Sampling distribution and CLT practice problems

By Jude Wallis · Published

This set has eight problems on sampling distributions and the central limit theorem. You practice the mean and standard deviation of the sample mean and sample proportion, probabilities from the normal model, and checking when that model applies. Work each one before opening the solution.

AP Statistics: Unit 2 (topics 2.12 Sampling Distributions and the Central Limit Theorem, 3.2 Sampling Distributions for Sample Proportions, 4.1 Sampling Distributions for Sample Means). These problems cover the sampling-distribution content introduced with the central limit theorem in Unit 2 (topic 2.12), then applied to proportions in Unit 3 (topic 3.2) and to means in Unit 4 (topic 4.1) of the Fall 2026 AP Statistics course.

What these problems build

These eight problems train the core sampling-distribution skills tested in AP Statistics: finding the mean and standard deviation of the sample mean xˉ\bar{x} (x-bar) and the sample proportion p^\hat{p} (p-hat), computing probabilities from the normal model, and checking whether that normal model even applies.

Work each problem on paper before opening its solution, then compare every arithmetic step. If a formula feels shaky, review sampling distributions explained and the central limit theorem guide; to check a probability, use the normal distribution calculator.

Problems 1 and 2 are routine center-and-spread calculations. The later ones layer on normal probabilities, condition checks, and an AP-style percentile question.

Problem 1

A coffee roaster fills bags on an automatic line. The filled weights have population mean μ=340\mu = 340 grams (the Greek letter mu is the population mean) and population standard deviation σ=8\sigma = 8 grams (sigma is the population standard deviation). Quality control pulls random samples of n=16n = 16 bags and records the sample mean weight xˉ\bar{x} (x-bar). Find the mean and standard deviation of the sampling distribution of xˉ\bar{x}.

Show the worked solution

  1. The mean of the sampling distribution of xˉ\bar{x} equals the population mean: μxˉ=μ=340\mu_{\bar{x}} = \mu = 340 grams.

  2. The standard deviation of the sampling distribution is σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}.

  3. Substitute σ=8\sigma = 8 and n=16n = 16: σxˉ=816\sigma_{\bar{x}} = \frac{8}{\sqrt{16}}.

  4. Evaluate the root: 16=4\sqrt{16} = 4, so σxˉ=84=2\sigma_{\bar{x}} = \frac{8}{4} = 2 grams.

The sampling distribution of xˉ\bar{x} has mean 340340 grams and standard deviation 22 grams.

Problem 2

In a mobile puzzle game, the true proportion of players who clear level 5 on their first attempt is p=0.25p = 0.25. A designer takes a random sample of n=400n = 400 players and records the sample proportion p^\hat{p} (p-hat) who cleared it on the first try. Find the mean and standard deviation of the sampling distribution of p^\hat{p}.

Show the worked solution

  1. The mean of the sampling distribution of p^\hat{p} equals the population proportion: μp^=p=0.25\mu_{\hat{p}} = p = 0.25.

  2. The standard deviation is σp^=p(1−p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}.

  3. Compute the numerator inside the root: p(1−p)=0.25×0.75=0.1875p(1-p) = 0.25 \times 0.75 = 0.1875.

  4. Divide by n=400n = 400: 0.1875400=0.00046875\frac{0.1875}{400} = 0.00046875.

  5. Take the square root: 0.00046875≈0.0217\sqrt{0.00046875} \approx 0.0217.

The sampling distribution of p^\hat{p} has mean 0.250.25 and standard deviation about 0.02170.0217.

Problem 3

A town official believes p=0.08p = 0.08 of residents commute mainly by bicycle. A survey plans to take a random sample of n=100n = 100 residents and record the sample proportion p^\hat{p} (p-hat) who bike. (a) Can the sampling distribution of p^\hat{p} be modeled as approximately normal at this sample size? (b) What is the smallest sample size that would satisfy the large-counts condition?

