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AP Statistics – Chapter 9 – Sampling Distributions – Study Guide
1. The candy company claims that 10% of the M&M’s it produces are green. Suppose that the candies are
packaged at random in small bags containing about 50 M&M’s. A class of elementary school students
learning about percents opens several bags, counts the various colors of candies, and calculates the
proportion that are green.
a. If we plot a histogram showing the proportions of green candies in various bags, what shape
would you expect it to have?
b. Where should the center of the histogram be?
c. Can that histogram be approximated by a Normal model? Explain.
d. What should the standard deviation of the proportion be?
e. How many green M&M’s would you expect to find?
f. What is the standard deviation of the green M&M’s
g. What is the probability that 10 of the M&M’s are green?
h. What is the probability that less than 5 of the 50 M&M’s are green?
2. In a really large bag of M&M’s the students found 500 candies, and 12% of them were green. Is this an
unusually large proportion of green M&M’s? Explain.
3. Based on past experience, a bank believes that 7% of the people who receive loans will not make
payments on time. The bank has recently approved 200 loans.
a. What are the mean and standard deviation of the proportion of clients in this group who may not
make timely payments?
b. What assumptions underlie your model? What are the conditions met? Explain.
c. What’s the probability that over 10% of these clients will not make timely payments?
4. A sample is chosen randomly from a population that is strongly skewed to the left.
a. Describe the sampling distribution model for the sample mean if the sample size is small.
b. If we make the sample larger, what happens to the sampling dist. model’s shape, center, spread?
c. As we make the sample larger, what happens to the expected distribution of other data in the
sample?
5. Assume the duration of human pregnancies can be described by a Normal model with mean 266 days
and standard deviation 16 days.
a. What percentages of pregnancies should last between 270 and 280 days?
b. At least how many days should the longest 25% of all pregnancies last?
c. Suppose a certain obstetrician is currently prenatal care to 60 pregnant women. Let x represent
the mean length of their pregnancies. According to the CLT, what’s the distribution of the
sample mean, x ? Specify the shape, mean, and standard deviation.
d. What’s the probability that the mean duration of these patients’ pregnancies will be less than 260
days?
e. What’s the probability that the mean duration of these patients’ (the 60 women in part d)
pregnancies will be more than 270 days?
f. If the correct model is in fact skewed to the left, does that change your answers to parts, a, b, and
c ? Explain why or why not for each.
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6. Suppose that 47% of all adult women think they do not get enough time for themselves. An opinion poll
interviews 1025 randomly chosen women and records the sample proportions who feel they don’t get
enough time for themselves. Show your work.
a. Describe the sampling distribution of p . Explain why it is appropriate to use a Normal model.
b. In general, how would this model change if the poll interviewed four times as many people?
Specific?
c. In what range will the middle 95% of all sample result fall?
d. What is the probability that the poll gets a sample in which fewer than 45% say they do not get
enough time for themselves? Show your method.
e. What is the probability that the poll gets a sample that is between 45% and 50%?
7. 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 of 100 and standard
deviation 15. Show your work.
a. What is the probability that a randomly chosen individual has a WAIS score of 105 or higher? .
b. What are the mean and standard deviation of the sampling distribution of the average WAIS
score for an SRS of 60 people?
c. What is the probability that the average WAIS score of an SRS of 60 people is 105 or higher?
d. What is the probability that the average WAIS of 100 people is less than 95?
8. About 8% of males are colorblind. A researcher needs some colorblind subjects for an experiment and
begins checking potential subjects.
a. On average, how many men should the researcher expect to check to find one who is colorblind?
b. What’s the probability that she won’t find anyone colorblind among the first 4 men she checks?
c. What’s the probability that the first colorblind man found will be the sixth person checked?
d. What’s the probability that she finds someone who is colorblind before checking the 10th man?
9. The College Board reported the score distribution shown in the table for all students who took the 2004
AP Statistics Exam.
Score
% of Students
1
20.4
2
19.8
3
24.8
4
22.5
5
12.5
a. Find the mean and standard deviation of the scores.
b. If we select a random sample of 40 AP Statistics students, would you expect their scores to
follow a Normal model? Explain.
c. Consider the mean scores of random samples of 40 AP Statistics students. Describe the
sampling model for these means (shape, center, spread).
d. Find the probability that the average of 40 AP Statistics students score a 3 or higher?
10. Suppose that IQs of East State University’s students can be described by a Normal model with mean 130
and standard deviation 8 points. Also suppose that IQs of students from West State University can be
described by a Normal model with mean 120 and standard deviation 10.
a. We select 1 student at random from East State. Find the probability that this student’s IQ is at
least 125 points?
b. We select 1 student at random from each school. Find the probability that the East State
student’s IQ is at least 5 points higher than the West State student’s IQ.
c. We select 3 West State students at random. Find the probability that the this group’s average IQ
is at least 125 points.
d. We also select 3 East State students at random. What’s the probability that their average IQ is at
least 5 points higher than the average for the 3 West State students?
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AP Statistics – Chapter 9 – Sampling Distributions – Study Guide – Answers
1a. Skewed right.
1b. p = .10
1c. Not normal because np  50(.1)  5  10 .
pq
(.1)(.9)
1e.  x  np  50(.1)  5 1f.  x  npq  50(.1)(.9)  212

