than 5
pounds per day?
n = 45 > 30, Central Limit Theorem can be used to verify Normality.
10(45) = 450; there are more than 450 people
1.2
𝜇𝜇𝑥𝑥̅ = 4.5;
𝜎𝜎𝑥𝑥̅ =
= .178885
N(4.5, .178885)
√45
normCdf(5, 9E999, 4.5, .178885); 𝑃𝑃(𝑥𝑥̅ > 5) = .002594.