A beverage filling line fills containers whose volumes are approximately normal with standard deviation σ = 12 mL. Engineers monitor the line with the sample mean x-bar of n randomly chosen containers. (a) Find the smallest n for which the standard deviation of x-bar is at most 2 mL. (b) Find the smallest n for which x-bar lands within 3 mL of the true mean at least 95% of the time.

(a) n ≥ 36, so 36 containers. (b) sqrt(n) ≥ (1.96 × 12)/3 = 7.84, so n ≥ 61.4656 and the answer is n = 62. Checking: n = 62 gives probability 0.9510 and n = 61 gives 0.9491.

A cafe chain's records show that the mean wait at its drive-through is μ = 4.0 minutes with σ = 1.2 minutes, and that wait times are right-skewed. A reporter times a random sample of 50 customers across the chain and gets x-bar = 4.5 minutes. (a) Describe the sampling distribution of x-bar assuming the chain's figures are correct. (b) Find P(x-bar ≥ 4.5). (c) Interpret that probability in context and say whether the sample is convincing evidence against the chain's claimed mean.

(a) Approximately normal by the central limit theorem (n = 50 ≥ 30), with μ_x-bar = 4.0 minutes and σ_x-bar = 0.1697 minutes. (b) z = 2.95, so P(x-bar ≥ 4.5) ≈ 0.0016. (c) Only about 0.16% of samples of 50 would average 4.5 minutes or more if the claim were true, which is convincing evidence the true mean wait exceeds 4.0 minutes.

Sampling distribution of x-bar: 8 practice problems

Q: How large does n have to be for the central limit theorem?

The AP course uses n ≥ 30 as its working rule, and that is the threshold to cite in a response. It is a convention rather than a theorem: mildly skewed populations settle down well before 30, and severely skewed ones with extreme outliers can need more. When the population is stated to be normal you do not invoke the theorem at all, since x-bar is then exactly normal at every sample size. Read the central limit theorem guide.

Q: If the population is skewed and n is small, is everything wrong?

No. Only the shape claim fails. The mean μ_x-bar = μ and the standard deviation σ_x-bar = σ/sqrt(n) are still exactly right, because neither depends on the shape of the population. What you lose is permission to compute areas from a normal curve, which means no normal probability, no confidence interval, and no p-value. Problem 4 works through that case.

Q: When do I use sigma over root n instead of s over root n?

Use σ/sqrt(n) only when the population standard deviation σ is given, which is standard in sampling-distribution problems like these and rare in real inference. Once you estimate the spread from the sample with s, the quantity s/sqrt(n) is a standard error, it carries its own sampling error, and the reference curve becomes a t distribution rather than the standard normal. See t test vs z test.

By Jude Wallis · Published July 31, 2026

This set has 8 problems on the sampling distribution of the sample mean. You find its center and its standard deviation, decide when a normal model is justified, compute probabilities about x-bar, and solve for the sample size a stated precision needs. Work each one before opening the steps.

AP Statistics: Unit 4 (topics 4.1 Sampling Distributions for Sample Means). These problems cover Unit 4 topic 4.1 of the Fall 2026 AP Statistics course: the mean, the standard deviation, and the shape of the sampling distribution of the sample mean, plus probabilities computed from it. The central limit theorem itself is introduced earlier, in Unit 2 topic 2.12. Unit 4 is 10 to 20% of the multiple-choice section, and every t procedure later in the unit rests on the three facts practiced here.

What these problems build

Every inference procedure for a mean rests on one fact: if you take random samples of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$ , the sample mean $\bar{x}$ (read 'x-bar') varies from sample to sample in a predictable way. Three facts describe that variation.

Center. $\mu_{\bar{x}} = \mu$ . The sample mean is an unbiased estimator, so it is centered on the parameter it estimates at every sample size and for every population shape.
Spread. $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ , provided the sample is less than 10% of the population when sampling without replacement.
Shape. Normal in two situations, and only those two: the population itself is normal, in which case $\bar{x}$ is exactly normal at any $n$ , or $n$ is large enough for the central limit theorem, which makes it approximately normal and which this course treats as $n \ge 30$ .

