Central Limit Theorem Calculator

The central limit theorem says that the mean of a large random sample is approximately normally distributed, whatever the shape of the population it was drawn from. The curve is centred on the population mean μ and its spread, the standard error, is the population standard deviation σ divided by √n. The sum of the sample is approximately normal as well, with mean nμ and standard deviation σ√n.

This calculator builds that normal curve from μ, σ and n and then answers the question you came with: how likely is a sample mean or sum below a value, above it or between two values, and which value marks a chosen lower-tail probability. Every result comes with the z-scores, the working, and a picture of the curve with the answer shaded.

• Enter the population mean μ, the population standard deviation σ (above 0) and the sample size n, a whole number.
• Choose the statistic: the sample mean or the sample sum. The sum is the mean multiplied by n, so the two answer the same questions on different scales.
• Leave the population size N empty when the population is large or the draws are made with replacement. Enter it when you sample without replacement from a finite population and n is a noticeable share of N, roughly above 5%. The spread is then multiplied by the finite population correction, and n must be smaller than N.
• Choose the question: the probability below a value, above a value, between two values, or the value that has a chosen probability to its left (a percentile of the sample mean or sum).
• Each probability is read from its own tail of the normal curve, so a very small probability keeps its digits. One below 1e-300, the smallest the calculator resolves, is shown as "< 1e-300".

μ and σ describe one draw from the population. With a finite population, multiply the standard error or the standard deviation of the sum by the correction; it is 1 when n = 1 and falls toward 0 as n approaches N. The population standard deviation σ here is the one for the whole population, with N in the divisor. If you have the version with N − 1 in the divisor, multiply it by √((N − 1)/N) first.

The mean of a normal population is exactly normal for every n. For any other population with a finite variance the normal curve is an approximation that gets better as n grows, and how large n has to be depends on the shape of the population (see below).

Choose Load example. A population has mean μ = 100 and standard deviation σ = 15. What is the chance that the mean of a random sample of n = 36 exceeds 104?

1. The sample mean has mean 100.
2. Its standard error is σ / √n = 15 / √36 = 15 / 6 = 2.5.
3. z = (104 − 100) / 2.5 = 1.6.
4. P(sample mean > 104) = P(Z > 1.6) = 0.054799, about 5.48%. The complementary probability is 0.945201.

A single draw is far more variable: with σ = 15 the chance that one observation exceeds 104 is 0.394863, about 39.5%. Averaging 36 draws makes a result that far above the mean rare. Choose Probability between two values with a = 95 and b = 105 to find that 95.45% of sample means fall within 5 of the mean, and ask for the value at a lower-tail probability of 0.975 to see that 97.5% of sample means are below 104.89991.

A lift carries 25 adults whose weights have mean 75 kg and standard deviation 12 kg. What is the chance that their total weight exceeds 1,950 kg? Choose the sample sum, enter μ = 75, σ = 12 and n = 25, and ask for the probability above 1950.

1. The total has mean nμ = 25 × 75 = 1,875 kg.
2. Its standard deviation is σ√n = 12 × √25 = 60 kg.
3. z = (1,950 − 1,875) / 60 = 1.25.
4. P(total > 1,950) = P(Z > 1.25) = 0.10565, about 10.6%.

The same event as a mean: a total above 1,950 kg is an average weight above 78 kg, and with the sample mean (standard error 12 / √25 = 2.4) the z-score is (78 − 75) / 2.4 = 1.25 again. The mean and the sum always give the same probability for the same event.

A test was taken by all 200 employees of a company, with mean 70 and standard deviation 10. A manager reads 50 of the papers, chosen without replacement. What is the chance that the sample mean exceeds 72? Enter μ = 70, σ = 10, n = 50 and N = 200.

The finite population correction is √((200 − 50) / (200 − 1)) = 0.8682. Leaving it out overstates the chance by about half: 0.07865 instead of 0.051667. The sample is a quarter of the population, so it carries more information than 50 draws from an unlimited population would.

The standard error falls with the square root of n, so four times as many draws are needed to halve it. For a population with σ = 15:

To reach a standard error of 1 with σ = 15 you need n = (15 / 1)² = 225 draws, and 900 for 0.5. To plan a sample from a target margin of error instead, use the sample size calculator.

