Sampling Distribution of a Sample Proportion: Mean, Standard Error and the Large Counts Condition

In 30 seconds

A sample proportion $\hat{p}$ changes from sample to sample. Its sampling distribution describes that variation.
The mean of $\hat{p}$ is the true proportion $p$, so $\hat{p}$ is unbiased. Its standard deviation (the standard error) is $\sqrt{p(1-p)/n}$.
When the Large Counts condition holds, $np \ge 10$ and $n(1-p) \ge 10$, the distribution of $\hat{p}$ is approximately normal.
The case answer: if 35% of a town’s adults are vaccinated, a random sample of 200 shows 28% or less only about 1.9% of the time, so a 28% result is a real warning sign.
Exam point: check the 10% condition (sample under a tenth of the population) and the Large Counts condition separately, and name them both.

The case: is the clinic’s vaccination rate really lower?

A regional health office knows from its records that 35% of adults in the town had a flu vaccination last winter. (The figures are illustrative.) A community clinic in one neighbourhood takes a simple random sample of 200 of its adult patients and finds that only 56 of them, 28%, were vaccinated.

The clinic manager has a practical question. Is 28% just the normal wobble you expect from a sample of 200, or is it evidence that her neighbourhood really lags behind the town? If it is just noise, a costly outreach campaign would be wasted. If it is real, people are missing protection.

To answer, she needs to know how much a sample proportion naturally varies when the true rate is 35%. That is exactly what the sampling distribution of the sample proportion tells us. It is also the foundation of every confidence interval and significance test for a proportion that comes later.

The concept: the sampling distribution of the sample proportion

Imagine taking a random sample of 200 people, recording the proportion vaccinated, then putting them back and drawing a fresh sample of 200, again and again. Each sample gives a slightly different $\hat{p}$: 0.33, 0.37, 0.31, 0.36 and so on. If you plotted thousands of these values you would see a clear pattern. That pattern is the sampling distribution.^[1]

Three features describe it: its centre, its spread and its shape. The centre and spread follow from the binomial distribution. If $X$ is the number of “successes” in the sample, then $\hat{p} = X/n$. Dividing a binomial count by $n$ divides its mean by $n$ and its variance by $n^2$, which gives the formulas below.

The shape needs a condition. A proportion near 0 or 1 in a small sample gives a lopsided distribution, because $\hat{p}$ cannot go below 0 or above 1. With enough expected successes and failures, the shape becomes close to normal.

Key termSampling distribution of $\hat{p}$: the distribution of the values of the sample proportion $\hat{p}$ over all possible random samples of size $n$ from the same population. Its standard deviation is called the standard error of $\hat{p}$ when it is estimated from data.

\[ \mu_{\hat{p}} = p, \qquad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}, \qquad \hat{p} \approx N\!\left(p,\ \frac{p(1-p)}{n}\right) \text{ if } np \ge 10 \text{ and } n(1-p) \ge 10 \]

Here $p$ is the true population proportion (0.35 in the case) and $\hat{p}$ (“p-hat”) is the proportion in one sample. $n$ is the sample size. $\mu_{\hat{p}}$ is the mean of all possible sample proportions, and $\sigma_{\hat{p}}$ is their standard deviation. The standard deviation formula assumes independent observations; when sampling without replacement, the usual rule is that the sample is less than 10% of the population. The condition $np \ge 10$ and $n(1-p) \ge 10$ is the Large Counts condition: it asks for at least 10 expected successes and 10 expected failures. Some textbooks use 5 instead of 10; 10 is the safer and more common choice in AP Statistics.^[2]

Statistics Made Simple explains sampling distributions of proportions with worked examples, diagrams and tables.Get the textbook · $12

Why this method fits the case

The clinic has one sample proportion and a claimed population value. To judge whether the sample is surprising, we need the distribution of $\hat{p}$ assuming the claim is true. Then we can ask how often a value as low as 0.28 would occur by chance.