Show the worked solution

  1. The large-counts condition for a sample proportion requires both np≥10np \geq 10 and n(1−p)≥10n(1-p) \geq 10, meaning at least 10 expected successes and 10 expected failures.

  2. Expected successes at n=100n = 100: np=100×0.08=8np = 100 \times 0.08 = 8.

  3. Expected failures at n=100n = 100: n(1−p)=100×0.92=92n(1-p) = 100 \times 0.92 = 92.

  4. Since np=8<10np = 8 < 10, the condition fails, so the sampling distribution is not approximately normal at n=100n = 100.

  5. For part (b), set np≥10np \geq 10: n≥100.08=125n \geq \frac{10}{0.08} = 125. Check the other condition: n≥100.92≈10.9n \geq \frac{10}{0.92} \approx 10.9, which is easier to meet.

  6. The binding requirement is n≥125n \geq 125, so a sample of at least 125 residents is needed.

(a) No, because np=8<10np = 8 < 10. (b) At least n=125n = 125 residents.

Problem 4

The battery life of a wireless earbud model is normally distributed with mean μ=30\mu = 30 hours and standard deviation σ=4\sigma = 4 hours. A reviewer tests a random sample of n=25n = 25 earbuds and records the sample mean battery life xˉ\bar{x} (x-bar). Find the probability that the sample mean exceeds 31 hours.

Show the worked solution

  1. Because the population is normal, the sampling distribution of xˉ\bar{x} is normal for any sample size, centered at μxˉ=30\mu_{\bar{x}} = 30 hours.

  2. Standard deviation of xˉ\bar{x}: σxˉ=σn=425=45=0.8\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 hours.

  3. Convert 31 hours to a z-score: z=31−300.8=10.8=1.25z = \frac{31 - 30}{0.8} = \frac{1}{0.8} = 1.25.

  4. Find the upper-tail area: P(Z>1.25)=1−P(Z<1.25)=1−0.8944=0.1056P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056.

P(xˉ>31)≈0.1056P(\bar{x} > 31) \approx 0.1056.

Problem 5

Time spent at a store's self-checkout is right-skewed with mean μ=4.5\mu = 4.5 minutes and standard deviation σ=2.4\sigma = 2.4 minutes. A manager records a random sample of n=36n = 36 self-checkout sessions and computes the sample mean time xˉ\bar{x} (x-bar). (a) Justify modeling xˉ\bar{x} with a normal distribution. (b) Find the probability that the sample mean exceeds 5.1 minutes.

Show the worked solution

  1. For part (a), the population is skewed, but the sample size n=36≥30n = 36 \geq 30, so by the central limit theorem the sampling distribution of xˉ\bar{x} is approximately normal.

  2. Center: μxˉ=μ=4.5\mu_{\bar{x}} = \mu = 4.5 minutes.

  3. Standard deviation: σxˉ=σn=2.436=2.46=0.4\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{36}} = \frac{2.4}{6} = 0.4 minutes.

  4. Convert 5.1 minutes to a z-score: z=5.1−4.50.4=0.60.4=1.5z = \frac{5.1 - 4.5}{0.4} = \frac{0.6}{0.4} = 1.5.

  5. Upper-tail area: P(Z>1.5)=1−P(Z<1.5)=1−0.9332=0.0668P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668.

(a) n=36≥30n = 36 \geq 30, so the CLT makes xˉ\bar{x} approximately normal. (b) P(xˉ>5.1)≈0.0668P(\bar{x} > 5.1) \approx 0.0668.

Problem 6

A streaming service finds that p=0.20p = 0.20 of free-trial users convert to a paid plan. An analyst takes a random sample of n=100n = 100 trial users and records the sample proportion p^\hat{p} (p-hat) who convert. Find the probability that fewer than 13% of the sample convert.

Show the worked solution

  1. Check the large-counts condition: np=100×0.20=20≥10np = 100 \times 0.20 = 20 \geq 10 and n(1−p)=100×0.80=80≥10n(1-p) = 100 \times 0.80 = 80 \geq 10, so the sampling distribution of p^\hat{p} is approximately normal.

