What is a sampling distribution? Mean and proportion

Q: Why is the standard deviation of the sample mean sigma over the square root of n?

Averaging n independent values cancels some of the random ups and downs, so the mean varies less than a single observation. The math works out to σ/sqrt(n), which shrinks as n grows. Quadruple the sample size and you cut the spread in half.

By Jude Wallis · Published July 28, 2026

A sampling distribution is the distribution of a statistic, like the sample mean or sample proportion, across every possible sample of a fixed size from a population. It shows how much a statistic varies from sample to sample, which is the basis for confidence intervals and significance tests.

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). Sampling distributions are 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 is a sampling distribution?

A sampling distribution is the distribution of a statistic across all possible samples of a given size from a population. A statistic is any number you compute from a sample, such as the sample mean or the sample proportion.

Here is the key idea. Every time you take a new random sample, the statistic comes out a little different. Collect the statistic from every possible sample and you get a whole distribution of values. That distribution is the sampling distribution.

A sampling distribution is not the distribution of the raw data, and it is not any single sample. It is the distribution of one summary number computed over and over across many samples.

You can approximate a sampling distribution by simulation: draw many random samples, record the statistic each time, and graph the results. The sampling distribution and CLT interactive lets you do exactly that.

Build a sampling distribution of the sample mean by hand

The cleanest way to see a sampling distribution is to build a small one where you can list every sample. Take the tiny population {2, 4, 6}. Its population mean, written $\mu$ (the Greek letter mu), is 4, and its population standard deviation, written $\sigma$ (sigma), is about 1.633.

Now take every possible sample of size $n = 2$ , drawing with replacement so the two picks are independent. There are $3 \times 3 = 9$ such samples. For each one, compute the sample mean, written $\bar{x}$ (x-bar). The nine sample means are 2, 3, 4, 3, 4, 5, 4, 5, and 6.

Group those values and you have the sampling distribution of the sample mean $\bar{x}$ : the value 4 shows up most often, while 2 and 6 are rare. The full calculation, including the mean and standard deviation of this sampling distribution, is worked out below.

Center, spread, and shape of the sample mean

For a population with mean $\mu$ and standard deviation $\sigma$ , when the sampled values are independent, the sampling distribution of the sample mean $\bar{x}$ has three describable features:

Center (mean): $\mu_{\bar{x}} = \mu$ . The sample mean is centered on the population mean, so on average it neither overshoots nor undershoots.
Spread (standard deviation): $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ . A larger sample gives a smaller spread, so bigger samples produce sample means that cluster tightly around $\mu$ .
Shape: if the population is normal, the sampling distribution of $\bar{x}$ is normal for any $n$ . If the population is not normal, the central limit theorem says the sampling distribution of $\bar{x}$ becomes approximately normal once $n$ is large enough, usually $n \geq 30$ .

Here $\mu_{\bar{x}}$ is the mean of the sampling distribution and $\sigma_{\bar{x}}$ is its standard deviation. When you replace the unknown $\sigma$ with the sample standard deviation $s$ , the estimate $s/\sqrt{n}$ is called the standard error; see standard error vs standard deviation for the distinction. The central limit theorem guide covers the shape rule in more detail.

Center, spread, and shape of the sample proportion

The same three questions have clean answers for a categorical variable. Suppose a population has population proportion $p$ , the true fraction with some trait. Take a random sample of size $n$ and record the sample proportion $\hat{p}$ (p-hat), the fraction of your sample with the trait. When the sampled values are independent:

Center: $\mu_{\hat{p}} = p$ . The sample proportion is centered on the true population proportion.
Spread: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ . As with the mean, a larger $n$ shrinks the spread.
Shape: the sampling distribution of $\hat{p}$ is approximately normal when the expected counts are large enough, specifically $np \geq 10$ and $n(1-p) \geq 10$ , meaning at least 10 expected successes and 10 expected failures.

Here $\mu_{\hat{p}}$ is the mean of the sampling distribution of $\hat{p}$ and $\sigma_{\hat{p}}$ is its standard deviation. Notice that both the mean and the spread depend only on $p$ and $n$ , so once you know those two numbers you can describe the whole sampling distribution.

Why inference depends on the sampling distribution

Inference means using one sample to make a claim about a population you cannot fully measure. That only works if you know how your statistic behaves across samples, and that behavior is exactly the sampling distribution.

Two procedures show why:

A confidence interval is built as statistic $\pm$ (critical value)(standard error), and that standard error comes from the standard deviation of the sampling distribution. Without the spread $\sigma/\sqrt{n}$ or $\sqrt{p(1-p)/n}$ , you would have no basis for the width.
A significance test asks how unusual your statistic is if a claim about the parameter were true. You answer that by locating your statistic on the sampling distribution and reading off how far into the tail it falls.

Both procedures assume the sampling distribution is approximately normal, so checking the shape conditions is not busywork. It is what makes the normal-based math valid.

Conditions to check before using the normal model

Before you treat a sampling distribution as normal, verify the conditions the AP course lists.