Check the conditions:

Random: the clinic drew a simple random sample of its patients. Good.
10% condition: 200 patients is far less than 10% of the adults on the clinic’s list (assume the list has several thousand). So independence is reasonable.
Large Counts: $np = 200 \times 0.35 = 70$ and $n(1-p) = 130$. Both are well above 10, so a normal model is fine.

When is this not the right tool?

When the sample is not random. A sample of patients who happened to visit on one morning, or volunteers who answered an online poll, can be biased. No formula fixes bias; the standard error only measures random variation.
When the counts are small. With $n = 50$ and $p = 0.08$, $np = 4$, so the normal model is poor. Use exact binomial probabilities instead.
When you are dealing with a mean, not a proportion. For averages of measurements, use the sampling distribution of $\bar{x}$, with standard error $\sigma/\sqrt{n}$.

Setting it up

Quantity	Symbol	Value
Population proportion (claimed)	$p$	0.35
Sample size	$n$	200
Number vaccinated in sample	$X$	56
Sample proportion	$\hat{p}$	$56/200 = 0.28$
Expected successes / failures	$np$, $n(1-p)$	70, 130
Standard deviation of $\hat{p}$	$\sigma_{\hat{p}}$	$\sqrt{0.35 \times 0.65 / 200} \approx 0.0337$

Use the claimed $p$, not $\hat{p}$, in the standard deviation here. We are asking what the sampling distribution would look like if the town-wide rate applied to the clinic.

Solving it

Worked example

Assume the true proportion vaccinated is $p = 0.35$. For a random sample of $n = 200$, find $P(\hat{p} \le 0.28)$.

Step 1Centre: $\mu_{\hat{p}} = 0.35$.

Step 2Spread: $\sigma_{\hat{p}} = \sqrt{0.35 \times 0.65 / 200} = \sqrt{0.0011375} \approx 0.03373$.

Step 3Shape: $np = 70 \ge 10$ and $n(1-p) = 130 \ge 10$, so $\hat{p}$ is approximately normal.

Step 4Standardise: $z = (0.28 – 0.35)/0.03373 \approx -2.08$.

Step 5$P(\hat{p} \le 0.28) = P(Z \le -2.08) \approx 0.019$.

If the clinic’s patients were like the town, a result this low would happen in only about 1.9% of samples. That is strong evidence the clinic’s rate is lower.

The second example looks at the middle of the distribution: how close to the truth a sample usually lands.

Worked example

Still with $p = 0.35$ and $n = 200$, find the probability that a sample proportion lands within 5 percentage points of the truth, that is between 0.30 and 0.40.

Step 1Standardise the upper limit: $z = (0.40 – 0.35)/0.03373 \approx 1.48$.

Step 2By symmetry the lower limit gives $z \approx -1.48$.

Step 3$P(-1.48 \le Z \le 1.48) \approx 0.861$.

About 86% of samples of 200 fall within 5 points of the true 35%; about 14% miss by more than that.

The third example shows what happens when the Large Counts condition only just holds, and compares the normal answer with the exact binomial one.

Worked example

A phone maker says 8% of its handsets need a repair in the first year. (The figures are illustrative.) A retailer sells 150 of these handsets to random customers. Find the probability that 12% or more of them need repair, and check the conditions.

Step 1Large Counts: $np = 150 \times 0.08 = 12$ and $n(1-p) = 138$. Both are at least 10, so the normal model is acceptable but not generous.

Step 2$\sigma_{\hat{p}} = \sqrt{0.08 \times 0.92 / 150} \approx 0.02215$.

Step 3$z = (0.12 – 0.08)/0.02215 \approx 1.81$, so $P(\hat{p} \ge 0.12) \approx P(Z \ge 1.81) \approx 0.035$.

Step 4Exact check: 12% of 150 is 18 handsets, and the binomial probability $P(X \ge 18)$ is about 0.031.