  2. Center: μp^=p=0.20\mu_{\hat{p}} = p = 0.20.

  3. Standard deviation: σp^=0.20×0.80100=0.16100=0.0016=0.04\sigma_{\hat{p}} = \sqrt{\frac{0.20 \times 0.80}{100}} = \sqrt{\frac{0.16}{100}} = \sqrt{0.0016} = 0.04.

  4. Convert p^=0.13\hat{p} = 0.13 to a z-score: z=0.13−0.200.04=−0.070.04=−1.75z = \frac{0.13 - 0.20}{0.04} = \frac{-0.07}{0.04} = -1.75.

  5. Lower-tail area: P(Z<−1.75)=0.0401P(Z < -1.75) = 0.0401.

P(p^<0.13)≈0.0401P(\hat{p} < 0.13) \approx 0.0401.

Problem 7

A bakery's sourdough loaves have weights that are approximately normal with mean μ=500\mu = 500 grams and standard deviation σ=20\sigma = 20 grams. A random sample of n=25n = 25 loaves is weighed and the sample mean xˉ\bar{x} (x-bar) is recorded. (a) Give the mean and standard deviation of the sampling distribution of xˉ\bar{x}. (b) Find the probability that the sample mean is below 492 grams. (c) Interpret that probability in context.

Show the worked solution

  1. Part (a) center: μxˉ=μ=500\mu_{\bar{x}} = \mu = 500 grams.

  2. Part (a) standard deviation: σxˉ=σn=2025=205=4\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4 grams.

  3. Part (b): the population is normal, so xˉ\bar{x} is normal. Convert 492 grams to a z-score: z=492−5004=−84=−2z = \frac{492 - 500}{4} = \frac{-8}{4} = -2.

  4. Lower-tail area: P(Z<−2)=0.0228P(Z < -2) = 0.0228.

  5. Part (c): about 2.28% of random samples of 25 loaves would have a mean weight below 492 grams.

(a) Mean 500500 grams, standard deviation 44 grams. (b) P(xˉ<492)≈0.0228P(\bar{x} < 492) \approx 0.0228. (c) Roughly 2.28% of such samples average below 492 grams.

Problem 8

A bike courier's delivery times are right-skewed with mean μ=45\mu = 45 minutes and standard deviation σ=12\sigma = 12 minutes. A dispatcher studies random samples of n=64n = 64 deliveries and records each sample mean xˉ\bar{x} (x-bar). (a) Explain why xˉ\bar{x} can be modeled as approximately normal. (b) Find the standard deviation of the sampling distribution of xˉ\bar{x}. (c) The dispatcher flags a day when the sample mean lands in the slowest 5% of the sampling distribution. Find the sample mean that marks the 95th percentile.

Show the worked solution

  1. Part (a): the population is skewed, but n=64≥30n = 64 \geq 30, so the central limit theorem makes the sampling distribution of xˉ\bar{x} approximately normal, centered at μxˉ=45\mu_{\bar{x}} = 45 minutes.

  2. Part (b): σxˉ=σn=1264=128=1.5\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{64}} = \frac{12}{8} = 1.5 minutes.

  3. Part (c): the 95th percentile of a normal distribution sits at z=1.645z = 1.645 standard deviations above the mean.

  4. Convert that z-score back to minutes: xˉ=μxˉ+z⋅σxˉ=45+1.645×1.5\bar{x} = \mu_{\bar{x}} + z \cdot \sigma_{\bar{x}} = 45 + 1.645 \times 1.5.

  5. Compute the product: 1.645×1.5=2.46751.645 \times 1.5 = 2.4675.

  6. Add to the center: 45+2.4675=47.4675≈47.4745 + 2.4675 = 47.4675 \approx 47.47 minutes.

(a) n=64≥30n = 64 \geq 30, so the CLT applies. (b) 1.51.5 minutes. (c) About 47.4747.47 minutes.