The Sample Mean and the Central Limit Theorem

The mean of a random sample is itself a random variable. It is centred on $\mu$ , its spread shrinks like $\dfrac{\sigma}{\sqrt n}$ , and for large samples it is approximately normal whatever the population looks like.

Cambridge 9709 Paper 6Edexcel YMA01 WST03Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Treat $\bar X$ as a random variable with $E(\bar X)=\mu$ and $\text{Var}(\bar X)=\dfrac{\sigma^2}{n}$ .
Use $\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}n\right)$ exactly when $X$ is normal.
Use the Central Limit Theorem for large samples from any population.
Say when the CLT is needed and when it is not.

Pause and think

A single throw of a fair die is equally likely to be $1,2,\ldots,6$ : a flat distribution. What do you think the distribution of the average of $30$ throws looks like?

Section 2 of 14

Core idea

Start hereAverages are calmer than individuals

One die roll could be 1 or 6. The average of 50 rolls is almost certainly near 3.5. Averages vary less than single values, and, remarkably, their distribution is bell-shaped whatever the original distribution. That's the Central Limit Theorem.

The big idea

$\bar X$ has mean $\mu$ and variance $\frac{\sigma^2}n$ . If $X$ is normal, $\bar X\sim N(\mu,\frac{\sigma^2}n)$ exactly; otherwise approximately for large $n$ (CLT).

1. Distribution of the mean

Mean and variance of

\bar X

$\bar X=\tfrac1n(X_1+\cdots+X_n)$
Definition.
$E(\bar X)=\tfrac1n(n\mu)=\mu$
Expectation rules.
$\operatorname{Var}(\bar X)=\tfrac1{n^2}(n\sigma^2)=\tfrac{\sigma^2}n$
Independent: variances add; $\frac1n$ squared.

Distribution of the sample mean as n grows (μ = 50, σ = 10)

sample size n = 1

The standard deviation of

\bar X

\dfrac{\sigma}{\sqrt n}

: four times the sample size halves the spread.

Worked example $\mu=50$ , $\sigma=10$ , $n=25$

$P(\bar X>53)$ ?

Goal: Probability.
Strategy: $\bar X\sim N(50,4)$ .

$\text{SD}=\tfrac{10}{5}=2$
$\frac\sigma{\sqrt n}$ .
$z=1.5\Rightarrow1-0.9332=0.0668$
Standardise.

Answer

0.0668

✓ CheckA single value above 53 has probability

0.38

: averages are much less extreme ✓.

Worked exampleCLT with a Poisson population

$X\sim\operatorname{Po}(4)$ , $n=50$ . $P(\bar X<3.8)$ ?

Goal: Probability.
Strategy: CLT: $\bar X\approx N(4,\frac4{50})$ .

$\text{SD}=\sqrt{0.08}=0.283$
Variance $\frac\lambda n$ .
$z=\tfrac{-0.2}{0.283}=-0.707\Rightarrow0.240$
Standardise.

Answer

0.240

✓ CheckPopulation not normal, but

n=50

is large ✓.

Pause and think

To halve the SD of $\bar X$ ?

Summary

Mean $\mu$ , variance $\sigma^2/n$ .
CLT for large $n$ .

Central Limit Theorem: Sample means are approximately normal for large n.

X ~ N(50, 10²) and the mean of 25 observations

\bar X\sim N\!\left(\mu,\tfrac{\sigma^2}{n}\right)

: the mean of 25 values has the same centre but a standard deviation five times smaller, so it is far more predictable.

HistoryLaplace's theorem. Pierre-Simon Laplace proved an early version of the Central Limit Theorem in 1810, generalising de Moivre's result for coin tosses to sums of many kinds of random variable. The name "central limit theorem" was coined by George Pólya in 1920, "central" because of its central importance in probability.

Fun factWhy so many things are normal. Your height is the result of many small, roughly independent effects: hundreds of genes, nutrition, sleep and more. The CLT says sums and averages of many small effects tend to be normal, which is why heights, measurement errors and exam totals all form bell curves.

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