The normal answer (0.035) is close to the exact one (0.031). With smaller counts the gap would grow.

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What the answer tells us

A sample proportion of 0.28 is about 2.1 standard errors below 0.35. Results that far out happen only about 2 times in 100 by chance alone. So the clinic manager has good reason to believe her neighbourhood really is behind the town average. That does not tell her why, but it tells her the gap is probably not a fluke, and outreach is worth planning.

Reasonableness check. The standard error of 0.034 means a typical sample of 200 misses the truth by about 3 to 4 percentage points. A 7-point gap is about twice that, which feels unusual but not impossible, matching a probability of about 2%. The binomial gives a similar answer: $P(X \le 56)$ when $X \sim B(200, 0.35)$ is about 0.021.

Sensitivity to sample size. Sample size is the main lever. Figure 1 shows the sampling distribution for three sample sizes. With only 50 patients the standard error is 0.067, and a result of 0.28 or lower would happen about 15% of the time, so it would prove nothing. With 200 it happens about 1.9% of the time. With 800 the standard error is 0.017, and 0.28 would be over 4 standard errors away, with a probability of about 0.00002. Because the standard error is proportional to $1/\sqrt{n}$, you need four times the sample to halve it.

Sampling distribution of p̂ when p = 0.35 density

n = 50 (SE 0.067)n = 200 (SE 0.034)n = 800 (SE 0.017)

Figure 1. All three curves are centred on the true proportion 0.35, because p̂ is unbiased. Larger samples give narrower curves, so a result of 0.28 becomes more and more surprising. Illustrative figures.

The value of $p$ matters too. The product $p(1-p)$ is largest at $p = 0.5$, so proportions near one-half are hardest to pin down. A proportion near 0.05 or 0.95 has a smaller standard error, but it also needs a larger sample to satisfy the Large Counts condition.^[3]

Common mistakeUsing the sample size $n$ instead of $\sqrt{n}$ in your thinking, or forgetting the square root in the formula. $\sqrt{p(1-p)/n}$ with $p = 0.35, n = 200$ is 0.0337, not 0.0011. If your standard error is tiny compared with typical sampling error of a few percentage points, you have probably forgotten the root.

Common mistakeThinking a bigger population needs a bigger sample. As long as the sample is under 10% of the population, the population size does not appear in the formula. A random sample of 1,000 is about as precise for a city of 100,000 as for a country of 100 million.

Exam tipWhen asked to justify a normal model for $\hat{p}$, show the numbers: “$np = 70 \ge 10$ and $n(1-p) = 130 \ge 10$, so the Large Counts condition is met.” Then state the 10% condition in context: “200 is less than 10% of all adult patients.” Sampling distributions are a core topic in AP Statistics, and a bare “conditions are met” usually earns little.^[4] Our statistics guide links this step to confidence intervals and tests that follow.

The Statistics Made Simple textbook builds sampling distributions from simulations up to formulas, so you can see where $\sqrt{p(1-p)/n}$ comes from. For practice, the Statistics Made Simple Practice Questions book includes many condition-checking and probability questions for $\hat{p}$.

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Practice questions

Use the normal model where the conditions allow. Round standard errors to 4 decimal places.

EasyQ1. A fair coin is tossed 100 times. Find the mean and standard deviation of the proportion of heads. Show worked answer

$\mu_{\hat{p}} = 0.5$.
$\sigma_{\hat{p}} = \sqrt{0.5 \times 0.5/100} = \sqrt{0.0025} = 0.05$.

EasyQ2. 20% of a city’s households own an electric car (illustrative). For random samples of 400 households, state the mean and standard deviation of $\hat{p}$, and check the Large Counts condition. Show worked answer

$\mu_{\hat{p}} = 0.20$; $\sigma_{\hat{p}} = \sqrt{0.2 \times 0.8/400} = 0.02$.
$np = 80 \ge 10$, $n(1-p) = 320 \ge 10$: condition met.

