CAIE A-Level · Mathematics 9709 · Sampling and Estimation

Distribution of the Sample Mean (PS2 §6.4)

10 min readSyllabus 6.4PreviewBy Uzair Khan

Syllabus objective

Recognise that a sample mean can be regarded as a random variable, and use the facts that E(X̄) = μ and that Var(X̄) = σ²/n; use the fact that X̄ has a normal distribution if X has a normal distribution; and use the Central Limit Theorem where appropriate. Only an informal understanding of the Central Limit Theorem is required; for large sample sizes, the distribution of a sample mean is approximately normal.

Introduction

When data are collected, we rarely have access to an entire population — instead, we work with a sample. A key question in statistics is: what can a sample tell us about the population? One of the most important tools for answering this is the distribution of the sample mean.

Rather than thinking of the sample mean Xˉ\bar{X} as a fixed number, we recognise that it varies from sample to sample. This makes Xˉ\bar{X} a random variable with its own probability distribution. Understanding this distribution is essential for constructing confidence intervals (coming later in PS2) and for interpreting results in hypothesis tests. Expect to use the results in this topic in nearly every estimation question on the 9709 PS2 paper.


Core Concept

The Sample Mean as a Random Variable

Suppose X1,X2,,XnX_1, X_2, \ldots, X_n is a random sample of size nn drawn from a population with mean μ\mu and variance σ2\sigma^2. Each XiX_i is an independent observation from the same distribution. The sample mean is defined as:

Xˉ=X1+X2++Xnn\bar{X} = \frac{X_1 + X_2 + \cdots + X_n}{n}

Because each XiX_i is random, Xˉ\bar{X} is itself a random variable. Its distribution — how it behaves across all possible samples of size nn — is called the sampling distribution of the mean.

Result 1: The Mean of Xˉ\bar{X}

Since each XiX_i has mean μ\mu:

E(Xˉ)=μ\mathrm{E}(\bar{X}) = \mu

The sample mean is an unbiased estimator of the population mean — on average, it hits the right target.

Result 2: The Variance of Xˉ\bar{X}

Since the XiX_i are independent and each has variance σ2\sigma^2:

Var(Xˉ)=σ2n\mathrm{Var}(\bar{X}) = \frac{\sigma^2}{n}

As nn increases, the variance of Xˉ\bar{X} decreases: larger samples produce sample means that cluster more tightly around μ\mu.

Result 3: Exact Normality (when XX is Normal)

If the population itself follows a normal distribution, i.e.\ XN(μ,σ2)X \sim \mathrm{N}(\mu, \sigma^2), then for any sample size nn:

XˉN ⁣(μ, σ2n)\bar{X} \sim \mathrm{N}\!\left(\mu,\ \frac{\sigma^2}{n}\right)

This is an exact result, holding for all n1n \geq 1.

Result 4: The Central Limit Theorem (CLT)

If the population distribution is not normal (or is unknown), the CLT tells us that for large nn:

XˉN ⁣(μ, σ2n)\bar{X} \approx \mathrm{N}\!\left(\mu,\ \frac{\sigma^2}{n}\right)

The approximation improves as nn grows. For 9709, you need only an informal understanding: a large sample size makes the distribution of Xˉ\bar{X} approximately normal, regardless of the shape of the underlying population. As a working rule, n30n \geq 30 is generally considered sufficient.


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