Problem 1 is the routine center-and-spread calculation, problems 2 through 7 add probabilities, sample-size algebra, and a shape argument, and problem 8 uses the sampling distribution the way a hypothesis test does. Every probability below is a normal-curve area, so review how to find a z score if the standardizing step is shaky, read sampling distributions explained for the concept, and check your areas against the normal distribution calculator or the z-table.

The shape question, which is where points are lost

Center and spread are formula work. Shape is an argument, and it is the part a rubric reads. Two separate routes make $\bar{x}$ normal enough to use, and a response has to name the one it is using.

Route	What you must state	Sample size needed
Normal population	The problem says the population is normal or approximately normal	Any $n$ , even $n = 4$
Central limit theorem	The population shape is unknown or skewed	$n \ge 30$

A skewed population with $n = 12$ satisfies neither, so no normal probability may be computed at all. Notice what does not break: $\mu_{\bar{x}} = \mu$ and $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ are still correct in that case, because neither depends on shape. Only the area under a normal curve becomes unavailable. Problem 4 is built on exactly that distinction. For the theorem itself, read the central limit theorem guide, and to watch the shape settle down as $n$ grows, open the sampling distribution simulator. More sets are on the practice page.

Frequently asked questions

Why is the standard deviation of x-bar sigma over root n rather than sigma?

Because averaging cancels variation. High values in a sample tend to be offset by low ones, so a mean of $n$ observations bounces around far less than a single observation does. The variance of the mean is $\frac{\sigma^2}{n}$ , and taking the square root gives $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ . The square root is why quadrupling the sample size only halves the spread. See standard error vs standard deviation.

How large does n have to be for the central limit theorem?

The AP course uses $n \ge 30$ as its working rule, and that is the threshold to cite in a response. It is a convention rather than a theorem: mildly skewed populations settle down well before 30, and severely skewed ones with extreme outliers can need more. When the population is stated to be normal you do not invoke the theorem at all, since $\bar{x}$ is then exactly normal at every sample size. Read the central limit theorem guide.

If the population is skewed and n is small, is everything wrong?

No. Only the shape claim fails. The mean $\mu_{\bar{x}} = \mu$ and the standard deviation $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ are still exactly right, because neither depends on the shape of the population. What you lose is permission to compute areas from a normal curve, which means no normal probability, no confidence interval, and no p-value. Problem 4 works through that case.

When do I use sigma over root n instead of s over root n?

Use $\frac{\sigma}{\sqrt{n}}$ only when the population standard deviation $\sigma$ is given, which is standard in sampling-distribution problems like these and rare in real inference. Once you estimate the spread from the sample with $s$ , the quantity $\frac{s}{\sqrt{n}}$ is a standard error, it carries its own sampling error, and the reference curve becomes a t distribution rather than the standard normal. See t test vs z test.

Problem 1

A utility reports that daily water use per household in a town of 5000 households is right-skewed with mean $\mu = 310$ gallons and standard deviation $\sigma = 84$ gallons. An auditor takes a simple random sample of $n = 49$ households and records the sample mean $\bar{x}$ . (a) Find the mean and the standard deviation of the sampling distribution of $\bar{x}$ . (b) Check the 10% condition and say what it buys you.

Show the worked solution

(a) Center. The sample mean is an unbiased estimator of the population mean, so $\mu_{\bar{x}} = \mu = 310$ gallons. This holds for any sample size and any population shape, so the right skew does not affect it.
(a) Spread. $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{84}{\sqrt{49}} = \frac{84}{7} = 12$ gallons. Sample means vary about seven times less than individual households do, since $\sqrt{49} = 7$ .
(b) Check the condition. Sampling is without replacement, so the requirement is $n < 0.10N$ . Here $0.10 \times 5000 = 500$ , and $49 < 500$ , so the condition is satisfied.
(b) What it buys. The formula $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ assumes the observations are independent. Removing one household changes the pool for the next draw, so they are not exactly independent, but with the sample under 10% of the population the dependence is small enough that the formula stays accurate. Without the condition, $\frac{\sigma}{\sqrt{n}}$ would overstate the true variability of $\bar{x}$ .
Note what the problem does not ask. Nothing here requires the sampling distribution to be normal, so the skew is irrelevant to both answers.

