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Statistics

Sampling distribution of the sample proportion calculator

Enter the population proportion and the sample size to get the sampling distribution of p̂: its mean, standard error, whether the normal approximation holds, and the probability that the sample proportion falls in a range.

Decimal, fraction or percentage.

P(p̂ > 0.65) 0.1537
Mean of p̂, μp̂ = p0.6
Standard error σp̂0.0489898
np60
n(1 − p)40
Normal approximation OK?Yes, both ≥ 10
z-score1.021
0.65 0.50.60.7

The distribution of p̂ over repeated samples of 100: centre 0.6, standard error 0.04899. Shaded: P(p̂ > 0.65) = 0.1537.

Rangep̂ fromp̂ toShare of samples
±1 SE0.5510.64968.27%
±2 SE0.5020.69895.45%
±3 SE0.4530.74799.73%
Show the working, step by step

  1. The mean of the sample proportion equals the population proportion.

    μp̂ = p = 0.6

  2. Its standard error:

    σp̂ = √(p(1 − p) ÷ n) = √(0.6 × 0.4 ÷ 100) = √0.0024 = 0.0489898

  3. Check the success–failure condition for the normal shape.

    np = 100 × 0.6 = 60, n(1 − p) = 100 × 0.4 = 40 (both ≥ 10)

  4. Standardise the value(s):

    z = (0.65 − 0.6) ÷ 0.0489898 = 1.0206

  5. Probability from the standard normal distribution:

    P(p̂ > 0.65) = P(Z > 1.0206) = 1 − 0.846283 = 0.153717

The formulas

μp̂ = p σp̂ = √(p(1 − p) ÷ n) z = (p̂ − p) ÷ σp̂ Normal shape OK when np ≥ 10 and n(1 − p) ≥ 10

The sample proportion is an average of 0s and 1s, so the central limit theorem applies to it just as it does to a sample mean. The population standard deviation of a 0/1 variable is √(p(1 − p)), which is where the numerator comes from.

A worked example

The defaults: 60% of a large population support a proposal (p = 0.6), and you poll a random sample of 100 people. How likely is it that more than 65% of the sample say yes?

  1. The sampling distribution is centred on μp̂ = 0.6.
  2. Standard error: √(0.6 × 0.4 ÷ 100) = √0.0024 = 0.04899.
  3. Condition: np = 60 and n(1 − p) = 40, both at least 10, so the normal model is fine.
  4. z = (0.65 − 0.6) ÷ 0.04899 = 1.0206.
  5. P(p̂ > 0.65) = 1 − 0.846283 = 0.1537.

About 15% of samples of 100 would show more than 65% support, even though the true figure is 60%. That is a reminder of how much a sample of 100 can wander: two standard errors either side of 0.6 runs from 0.502 to 0.698, and about 95% of sample proportions land in that range.

Choose "between" with 0.65 and 0.7 and the calculator gives 0.1331, the share of samples with support between 65% and 70%.

What changes the spread

Only two things: p and n. Quadrupling the sample to 400 halves the standard error to 0.0245. Moving p away from 0.5 also narrows it, since p(1 − p) falls. The population size does not appear at all, as long as the sample is a small fraction of it; a poll of 1,000 people is about as precise in a country of 300 million as in a city of 1 million. When the sample is more than about 5% of the population, use the sampling distribution calculator with the finite population correction.

When the condition fails

With p = 0.05 and n = 50, np is only 2.5. The sampling distribution is then lumpy and skewed to the right, and a normal probability can be badly wrong in the tails. The calculator still shows the normal answer but flags it; the binomial distribution calculator gives the exact probability for the count x = n·p̂.

Common mistakes

  • Entering percentages as whole numbers. Type 0.65, not 65, for the bound on p̂.
  • Using p̂ from your sample in the standard error when the question gives the population p.
  • Skipping the success–failure check, especially for rare events.
  • Treating 1.96 × σp̂ as a probability. It is a margin; the probability comes from the z-score.
Sampling distribution of the sample proportion calculator: the worked example on this page, with its result and chart
Sampling distribution of the sample proportion calculator: the worked example above, at a glance.

Common questions

What is the sampling distribution of the sample proportion?

It describes how the proportion of successes p̂ = x/n varies from one random sample to the next. Its mean is the population proportion p, its standard deviation (the standard error) is √(p(1 − p)/n), and for large enough samples its shape is approximately normal.

What is the success–failure condition?

A check that the normal approximation is reasonable: the expected number of successes np and of failures n(1 − p) should both be at least 10. Some textbooks use 5 or 15 instead. The condition fails most often when p is close to 0 or 1, because the sampling distribution is then squashed against one end and skewed.

Should I use p or p̂ in the standard error?

Use the population value p when it is known or assumed, as in a probability question or a hypothesis test. When p is unknown and you are building a confidence interval, you have to plug in the sample proportion p̂ instead, and the result is an estimated standard error.

Why does the standard error peak at p = 0.5?

p(1 − p) is largest when p = 0.5, where it equals 0.25. That is why pollsters plan sample sizes with p = 0.5: it gives the widest, most cautious margin. At p = 0.1 the product is only 0.09, so the standard error is 60% of its value at 0.5.

Should I use a continuity correction?

It helps with small samples, because p̂ can only take the values 0, 1/n, 2/n and so on. The correction moves each bound by 1/(2n). With n = 100 that is 0.005, which changes the default answer only slightly. The continuity correction calculator shows the effect on counts.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.