Sampling Distribution of the Sample Proportion Calculator

What is the sampling distribution of the sample proportion?

The sampling distribution of the sample proportion is the distribution of all the values p-hat could take over every possible random sample of the same size n from a population with proportion p. Each sample gives a different p-hat, and the collection of those values forms the distribution. Its mean is p, so the distribution is centred on the population proportion, and its spread is the standard error SE = √(p(1 − p)/n).

For example, with p = 0.5 and n = 100, the mean of the sampling distribution is 0.5 and the standard error is 0.05. Most sample proportions fall within 0.10 of 0.5 (two standard errors), and the probability that p-hat is below 0.55 is 0.8413 — the worked result at the top of this page. The probability here is a plain area under the curve, not a test result — what a p-value is explains the difference. If you are testing a proportion against a claim rather than finding a probability, use a one-proportion z-test instead.

How to find the probability for a sample proportion

Use the calculator above, or work by hand in five steps. If you are rusty on the conversion, calculating a z-score by hand walks through it.

1. Enter the population proportion p and the sample size n.
2. Choose below, above or between: do you want P(p-hat below a value), P(p-hat above a value) or P(p-hat between two values)?
3. Enter the p-hat value, or the lower and upper p-hat values for a "between" probability.
4. Compute the mean μ = p, the standard error SE = √(p(1 − p)/n) and the z-score z = (p-hat − p) ÷ SE (one z-score for each value in a "between" calculation).
5. Read the probability from Φ(z), the standard normal curve: Φ(z) for "below", 1 − Φ(z) for "above", or Φ(z2) − Φ(z1) for "between". Then check that np ≥ 10 and n(1 − p) ≥ 10, so the normal approximation is valid.

Worked example: p = 0.5, n = 100, P(p-hat < 0.55)

The mean is μ = p = 0.50 and the standard error is SE = √(0.5 × 0.5 ÷ 100) = 0.05. The z-score is z = (0.55 − 0.50) ÷ 0.05 = 1.00, so P(p-hat < 0.55) = Φ(1.00) = 0.8413. About 84.13% of samples of size 100 would give a sample proportion below 0.55. Try it with the calculator's first preset above.

Worked example: p = 0.3, n = 200, P(p-hat > 0.35)

Here SE = √(0.3 × 0.7 ÷ 200) = 0.0324 and z = (0.35 − 0.30) ÷ 0.0324 = 1.54, so P(p-hat > 0.35) = 1 − Φ(1.54) = 0.0614 — only about 6.14% of samples of 200 would give p-hat above 0.35.

Worked example: p = 0.4, n = 50, P(0.35 < p-hat < 0.45)

Here SE = √(0.4 × 0.6 ÷ 50) = 0.0693. The two z-scores are z1 = (0.35 − 0.40) ÷ 0.0693 = −0.72 and z2 = (0.45 − 0.40) ÷ 0.0693 = 0.72, so P(0.35 < p-hat < 0.45) = Φ(0.72) − Φ(−0.72) = 0.5295. About 52.95% of samples of 50 fall in that range.

Sampling distribution of the sample proportion formula

The calculator uses three formulas:

Quantity	Formula	What it means
Mean of the sampling distribution	μ = p	The distribution is centred on the population proportion.
Standard error	SE = √(p(1 − p)/n)	The spread of p-hat across samples; it shrinks as n grows.
z-score	z = (p-hat − p) ÷ SE	How many standard errors p-hat sits from p.

The probability then comes from the standard normal curve: P(p-hat < a value) = Φ(z), P(p-hat > a value) = 1 − Φ(z), and P(lower < p-hat < upper) = Φ(z2) − Φ(z1). To check a probability by hand, find the same tail area with a normal distribution calculator — or convert the z-score with a z-score calculator.

When is the normal approximation valid? (CLT conditions)

The normal curve is a good approximation to the sampling distribution only when three conditions hold:

• Random sample — the data come from a random sample (or a randomised experiment).
• Independence — the observations are independent, which is usually taken to mean the sample is no more than 10% of the population (the 10% condition).
• Success–failure condition — np ≥ 10 and n(1 − p) ≥ 10, so there are at least 10 expected successes and 10 expected failures.

When both counts are at least 10, the sampling distribution is close enough to normal for the probabilities on this page to be trustworthy. The calculator checks both counts for you and shows a warning when either one fails. A larger sample gives a smaller standard error: the margin of error for a proportion is roughly 1.96 × SE, and what the margin of error means explains the same idea for an estimate.

A warning case: p = 0.05 and n = 20 give np = 1, well below 10, so the normal approximation should not be used. The page will warn you, and the right move is an exact binomial calculation instead.

Why is the sampling distribution centred on p?

Because p-hat is an unbiased estimator of p: its expected value is E(p-hat) = p. If you took every possible sample of size n and averaged their p-hat values, you would get exactly p — the sample proportions balance around the population proportion, with no systematic tendency to fall above or below it. The standard error then measures how far a single sample's p-hat typically falls from p. The same standard error appears when you build confidence intervals from a mean and standard deviation: that interval is centred on the sample mean in the same way this distribution is centred on p.

Sampling distribution of the sample proportion FAQs

What is the sampling distribution of the sample proportion?

The sampling distribution of the sample proportion is the distribution of p-hat over all possible samples of size n from a population with proportion p. Its mean is p and its standard error is SE = √(p(1 − p)/n). When np ≥ 10 and n(1 − p) ≥ 10 it is approximately normal. For p = 0.5 and n = 100 the mean is 0.5 and the SE is 0.05.

What is the formula for the standard error of p-hat?

The formula is SE = √(p(1 − p)/n). For p = 0.4 and n = 50, SE = √(0.4 × 0.6 ÷ 50) = 0.0693.

How do you find the probability that p-hat is less than a value?

Convert p-hat to a z-score, z = (p-hat − p) ÷ SE, then read Φ(z), the area to the left of z under the standard normal curve. For p = 0.5, n = 100 and p-hat = 0.55, z = 1.00 and P(p-hat < 0.55) = 0.8413.

How do you find the probability that p-hat is between two values?

Find the z-score for each value, z1 and z2, then take Φ(z2) − Φ(z1). For p = 0.4, n = 50 and values 0.35 and 0.45, z = ±0.72 and P(0.35 < p-hat < 0.45) = 0.5295.

What are the conditions for the normal approximation to hold?

The sample must be random, the observations independent, and np ≥ 10 and n(1 − p) ≥ 10. These are the success–failure conditions for the central limit theorem. If they fail, the normal curve is a poor fit for the sampling distribution.

Why is the sampling distribution centred on p?

Because p-hat is an unbiased estimator of p: its expected value is E(p-hat) = p. Over many samples of the same size, the sample proportions average out to the population proportion, so the sampling distribution is centred on p, with spread given by the standard error.

What happens when np < 10 or n(1 − p) < 10?

The normal approximation becomes unreliable, so the probability it gives can be far from the true binomial probability. For example, p = 0.05 and n = 20 give np = 1, well below 10. In that case use an exact binomial calculation instead of the normal curve.

Can I use this calculator for a small sample?

You can use it for the mean and the standard error, but trust the probability only when np ≥ 10 and n(1 − p) ≥ 10. With a small sample that fails either condition, the sampling distribution is skewed and an exact binomial calculation is the right choice.

Keep going: you can browse all the calculator tools on the tools page. The normal-approximation rule used here follows the NIST/SEMATECH e-Handbook of Statistical Methods.