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AP Statistics 5.7 Sampling Distributions for Sample Means - MCQs - Exam Style Questions

Question

The graph shows the population distribution of random variable \(X\) with mean \(85\) and standard deviation \(18\).
Which of the following graphs is a sampling distribution of the sample mean \(\overline{X}\) for samples of size \(40\) taken from the population?
▶️ Answer/Explanation
Detailed solution

For \(n=40\), the sampling distribution of \(\overline{X}\) is approximately normal by the Central Limit Theorem (sample is sufficiently large).
Mean: \(\mu_{\overline{X}}=\mu=85\).
Standard deviation (standard error): \(\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{18}{\sqrt{40}}\approx 2.85\).
Therefore the correct graph should be approximately normal, centered at \(85\), with spread about \(2.85\).
Among the choices, graph **B** is roughly normal, centered near \(85\), and has an appropriate spread (ticks around \(72\)–\(96\), consistent with a few standard errors).
Answer: (B)

Question

At a certain restaurant, the distribution of wait times between ordering a meal and receiving the meal has mean \(11.4\) minutes and standard deviation \(2.6\) minutes. The restaurant manager wants to find the probability that the mean wait time will be greater than \(12.0\) minutes for a random sample of \(84\) customers. Assuming the wait times among customers are independent, which of the following describes the sampling distribution of the sample mean wait time for random samples of size \(84\)?
(A) Approximately normal with mean \(11.4\) minutes and standard deviation \(2.6\) minutes
(B) Approximately normal with mean \(11.4\) minutes and standard deviation \(\dfrac{2.6}{\sqrt{84}}\) minute
(C) Approximately normal with mean \(12.0\) minutes and standard deviation \(2.6\) minutes
(D) Binomial with mean \(84(0.41)\) minutes and standard deviation \(\sqrt{84(0.41)(0.59)}\) minutes
(E) Binomial with mean \(84(0.5)\) minutes and standard deviation \(\sqrt{84(0.5)(0.5)}\) minutes
▶️ Answer/Explanation
Detailed solution

For sample means with independent observations and sufficiently large \(n\) (\(n=84\)), the sampling distribution of \(\bar{X}\) is approximately normal.
Mean: \(\mu_{\bar{X}}=\mu=11.4\) minutes.
Standard deviation (standard error): \(\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{2.6}{\sqrt{84}}\) minute.
Therefore the correct description is approximately normal with mean \(11.4\) and standard deviation \(\dfrac{2.6}{\sqrt{84}}\).
Answer: (B)

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