The Central Limit Theorem and large samples

The theorem

If X1, ..., Xn is a random sample from any distribution with mean μ and variance σ², then for large n

X̄ ≈ N(μ, σ²/n) (approximately).

The larger n is, the better the approximation. The population can be discrete (binomial, Poisson) or continuous (uniform, exponential-shaped).

If the population is already normal, X̄ is exactly normal for any n and the theorem is not needed.

Using it

Find μ and σ² of the population first:

• X ~ Po(4): μ = 4, σ² = 4, so for n = 50, X̄ ≈ N(4, 0.08).
• X ~ U[0, 10]: μ = 5, σ² = 100/12, so for n = 40, X̄ ≈ N(5, 100/480).

Then standardise as usual. When justifying, examiners want a precise statement, such as 'n is large, so by the Central Limit Theorem X̄ is approximately normally distributed'; vague comments earn nothing.

Unknown variance, large sample

For a large sample, s² is a good estimate of σ², so use x̄ ± z s/sqrt(n) and z = (x̄ - μ0)/(s/sqrt(n)). State that you are using σ² ≈ s². (A t-distribution is not required in S3.)

Example: n = 80, x̄ = 12.4, s = 3.1. 95% interval: 12.4 ± 1.96 × 3.1/sqrt(80) = (11.72, 13.08).

Worked example

A claim says the mean waiting time is 15 minutes. A random sample of 60 waits has x̄ = 15.9 and s = 3.2. Test at the 5% level whether the mean is more than 15.

1. H0: μ = 15, H1: μ > 15.
2. n is large, so by the CLT X̄ is approximately normal, with σ estimated by s.
3. z = 0.9/(3.2/sqrt(60)) = 2.1786 > 1.6449

Answer: Reject H0: there is evidence that the mean waiting time is more than 15 minutes.

Check yourself

1. X ~ Po(4). Use the Central Limit Theorem to find P(X̄ > 4.3) for a random sample of 50. Give your answer to 4 decimal places.
   Show answer

   0.1444. X̄ ≈ N(4, 4/50); z = 0.3/sqrt(0.08) = 1.0607; P = 0.1444 (0.1446 using z = 1.06 in the tables).

2. X ~ U[0, 10]. Use the Central Limit Theorem to find P(X̄ < 4.5) for a random sample of 40. Give your answer to 4 decimal places.
   Show answer

   0.1367. Var(X) = 100/12; X̄ ≈ N(5, 100/480); z = -1.0954; P = 0.1367 (0.1357 using z = -1.10 in the tables).

3. A random sample of 80 has x̄ = 12.4 and s = 3.1. Find the lower limit of an approximate 95% confidence interval for μ. Give your answer to 3 significant figures.
   Show answer

   11.7. 12.4 - 1.96 × 3.1/sqrt(80) = 11.7207.

4. A random sample of 60 has x̄ = 15.9 and s = 3.2. Calculate the test statistic for H0: μ = 15. Give your answer to 3 significant figures.
   Show answer

   2.18. 0.9/(3.2/sqrt(60)) = 2.1786.

5. X ~ B(20, 0.3). Find the variance of the mean of a random sample of 50 observations of X.
   Show answer

   0.084. Var(X) = np(1 - p) = 4.2, so Var(X̄) = 4.2/50 = 0.084.

6. Why is the Central Limit Theorem needed to find P(X̄ > 4.3) when X ~ Po(4)?
   • X is normal, but its variance is unknown
   • X is not normally distributed, so X̄ is only approximately normal for large n
   • The sample is small
   • The Poisson distribution has no mean
   Show answer

   X is not normally distributed, so X̄ is only approximately normal for large n. The CLT gives approximate normality of X̄ for a non-normal population when n is large.

7. For which situation is the Central Limit Theorem not needed?
   • Finding a confidence interval from a large sample of an unknown distribution
   • Testing a mean using a large sample of Poisson data
   • Approximating the mean of 100 uniform observations
   • Finding the distribution of X̄ when the population is normal
   Show answer

   Finding the distribution of X̄ when the population is normal. If the population is normal, X̄ is exactly normal for any n.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit S3
• Pearson Edexcel IAL Statistics S3 (WST03) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.