MediumQ3. Using Q2, find the probability that a sample of 400 households shows 25% or more owning an electric car. Show worked answer

$z = (0.25 – 0.20)/0.02 = 2.5$.
$P(\hat{p} \ge 0.25) = P(Z \ge 2.5) \approx 0.0062$.
Such a sample would be very unusual if the true rate were 20%.

MediumQ4. An airline says 90% of its flights arrive on time. A random sample of 120 flights is taken. Find the probability that 85% or fewer arrive on time. Show worked answer

Check: $np = 108$, $n(1-p) = 12$, both $\ge 10$.
$\sigma_{\hat{p}} = \sqrt{0.9 \times 0.1/120} \approx 0.0274$.
$z = (0.85 – 0.90)/0.0274 \approx -1.83$; $P(Z \le -1.83) \approx 0.034$.

ExamQ5. In a large town, 60% of voters support a new park. A polling company takes a random sample of 500 voters. (a) Describe the sampling distribution of $\hat{p}$, justifying your answer. [4] (b) Find the probability that the sample shows at least 63% support. [2] (c) What sample size would make the standard deviation of $\hat{p}$ equal to 0.02? [2] Show worked answer

(a) Mean 0.60 [B1]; SD $= \sqrt{0.6 \times 0.4/500} \approx 0.0219$ [B1]. Approximately normal since $np = 300 \ge 10$, $n(1-p) = 200 \ge 10$ [B1]; 500 is less than 10% of voters, so independence is reasonable [B1].
(b) $z = (0.63 – 0.60)/0.0219 \approx 1.37$ [M1]; $P(Z \ge 1.37) \approx 0.085$ [A1].
(c) $0.6 \times 0.4/n = 0.02^2$ [M1], so $n = 0.24/0.0004 = 600$ [A1].

Frequently asked questions

Why is the sample proportion called unbiased?

Because the mean of its sampling distribution equals the true proportion: $\mu_{\hat{p}} = p$. Individual samples miss in both directions, but they do not miss systematically high or low. That property holds only for random samples; a biased sampling method can push every $\hat{p}$ the same way.

What is the difference between standard deviation and standard error here?

$\sqrt{p(1-p)/n}$ uses the true $p$ and is the standard deviation of the sampling distribution. In practice $p$ is unknown, so we plug in $\hat{p}$ and call $\sqrt{\hat{p}(1-\hat{p})/n}$ the standard error. Our post on standard deviation vs standard error explains the distinction in more detail.

Is this the central limit theorem?

It is a special case. A proportion is the mean of 0s and 1s, so the central limit theorem says it becomes approximately normal for large $n$. The Large Counts condition is the practical rule for “large enough” when the variable is a proportion. See our central limit theorem guide for the general idea.

What do I do if the Large Counts condition fails?

Use the binomial distribution directly, since $X = n\hat{p}$ is binomial. A calculator or software gives exact probabilities. For inference, methods such as the “plus four” interval or exact tests are designed for small counts.

References

Illowsky, B. and Dean, S. (2023) Introductory Statistics 2e. Houston, TX: OpenStax. https://openstax.org/details/books/introductory-statistics-2e
Starnes, D. S. and Tabor, J. The Practice of Statistics. New York: W. H. Freeman.
Agresti, A., Franklin, C. and Klingenberg, B. Statistics: The Art and Science of Learning from Data. Boston: Pearson.
College Board. AP Statistics Course and Exam Description. New York: College Board.

Sampling Distribution of a Sample Proportion: Mean, Standard Error and the Large Counts Condition

The case: is the clinic’s vaccination rate really lower?

The concept: the sampling distribution of the sample proportion

Why this method fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

Why is the sample proportion called unbiased?

What is the difference between standard deviation and standard error here?

Is this the central limit theorem?

What do I do if the Large Counts condition fails?

References

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