Randomization: the data should come from a random sample or a randomized experiment. This keeps the sampled values representative.
The 10% condition: when you sample without replacement, the sample size should be at most 10% of the population, written $n \leq 0.10N$ , where $N$ is the population size. This keeps the draws close enough to independent for the standard deviation formulas to hold.
Shape check: for a proportion, confirm $np \geq 10$ and $n(1-p) \geq 10$ . For a mean, confirm the population is normal or that $n \geq 30$ .

Meet these conditions and the mean, standard deviation, and normal shape all apply, so you can compute probabilities and build intervals on solid ground.

Sampling distribution of the sample mean from {2, 4, 6}

Population = {2, 4, 6}. List every sample of size n = 2 taken with replacement, find the sampling distribution of the sample mean x-bar, then compute its mean and standard deviation. Confirm they match mu and sigma/sqrt(n).

Find the population parameters. Population mean: $\mu = \frac{2 + 4 + 6}{3} = \frac{12}{3} = 4$ .
Population variance (divide by N = 3): $\frac{(2-4)^2 + (4-4)^2 + (6-4)^2}{3} = \frac{4 + 0 + 4}{3} = \frac{8}{3} \approx 2.6667$ . Population standard deviation: $\sigma = \sqrt{8/3} \approx 1.6330$ .
List all $3 \times 3 = 9$ samples of size 2 and their means: (2,2) gives 2, (2,4) gives 3, (2,6) gives 4, (4,2) gives 3, (4,4) gives 4, (4,6) gives 5, (6,2) gives 4, (6,4) gives 5, (6,6) gives 6.
Collect the means into the sampling distribution of $\bar{x}$ : value 2 with probability 1/9, value 3 with 2/9, value 4 with 3/9, value 5 with 2/9, value 6 with 1/9.
Mean of the sampling distribution: $\mu_{\bar{x}} = 2(\frac{1}{9}) + 3(\frac{2}{9}) + 4(\frac{3}{9}) + 5(\frac{2}{9}) + 6(\frac{1}{9}) = \frac{2 + 6 + 12 + 10 + 6}{9} = \frac{36}{9} = 4$ .
Variance of the sampling distribution: $\frac{(2-4)^2(1) + (3-4)^2(2) + (4-4)^2(3) + (5-4)^2(2) + (6-4)^2(1)}{9} = \frac{4 + 2 + 0 + 2 + 4}{9} = \frac{12}{9} = \frac{4}{3}$ . So $\sigma_{\bar{x}} = \sqrt{4/3} \approx 1.1547$ .
Check against the formulas: $\mu_{\bar{x}} = 4 = \mu$ , and $\frac{\sigma}{\sqrt{n}} = \frac{1.6330}{\sqrt{2}} = \frac{1.6330}{1.4142} \approx 1.1547$ , which matches the standard deviation computed directly.

The sampling distribution of $\bar{x}$ is centered at $\mu_{\bar{x}} = 4$ with standard deviation $\sigma_{\bar{x}} \approx 1.1547$ , exactly matching $\mu = 4$ and $\sigma/\sqrt{n} = \sqrt{4/3} \approx 1.1547$ .

Center, spread, and shape of a sample proportion

In a large population, 30% of adults have a library card, so the population proportion is p = 0.30. A researcher takes a random sample of n = 200 adults. Describe the sampling distribution of the sample proportion p-hat.

Center: $\mu_{\hat{p}} = p = 0.30$ .
Spread: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.30(0.70)}{200}} = \sqrt{\frac{0.21}{200}} = \sqrt{0.00105} \approx 0.0324$ .
Shape check (expected counts): $np = 200(0.30) = 60 \geq 10$ and $n(1-p) = 200(0.70) = 140 \geq 10$ , so the sampling distribution is approximately normal.
The 10% condition holds as long as the population has at least $10 \times 200 = 2000$ adults, which a large population easily satisfies.

The sampling distribution of $\hat{p}$ is approximately normal, centered at $\mu_{\hat{p}} = 0.30$ with standard deviation $\sigma_{\hat{p}} \approx 0.0324$ .

Frequently asked questions

What is the difference between a sampling distribution and the population distribution?

The population distribution is the spread of individual raw values, while the sampling distribution is the spread of a statistic like $\bar{x}$ computed from many samples. For the sample mean, the sampling distribution is narrower than the population, because averaging cancels out extremes: its standard deviation is $\sigma/\sqrt{n}$ , not $\sigma$ .

Why is the standard deviation of the sample mean sigma over the square root of n?

Averaging $n$ independent values cancels some of the random ups and downs, so the mean varies less than a single observation. The math works out to $\sigma/\sqrt{n}$ , which shrinks as $n$ grows. Quadruple the sample size and you cut the spread in half.

Does the central limit theorem apply to proportions?

Yes, indirectly. A sample proportion is a mean of 0s and 1s, so the same large-sample logic gives an approximately normal sampling distribution. For proportions the shape condition is stated as $np \geq 10$ and $n(1-p) \geq 10$ .

Do you ever actually collect a sampling distribution in practice?

Almost never. In a real study you take one sample, not all possible samples. The sampling distribution is a theoretical model of how your statistic would behave, and that model is what lets you attach a margin of error or a p-value to your single sample.