.042
.
n
50
50
1g. P( x  10)  ( )(.1)10 (.9) 40 .0152
1h. P( x  5)  P( x  1) ... P( x  4) .4312
10
2. For 500 candies, the sampling distribution model is Normal with  p  p .1 and
1d.  p 
pq
(.1)(.9)

.01342 . The sample proportion of p .12 is about 1.49 standard deviations above the
n
500
expected proportion, which is not at all unusual. According to the Normal model, we expect sample proportions
this high or higher about 6.8% of the time.
pq
(.07)(.93)
3a.  p  p .07
 p 

.018
n
200
3b. The sample size is large ( n  200 ). Each bank loan is independent. N  10n  10(200)  2000 . The
sampling distribution is approximately Normal because np  200(.07)  14  10 and nq  200(.93)  186  10 .
.1.07
)  P( z  1663
. ) .048
3c. P( p .10)  P( z 
(.07)(.93)
200
4a. The sampling distribution model for the sample mean will be skewed to the left as well, centered
 p 
 x   x and standard deviation of  x 
x
.
n
4b. When the sample size is increased, the sampling distribution model becomes more Normal in shape,
centered  x   x and standard deviation of  x 
x
.
n
4c. As we make the sample larger the distribution of data in the sample should more closely resemble the shape,
center, and spread of the population.
5a. P(270  x  280)  P(.25  z .875) .8106.5987 .212
x  266
.67 
5b. z .67
x  276.8 days
16
5c. The sample size is large ( n  60 ). Each pregnancy is independent. N  10n  10(60)  600 . The sampling
distribution will be normal because the population data is normal with a mean of  x   x  266 days and

16
standard deviation of  x  x 
 2.07
n
60
5d. P( x  260)  P( z  2.899) .002 According to the Normal model, the mean pregnancy for 60 women
lasting less than 260 days is .002.
270  266
)  P( z  194
. )  1.9738 .0262
5e. P( x  270)  P( z 
16
60
5f. We can no longer answer the questions posed in a and b. The Normal model is not appropriate for skewed
distributions. The answer to part c is still valid. The CLT guarantees that the sampling distribution model is
Normal when the sample size is large.
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6a. The sample size is larger ( n  1025 ). Each interview should be independent of each other.
N  10n  10(1025)  10,250 . The shape of the distribution should be approximately normal because
np  1025(.47)  48175
.  10 and nq  1025(.53)  543.25  10 . The center (mean) of the distribution is
pq
(.47)(.53)

.0156 .
n
1025
6b. General: The distribution would become more normal, and the standard deviation would get smaller.
Specific: The standard deviation would be cut in half.  .0078
6c. 95% of the data will fall between two standard deviations ( 2(.0156) .0312 ) or from .4388 to .5012.
. ) .1003
6d. P( p .45)  P( z  128
.  z  192
. ) .9726.1003 .8723
6e. P(.45  p .50)  P( 128

15
7a. P( x  105)  P( z  0.33) .3707
7b.  x   x  100
x  x 
 19365
.
n
60
95  100
7c. P( x  105)  P( z  2.582) .0049
7d. P( x  95)  P( z 
)  P( z  3.33) .0006
15
100
8. This is a geometric distribution. Bernoulli trials are satisfied.
1
1
8a.   
 12.5 We expect to examine 12.5 people until finding the first colorblind person.
p .08
8b. P(no colorblind men among the 1st four) = P( X  4)  (0.92) 4 .716
8c. P(first colorblind man is the 6th man checked) = P( X  6)  (0.92)5 (.08) .0527
8d. P(she find a colorblind man before the 10th man) = P(1  X  10)  1  (.92) 9 .528
 p  p .47 and standard deviation  p 
9a.    x  p( x)  2.869

 ( x )
2
p( x)  1312
.
9b. The distribution of scores for 40 randomly selected students would not follow a Normal model. The
distribution would follow the population, mostly uniform.
9c. The sample size is large ( n  40 ). Each student should be independent. N  10n  10(40)  400 . The
shape of the sampling distribution should be Normal (CLT). The mean should be  x   x  2.869 and standard
 1312
.
deviation of  x 

.2075 .
n
40
3  2.869
)  P( z .794) .214
9d. P( x  3)  P( z 
0165
.
10a. According to the Normal model, the probability that the IQ of a student from East State is at least 125 is
approximately 0.734. P( x  125)  P( z  .625) .734
 X Y  164  12.806
 2X Y   2X   Y2  64  100  164
10b.  X Y   X   Y  130  120  10
P( X  Y  5)  P( z  .3904)  0.652
10c. Small sample (n=3), each student is independent, N  10n  30 , because the population data is normal,

10
then the sampling distribution will be normal.  x   x  120
x  x 
 5.7735
n
3
P( x  125)  P( z .866) .193
10d.  x  y   x   y  130  120  10
 x  y   2x   2y  (
10 2
8
)  ( ) 2  7.3937
3
3
P( x  y  5)  P( z  0.6763) .751
According to the Normal model, the probability that the mean IQ of 3 ESU students is at least 5 points higher
than the mean IQ of 3 WSU students is approximately 0.751.