Sampling Distribution of the Sample Proportion Calculator

Enter a population proportion and sample size to get the mean and standard error of p̂, then compute the probability that a sample proportion falls below, above, or within a range using the normal approximation.

Quick Facts

Method
Mean = p, Standard error = √(p(1−p)/n)
Probabilities use the normal approximation (valid when np ≥ 10 and n(1−p) ≥ 10).

Your Results

Calculated
Standard error of p̂
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SE = √(p(1−p)/n)
Mean of p̂
-
Equals the population proportion p
Z-score of p̂
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(p̂ − p) / SE
Probability
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P(P̂ ≤ your p̂)

Ready

Enter a population proportion, sample size, and sample proportion, then calculate.

Understanding the Sampling Distribution of the Sample Proportion

If you draw a random sample of size n from a population in which a proportion p have some characteristic, the observed sample proportion p̂ = (number of successes) / n will not exactly equal p. Draw another sample and you get a slightly different p̂. The sampling distribution of the sample proportion is the probability distribution of all these possible p̂ values across every sample of size n. This calculator gives you its center, its spread, and the probability that p̂ lands in a range you care about.

The formulas

When observations are independent (large population, or sampling with replacement), the sampling distribution of p̂ has:

  • Mean: μp̂ = p. The sample proportion is an unbiased estimator, so on average it equals the true population proportion.
  • Standard error (standard deviation): SE = √(p(1−p)/n). This shrinks as n grows — quadrupling the sample size halves the standard error.
  • Shape: by the Central Limit Theorem, p̂ is approximately normal when the sample is large enough. The z-score of a particular sample proportion is z = (p̂ − p) / SE, and the tail probability comes from the standard normal distribution.

The success/failure condition

The normal approximation is trustworthy only when the expected counts of successes and failures are both reasonably large. The standard rule is np ≥ 10 and n(1−p) ≥ 10 (some introductory texts relax this to 5). When p is near 0 or 1, or n is small, the true distribution of p̂ is skewed and discrete, and you should use the exact binomial distribution instead of this normal approximation.

A worked example

Suppose 50% of voters support a measure (p = 0.5) and you poll n = 100 people. Then SE = √(0.5 × 0.5 / 100) = √(0.0025) = 0.05. The probability that your poll shows 60% support or less corresponds to z = (0.60 − 0.50) / 0.05 = 2.0, giving P(P̂ ≤ 0.60) ≈ 97.7%. So a poll landing above 60% would be a genuinely surprising result (about 2.3% of the time) even though the true support is only 50%.

The finite population correction

The formula SE = √(p(1−p)/n) assumes the population is effectively infinite or that you sample with replacement. If you sample without replacement from a small population of size N and n is more than about 5% of N, multiply the standard error by the finite population correction factor √((N−n)/(N−1)). This calculator uses the standard uncorrected formula, which is the right choice for the vast majority of survey and quality-control settings.

Frequently Asked Questions

What is the standard error of a sample proportion?
SE = √(p(1−p)/n), where p is the population proportion and n is the sample size. It is the standard deviation of the sampling distribution of p̂ and tells you how much a sample proportion typically differs from the true value. With p = 0.5 and n = 100, SE = √(0.25/100) = 0.05.
How is the standard error different from the standard deviation?
The standard deviation of the underlying Bernoulli variable is √(p(1−p)). The standard error of the sample proportion is that value divided by √n: SE = √(p(1−p))/√n = √(p(1−p)/n). The standard error describes the variability of the average (the proportion) across samples, not the variability of individual observations.
When can I use the normal approximation?
Use it when both np ≥ 10 and n(1−p) ≥ 10. If either expected count is small, the distribution of p̂ is skewed and you should switch to the exact binomial distribution. The calculator flags this condition automatically.
Why does the standard error shrink with larger samples?
Because n appears in the denominator under a square root: SE = √(p(1−p)/n). Larger samples average out random variation, so estimates cluster more tightly around the true p. To cut the standard error in half you need four times the sample size.

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