What is the sampling distribution?
Find the sampling distribution of a sample proportion p̂, a sample mean x̄ or a sample sum, and the probability that one sample gives a value below, above or between numbers you choose.
- Probability
- 0.8511
The probability is 0.8511, with a sampling distribution of mean 0.4 and standard error 0.034641.
- Mean of the sampling distribution
- 0.4
- Standard error
- 0.034641
- z of the lower value
- −1.4434
- z of the upper value
- 1.4434
- Normal approximation
- np and n(1 − p) are both more than 5, so the normal approximation is reasonable.
Probability: 0.8511. The probability is 0.8511, with a sampling distribution of mean 0.4 and standard error 0.034641.
The sampling distribution
How to calculate
Finds the sampling distribution of a sample proportion, mean or sum: its mean, its standard error, and the probability that a sample statistic falls in a range.
Example with the default inputs (Sampling distribution of Sample proportion (p̂), Population proportion (p) 0.4, Sample size (n) 200, Probability Between, Lower value 0.35, Upper value 0.45): The probability is 0.8511, with a sampling distribution of mean 0.4 and standard error 0.034641.
Method: p̂ ~ N(p, √(p(1 − p) ÷ n)); x̄ ~ N(μ, σ ÷ √n); Σx ~ N(nμ, √n σ); z = (value − mean) ÷ standard error; the probability is the normal area over the range.
- The samples are random and independent, and the normal shape comes from the central limit theorem.
- For p̂, the normal approximation is reasonable when np and n(1 − p) are both more than 5.
Worked examples
Each example is checked against the calculator on every build.
- Sampling distribution of Sample proportion (p̂), Population proportion (p) 0.4, Sample size (n) 200, Probability Between, Lower value 0.35, Upper value 0.45 gives Probability 0.851085, Mean of the sampling distribution 0.4, Standard error 0.034641, z of the lower value -1.443376, z of the upper value 1.443376.Source: OpenStax, Introductory Business Statistics, §7.3 The Central Limit Theorem for Proportions (mean p, standard deviation √(p(1 − p) ÷ n)), https://openstax.org/books/introductory-business-statistics/pages/7-3-the-central-limit-theorem-for-proportions (retrieved 2026-10-05)
- Sampling distribution of Sample mean (x̄), Population mean (μ) 100, Population standard deviation (σ) 12, Sample size (n) 36, Probability Below, Upper value 103 gives Probability 0.933193, Mean of the sampling distribution 100, Standard error 2, z of the upper value 1.5.Source: OpenStax, Introductory Statistics 2e, §7.1 The Central Limit Theorem for Sample Means (Averages) (x̄ ~ N(μ, σ ÷ √n); Example 7.1: μ = 90, σ = 15, n = 25, P(85 < x̄ < 92) = 0.6997), https://openstax.org/books/introductory-statistics-2e/pages/7-1-the-central-limit-theorem-for-sample-means-averages (retrieved 2026-10-05)
How it works
This page uses the same model as the central limit theorem calculator, and opens on a sample proportion. Pick the statistic, then give the population and the sample size n:
| Statistic | Mean of its distribution | Standard error |
|---|---|---|
| Sample proportion p̂ | p | √(p(1 − p) ÷ n) |
| Sample mean x̄ | μ | σ ÷ √n |
| Sample sum Σx | n × μ | √n × σ |
The statistic is treated as normal with that mean and standard error. For each value you type, z = (value − mean) ÷ standard error, and:
- Below a value b: P = Φ(z_b).
- Above a value a: P = Φ(−z_a), so a small tail keeps its digits.
- Between a and b: P = Φ(z_b) − Φ(z_a).
Rules:
- p is strictly between 0 and 1; μ is between −10¹² and 10¹²; σ is more than 0 and at most 10¹²; n is a whole number from 1 to 10⁹.
- The value minus the mean, n × μ and p(1 − p) ÷ n are exact on the decimals you type; only the square root and the normal area use floating point.
- A lower value above the upper value gives no answer. So does a z-score too large to show, with a message.
- The "Normal approximation" line says, for p̂, whether np and n(1 − p) are both more than 5; for x̄ and Σx, that the result is exact for a normal population and approximate otherwise.
Assumptions
- The samples are random, and the observations in a sample are independent.
Worked examples by hand
Sample proportion, p = 0.4, n = 200, between 0.35 and 0.45 (the default). The standard error is √(0.4 × 0.6 ÷ 200) = √0.0012 = 0.034641. z = −0.05 ÷ 0.034641 = −1.4434 and z = 1.4434. P = Φ(1.4434) − Φ(−1.4434) = 0.8511. np = 80 and n(1 − p) = 120, both more than 5.
Sample mean, μ = 100, σ = 12, n = 36, below 103. The standard error is 12 ÷ 6 = 2. z = 3 ÷ 2 = 1.5, and P = Φ(1.5) = 0.9332.
Other questions people ask
What is a sampling distribution?
It is the distribution of a statistic, such as a sample mean or a sample proportion, over all the random samples of the same size you could take. Each sample gives a slightly different value; the sampling distribution shows how those values spread.
What is the sampling distribution of the sample proportion?
For samples of size n from a population with proportion p, the sample proportion p̂ has mean p and standard error √(p(1 − p) ÷ n). For large samples it is close to normal. With p = 0.4 and n = 200, the standard error is 0.0346.
What is the sampling distribution of the sample mean?
The sample mean x̄ has mean μ and standard error σ ÷ √n, and by the central limit theorem it is close to normal for large n. With σ = 12 and n = 36, the standard error is 2.
How do I find a probability from a sampling distribution?
Subtract the mean of the sampling distribution from your value and divide by the standard error to get z. Then read the normal area. For p = 0.4 and n = 200, P(0.35 < p̂ < 0.45) = 0.8511.
When is the normal approximation good for a proportion?
A usual check is that np and n(1 − p) are both more than 5, so the sample is likely to hold enough of each kind of outcome. The page shows this check under the result.
How does sample size change the sampling distribution?
The standard error falls with the square root of n. Four times as many observations halve the spread, so sample statistics sit closer to the population value.