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AP Statistics Unit 7 Review
1. Sale of eggs that are contaminated with salmonella can cause food poisoning in consumers. A large egg producer
takes an SRS of 200 eggs from all the eggs shipped in one day. The laboratory reports that 9 of these eggs had
salmonella contamination. Unknown to the producer, 0.1% (one-tenth of 1%) of all eggs shipped had salmonella.
Identify the population, the parameter, the sample, and the statistic.
Population – all eggs shipped on that day
Parameter – p = 0.001
Sample – the 200 eggs randomly selected
Statistic – pˆ 
 0.045
2. A. Sketch a graph that displays the distribution of birth weights for this population.
Graph should appear approximately normal with mean 3668 and standard deviation 511
B. Sketch a possible graph of the distribution of birth weights for an SRS of size 5.
Students’ graph should vary but should include 5 points between 2600 and 4600
C. In the graph above, there is a dot at approximately 2750. Explain what this value represents.
The dot 2750 represents one SRS of size 5 from the population, where the sample range was 2750 grams
3. A. Is the sample range an unbiased estimator of the population range? Give evidence from the graph above to
support your answer.
The sample range is not an unbiased estimator of the population range because the sample range is not the same as
the population range.
B. Explain how we could decrease the variability of the sampling distribution of the sample range.
To decrease the variability of the sampling distribution, increase the sample size
4. The Gallup Poll once asked a random sample of 1540 adults, “Do you happen to jog?” Suppose that in fact 15% of all
adults jog.
A. What is the mean of the sampling distribution of p̂ ? Justify your answer.
 pˆ  p  0.15 the mean is an unbiased estimator
B. Find the standard deviation of the sampling distribution of p̂ . Check that the 10% condition is met.
It is reasonable to assume the population of adults > 10(1540) = 15400, therefore reasonable to use the
standard deviation formula.  pˆ 
p 1  p 
0.15  0.85 
 0.009
C. Is the sampling distribution of p̂ approximately Normal? Justify your answer.
Since np = 1540(0.15) = 231 > 10 and n(1 – p) = 1540(1 – 0.15) = 1309 > 10, the sampling distribution is
approximately normal.
D. Find the probability that between 13% and 17% of a random sample of 1540 adults are joggers. Show your
0.15  0.85   
P  0.13  pˆ  0.17   0.9721  using normalcdf  .13, .17, .15,
1540  
5. Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each
passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not
bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they
require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger
will be searched by Customs agents. A green light means “go ahead.” Customs agents claim that the proportion of
all travelers who will be stopped (red light) is 0.30 of showing red on any push of the button. To test this claim, a
concerned citizen watches a random sample of 100 travelers push the button. Only 20 get a red light.
A. Assume that the Customs agents’ claim is true. Find the probability that the proportion of travelers who get a
red light is as small as or smaller than the result in this sample. Show your work.
0.3  0.7   
P  x  0.2   0.0145  using normalcdf  9, 0.2, 0.3,
100  
B. Based on you results in part A, do you believe the Customs agents’ claim? Explain.
Since we would get a sample like this or more extreme only about 1% of the time, it is unlikely that the customs
agent’s claim is true.
6. 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 100 and standard deviation 15.
A. What is the probability that a randomly chosen individual has a WAIS score of 105 or higher? Show your work.
P  x  105  0.3694  using normalcdf 105, 9999, 100, 15 
B. Find the mean and standard deviation of the sampling distribution of the average WAIS score x for an SRS of 60
 x    100,  x 
 1.936
C. What is the probability that the average WAIS score of an SRS of 60 people is 105 or higher? Show your work.
P  x  105  0.0049  using normalcdf 105, 9999, 100, 1.936 
D. Would your answers to any of parts A, B, or C be affected if the distribution of WAIS scores in the adult
population were distinctly non-Normal? Explain.
The answer to part A would change but not the answers to parts B and C. According to the CLT, if the sample
size is large, the sampling distribution is approximately normal. Therefore, we can use normal calculations in
part B and C even if the population distribution is non-normal because our sample size is large. (60 > 30)
7. The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout
the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only 0.5,
but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard
deviation 0.7.
A. Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of x for SRSs
of size 50.
According to the CLT, the sampling distribution will be approximately normal when the sample size is large.
Since n = 50, the sample size is large (>30) and the sampling distribution is approximately normal.
B. Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6.
0.7  
P  x  0.6   0.1562  using normalcdf  0.6, 99, 0.5,
60  
C. In a recent month, the mean number of moths in an SRS of size 50 was 0.6. Based on this result, should the
state agriculture department be worried that the moth population is getting larger in their state? Explain.
The state agriculture department should not be concerned. They get this result or something more extreme
about 15% of the time. Since this is not unusual we think it happened because of sampling variability.