(a) $\mu_{\bar{x}} = 310$ gallons and $\sigma_{\bar{x}} = \frac{84}{7} = 12$ gallons. (b) $49 < 0.10 \times 5000 = 500$ , so the 10% condition holds and $\frac{\sigma}{\sqrt{n}}$ is a valid standard deviation for $\bar{x}$ .

Problem 2

Resting heart rates in a large clinic population are approximately normal with mean $\mu = 72$ beats per minute and standard deviation $\sigma = 12$ bpm. (a) Find the probability that one randomly chosen adult has a resting heart rate above 78 bpm. (b) Find the probability that a random sample of 36 adults has a sample mean above 78 bpm. (c) Explain why the two answers differ so much.

Show the worked solution

(a) One individual. The relevant standard deviation is the population value $\sigma = 12$ bpm, not a standard error. Standardize: $z = \frac{78 - 72}{12} = \frac{6}{12} = 0.50$ .
(a) Area. $P(Z > 0.50) = 1 - 0.6915 = 0.3085$ , so about 30.85% of individual adults are above 78 bpm.
(b) One sample mean. The population is normal, so $\bar{x}$ is normal for any $n$ , centered at $\mu_{\bar{x}} = 72$ with $\sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2$ bpm.
(b) Standardize and find the area. $z = \frac{78 - 72}{2} = \frac{6}{2} = 3.00$ , and $P(Z > 3.00) = 0.0013$ .
(c) Compare. The value 78 sits half a population standard deviation above the mean, but three standard errors above it. Averaging 36 people cancels out most of the individual variation, so a sample mean 6 bpm above the truth is far more extreme than one person 6 bpm above the truth. That is the whole reason inference works on means: $\frac{\sigma}{\sqrt{n}}$ shrinks while $\sigma$ does not.

(a) $z = 0.50$ , so $P(X > 78) \approx 0.3085$ . (b) $\sigma_{\bar{x}} = 2$ bpm, $z = 3.00$ , so $P(\bar{x} > 78) \approx 0.0013$ . (c) The same 6 bpm gap is $0.5$ population standard deviations but $3$ standard errors, because averaging 36 values divides the spread by $\sqrt{36} = 6$ .

Problem 3

A beverage filling line fills containers whose volumes are approximately normal with standard deviation $\sigma = 12$ mL. Engineers monitor the line with the sample mean $\bar{x}$ of $n$ randomly chosen containers. (a) Find the smallest $n$ for which the standard deviation of $\bar{x}$ is at most 2 mL. (b) Find the smallest $n$ for which $\bar{x}$ lands within 3 mL of the true mean at least 95% of the time.

Show the worked solution

(a) Set up the inequality. Require $\sigma_{\bar{x}} = \frac{12}{\sqrt{n}} \le 2$ , which rearranges to $\sqrt{n} \ge \frac{12}{2} = 6$ , so $n \ge 36$ . The smallest sample size is $n = 36$ .
(b) Translate the requirement. 'Within 3 mL at least 95% of the time' means $P(|\bar{x} - \mu| \le 3) \ge 0.95$ . For a normal sampling distribution, the middle 95% runs from $-1.96$ to $1.96$ standard errors, so the requirement is $1.96 \times \sigma_{\bar{x}} \le 3$ .
(b) Solve for $n$ . $1.96 \times \frac{12}{\sqrt{n}} \le 3$ gives $\sqrt{n} \ge \frac{1.96 \times 12}{3} = \frac{23.52}{3} = 7.84$ , so $n \ge 7.84^2 = 61.4656$ . Sample size must be a whole number and the inequality runs upward, so round up to $n = 62$ .
(b) Verify both sides of the cutoff. At $n = 62$ , $\sigma_{\bar{x}} = \frac{12}{\sqrt{62}} = 1.5240$ and $z = \frac{3}{1.5240} = 1.9685$ , giving $P(|\bar{x} - \mu| \le 3) = 0.9510 \ge 0.95$ . At $n = 61$ , $\sigma_{\bar{x}} = 1.5364$ and $z = \frac{3}{1.5364} = 1.9526$ , giving $0.9491 < 0.95$ . So 62 really is the smallest.
Note the direction of rounding. Rounding 61.4656 down to 61 would fail the requirement by a whisker, which is why sample-size answers always round up, never to the nearer integer.