Sampling without replacement removes each unit that has been drawn, so the sample mean varies less than it would with replacement. The effect is small while the sample is a small share of the population and large as it approaches the whole population. For a population of N = 1,000:

At 5% the correction is about 0.975, so it changes the standard error by only about 2.5%, and below that it is usually ignored. With a sample of the whole population there is no sampling variation at all, which is why the calculator refuses n = N.

The calculator always uses the normal curve. That is exact for a normal population and an approximation for any other, and the approximation is only as good as n allows. Take a strongly right-skewed population, the exponential distribution with mean 1 and standard deviation 1. The normal curve says that 5% of sample means lie above μ + 1.645 standard errors and 5% below μ − 1.645 standard errors. The exact shares, from the gamma distribution of the mean of n exponential draws (shape n, scale 1/n), are:

The error shrinks roughly in proportion to 1/√n but it is still visible at n = 30: the upper tail is 5.90% instead of 5% and the lower tail 3.72%. This is why n = 30 is a guideline and not a guarantee. A symmetric population needs far fewer draws. The sum of only 12 uniform (0, 1) draws has mean 6 and standard deviation 1, and P(sum > 7) is 0.158655 by the normal curve against an exact 0.160727.

The mean of exponential draws can be evaluated exactly with the gamma distribution calculator, and the exponential population itself with the exponential distribution calculator. If you cannot say anything about the shape of the population, the Chebyshev theorem calculator gives a bound that holds for every distribution, although a much weaker one.

A sample proportion is the mean of n draws that are 1 for a success and 0 for a failure, so the calculator handles it with μ = p and σ = √(p(1 − p)). For a fair coin, p = 0.5 and σ = 0.5. With n = 100 flips the standard error is 0.5 / √100 = 0.05, and the chance that the proportion of heads exceeds 0.6 is P(Z > 2) = 0.02275.

The number of successes is the sample sum: with the sample sum, μ = 0.5, σ = 0.5 and n = 100 the count of heads has mean 50 and standard deviation 5. The chance of 60 or more heads is 0.02275 by the calculator with 60 as the value. The count is a whole number, so the normal curve is better placed at 59.5 (the continuity correction): that gives 0.028717 against an exact binomial probability of 0.028444. For exact counts use the binomial distribution calculator, and to test a proportion the one-proportion z-test.

• Probability is the answer to the question you chose, and Complementary probability is the chance of the opposite event, computed from the other tail of the curve rather than as 1 minus the first.
• Mean of the sampling distribution is μ for the sample mean and nμ for the sum. Standard error is σ/√n for the sample mean; for the sum the card reads Standard deviation of the sum, σ√n. Both include the finite population correction when N is given.
• Z-score says how many standard errors the value lies from the mean. Values beyond about ±2 are unusual, and beyond ±3 rare, for a sample mean.
• For a percentile question the featured card is the x value: the value of the sample mean or sum that has your probability below it.
• A value more than 40 standard deviations from the mean is reported with its probability, but the curve is not drawn because nothing of it would be visible.

• The draws are independent. A random sample from a large population, or with replacement, qualifies. Sampling without replacement from a small population does not, which is what the finite population correction adjusts for. Clustered or serially correlated data need a different standard error.
• σ is the population value. If you only have a sample, its standard deviation is an estimate of σ that is reliable for large n. For small samples the standardized mean follows a t distribution rather than a normal one; use the t distribution calculator or a confidence interval built for it.
• The theorem is about the mean, not the data. Individual observations keep the shape of the population. The chance that one observation exceeds 104 is 0.394863 in the first example; the chance that a mean of 36 does is 0.054799.
• Standard deviation is not standard error. The standard deviation describes single observations and the standard error describes the sample mean. Using σ where σ/√n belongs makes every probability about a sample mean far too large.
• n = 30 is a guideline. A skewed population needs more, and a population without a finite variance, such as the Cauchy distribution, never gives a normal sample mean at any n.
• Do not apply the correction by habit. Leave N empty for draws with replacement and for populations so large that n is a tiny share of them.

Each line gives the chance that the mean of 36 draws from μ = 100, σ = 15 exceeds 104 (0.054799), or the 97.5th percentile of that mean (104.89991).

For a sum, use the mean nμ and the standard deviation σ√n in the same commands. The calculator adds the finite population correction, the z-scores and the working.

The z-test calculator and the z-score calculator use the same standardization as this page, applied to an observed sample mean or a single value.