(a) $n \ge 36$ , so 36 containers. (b) $\sqrt{n} \ge \frac{1.96 \times 12}{3} = 7.84$ , so $n \ge 61.4656$ and the answer is $n = 62$ . Checking: $n = 62$ gives probability $0.9510$ and $n = 61$ gives $0.9491$ .

Problem 4

Time to failure for a machine part is strongly right-skewed with mean $\mu = 1200$ hours and standard deviation $\sigma = 800$ hours. (a) A technician takes a random sample of $n = 12$ parts and wants to model $\bar{x}$ with a normal distribution. Explain why that is not justified, and state which of the three sampling-distribution facts still hold. (b) Would $n = 12$ be enough if failure times had been normal instead? (c) For a random sample of $n = 100$ parts, find $P(\bar{x} < 1050)$ .

Show the worked solution

(a) Check both routes to normality. The population is strongly right-skewed, so the normal-population route is closed. The central limit theorem route needs $n \ge 30$ , and $n = 12$ is far short of it. Neither route applies, so no normal probability may be computed for $\bar{x}$ at this sample size.
(a) What survives. Center and spread do not depend on shape, so $\mu_{\bar{x}} = 1200$ hours and $\sigma_{\bar{x}} = \frac{800}{\sqrt{12}} = \frac{800}{3.4641} = 230.94$ hours are both still correct. Only the shape claim fails, and with it any area under a normal curve. The sampling distribution of $\bar{x}$ for $n = 12$ would itself be right-skewed, though less so than the population.
(b) A normal population. Yes. If failure times were normal, $\bar{x}$ would be exactly normal for every sample size, including $n = 12$ and even $n = 2$ . The central limit theorem is only needed when the population is not normal.
(c) Justify the model at $n = 100$ . The population is skewed, but $n = 100 \ge 30$ , so the central limit theorem makes $\bar{x}$ approximately normal, centered at $\mu_{\bar{x}} = 1200$ hours with $\sigma_{\bar{x}} = \frac{800}{\sqrt{100}} = \frac{800}{10} = 80$ hours.
(c) Standardize and find the area. $z = \frac{1050 - 1200}{80} = \frac{-150}{80} = -1.875$ , and $P(Z < -1.875) = 0.0304$ .

(a) The population is strongly skewed and $n = 12 < 30$ , so neither route to normality applies; $\mu_{\bar{x}} = 1200$ hours and $\sigma_{\bar{x}} = 230.94$ hours are still correct, but no normal area may be computed. (b) Yes: a normal population makes $\bar{x}$ normal at every $n$ . (c) $\sigma_{\bar{x}} = 80$ hours, $z = -1.875$ , so $P(\bar{x} < 1050) \approx 0.0304$ .

Problem 5

Travel times on a city bus route are approximately normal with mean $\mu = 28$ minutes and standard deviation $\sigma = 6$ minutes. A supervisor records a random sample of $n = 16$ trips each day, and the route is logged as on schedule when the sample mean travel time falls between 26.5 and 31 minutes. Find the probability that a day is logged as on schedule.

Show the worked solution

Describe the sampling distribution. The population is stated to be approximately normal, so $\bar{x}$ is approximately normal for $n = 16$ with no appeal to the central limit theorem needed. Center: $\mu_{\bar{x}} = 28$ minutes.
Standard deviation. $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{16}} = \frac{6}{4} = 1.5$ minutes.
Standardize the lower bound. $z_1 = \frac{26.5 - 28}{1.5} = \frac{-1.5}{1.5} = -1.00$ .
Standardize the upper bound. $z_2 = \frac{31 - 28}{1.5} = \frac{3}{1.5} = 2.00$ .
Subtract the two areas. $P(-1.00 < Z < 2.00) = P(Z < 2.00) - P(Z < -1.00) = 0.97725 - 0.15866 = 0.81859$ , about $0.8186$ .
Read it back in context. About 81.9% of days produce a sample mean inside the window, so roughly 18% of days get flagged even when nothing about the route has changed.

$\sigma_{\bar{x}} = 1.5$ minutes, $z_1 = -1.00$ and $z_2 = 2.00$ , so $P(26.5 < \bar{x} < 31) = 0.97725 - 0.15866 \approx 0.8186$ .

Problem 6

A dairy fills yogurt cups whose net weights are approximately normal with mean $\mu = 150$ g and standard deviation $\sigma = 6$ g. Quality control weighs a random sample of $n = 9$ cups every hour and stops the line whenever the sample mean falls outside the two limits that capture the middle 99% of the sampling distribution of $\bar{x}$ . (a) Find those two limits. (b) Interpret what a stop means in terms of this probability.

Show the worked solution

Describe the sampling distribution. The population is approximately normal, so $\bar{x}$ is approximately normal at $n = 9$ . Center $\mu_{\bar{x}} = 150$ g and $\sigma_{\bar{x}} = \frac{6}{\sqrt{9}} = \frac{6}{3} = 2$ g.
(a) Find the critical value. The middle 99% leaves $0.005$ in each tail, so the limits sit at $z = \pm 2.576$ standard errors from the center. This is the same $z$ that a 99% confidence interval uses.
(a) Convert to grams. The distance from the center is $2.576 \times 2 = 5.152$ g.
(a) Build the limits. Lower: $150 - 5.152 = 144.848$ g. Upper: $150 + 5.152 = 155.152$ g. Rounded, the line stops if the hourly sample mean falls below 144.85 g or above 155.15 g.
(b) Interpret. If the process really is on target at 150 g, only 1% of hourly samples of 9 cups produce a mean outside those limits, so a stop happens about once every 100 hours by chance alone. When the line does stop, either that 1% event occurred or the process has genuinely drifted off 150 g. Narrower limits would catch real drift sooner but would also raise the false-alarm rate above 1%.

(a) $\sigma_{\bar{x}} = 2$ g and $z = \pm 2.576$ , so the limits are $150 \pm 5.152$ , that is $144.848$ g and $155.152$ g. (b) On a process that is truly on target, only about 1% of hourly samples fall outside, so a stop is either a 1-in-100 false alarm or a sign of real drift.

Problem 7

Lead concentration in a city's tap water is right-skewed with mean $\mu = 8.2$ ppb and standard deviation $\sigma = 3.0$ ppb. (a) For a random sample of $n = 36$ taps, find the probability that $\bar{x}$ lands within 0.5 ppb of $\mu$ . (b) Repeat for $n = 144$ taps. (c) The sample size was multiplied by 4. Did the probability double? Explain what actually changed.

Show the worked solution

Justify the model for both sizes. The population is skewed, but $n = 36$ and $n = 144$ both clear 30, so the central limit theorem makes $\bar{x}$ approximately normal in each case, centered at $\mu_{\bar{x}} = 8.2$ ppb.
(a) At $n = 36$ . $\sigma_{\bar{x}} = \frac{3.0}{\sqrt{36}} = \frac{3.0}{6} = 0.5$ ppb, so 0.5 ppb is exactly one standard error. Then $P(|\bar{x} - 8.2| \le 0.5) = P(-1 \le Z \le 1) = 0.84134 - 0.15866 = 0.68268$ , about $0.6827$ .
(b) At $n = 144$ . $\sigma_{\bar{x}} = \frac{3.0}{\sqrt{144}} = \frac{3.0}{12} = 0.25$ ppb, so the same 0.5 ppb is now two standard errors. Then $P(|\bar{x} - 8.2| \le 0.5) = P(-2 \le Z \le 2) = 0.97725 - 0.02275 = 0.95450$ , about $0.9545$ .
(c) Answer the doubling question. No. The probability went from $0.6827$ to $0.9545$ , a rise of about $0.27$ , not a doubling. A doubling was never possible, since $2 \times 0.6827 = 1.3654$ is greater than 1 and no probability can exceed 1.
(c) Explain what changed. Quadrupling $n$ halved the standard error, from $0.5$ to $0.25$ ppb, because $\sigma_{\bar{x}}$ shrinks like $\frac{1}{\sqrt{n}}$ and $\sqrt{144} = 2\sqrt{36}$ . A fixed 0.5 ppb window therefore stretched from 1 standard error to 2 standard errors on each side, and these are the two familiar empirical-rule areas, 68% and 95%. Precision improves with the square root of the sample size, and gains shrink as the probability approaches 1.

(a) $\sigma_{\bar{x}} = 0.5$ ppb, $z = \pm 1$ , probability $0.6827$ . (b) $\sigma_{\bar{x}} = 0.25$ ppb, $z = \pm 2$ , probability $0.9545$ . (c) No: quadrupling $n$ halved the standard error, so the same window widened from 1 to 2 standard errors, and the area rose from about 68% to about 95%.

Problem 8

A cafe chain's records show that the mean wait at its drive-through is $\mu = 4.0$ minutes with $\sigma = 1.2$ minutes, and that wait times are right-skewed. A reporter times a random sample of 50 customers across the chain and gets $\bar{x} = 4.5$ minutes. (a) Describe the sampling distribution of $\bar{x}$ assuming the chain's figures are correct. (b) Find $P(\bar{x} \ge 4.5)$ . (c) Interpret that probability in context and say whether the sample is convincing evidence against the chain's claimed mean.

Show the worked solution

(a) Center. If the chain's figures are correct, $\mu_{\bar{x}} = \mu = 4.0$ minutes.
(a) Spread. $\sigma_{\bar{x}} = \frac{1.2}{\sqrt{50}} = \frac{1.2}{7.0711} = 0.1697$ minutes. The 10% condition holds, since 50 customers is far less than 10% of everyone the chain serves.
(a) Shape. Wait times are right-skewed, so the normal-population route is closed, but $n = 50 \ge 30$ , so the central limit theorem makes $\bar{x}$ approximately normal. The sampling distribution is therefore approximately $N(4.0, 0.1697)$ .
(b) Standardize. $z = \frac{4.5 - 4.0}{0.1697} = \frac{0.5}{0.1697} = 2.95$ .
(b) Find the area. $P(Z \ge 2.95) = 1 - 0.9984 = 0.0016$ .
(c) Interpret the probability. If the true mean wait really were 4.0 minutes, only about 0.16% of random samples of 50 customers would average 4.5 minutes or longer. The reporter's sample is not the kind of result the chain's own figures produce by chance.
(c) Answer the question asked. Yes, this is convincing evidence that the true mean wait is longer than the claimed 4.0 minutes. Note the logic: you assumed the claim was true, found that the observed result would be rare under that assumption, and concluded against the claim. That is exactly what a significance test does, and the probability you computed plays the role of a p-value. One difference from a real test: here $\sigma$ was given, so a z procedure was possible. When only a sample standard deviation $s$ is available, the same reasoning runs through a t procedure instead.

(a) Approximately normal by the central limit theorem ( $n = 50 \ge 30$ ), with $\mu_{\bar{x}} = 4.0$ minutes and $\sigma_{\bar{x}} = 0.1697$ minutes. (b) $z = 2.95$ , so $P(\bar{x} \ge 4.5) \approx 0.0016$ . (c) Only about 0.16% of samples of 50 would average 4.5 minutes or more if the claim were true, which is convincing evidence the true mean wait exceeds 4.0 minutes.