What is the Central Limit Theorem?

Whatever the shape of the population, the distribution of the sample mean approaches a normal distribution as the sample size increases. In M1 it lets you treat the sample mean as normal when n is sufficiently large.

Why divide by n − 1 for the sample variance?

Because that makes the estimator unbiased: over many samples its average equals the population variance. Dividing by n gives values that are too small on average.

What is the difference between a parameter and a statistic?

A parameter describes the whole population (μ, σ²) and is usually unknown. A statistic is calculated from a sample (x̄, s²) and is used to estimate the parameter.

Sampling Distribution of the Sample Mean and Point Estimates (DSE M1)

If samples of size n are drawn from a population with mean μ and variance σ², the sample mean has E(X̄) = μ and Var(X̄) = σ²/n. If the population is normal, X̄ ~ N(μ, σ²/n); by the Central Limit Theorem this is approximately true for any population when n is large.

Solve Sampling Distribution and Point Estimates questions with AI

The grey curve is the population N(50, 8²). The red curve is the sample mean, which narrows as n grows; at n = 16 its standard deviation is 2, as in Example 1. Try it yourself →

Sampling Distribution Formulas

Mean of the sample meanE(X̄) = μ

Variance of the sample meanVar(X̄) = σ² / n

DistributionX̄ ~ N(μ, σ²/n)

Standardising a sample meanZ = (X̄ − μ) / (σ/√n)

Unbiased estimate of μx̄ = Σx / n

Unbiased estimate of σ²s² = Σ(x − x̄)² / (n − 1)

Worked Examples — Sampling Distribution and Point Estimates

Example 1: Probability about a sample mean

A population is normally distributed with mean 50 and standard deviation 8. A random sample of 16 is taken. Find the probability that the sample mean exceeds 53.

X̄ ~ N(50, 8²/16), so its standard deviation is 8/√16 = 2
z = (53 − 50) / 2 = 1.5
P(X̄ > 53) = P(Z > 1.5) = 0.5 − 0.4332
= 0.0668

Answer: 0.0668

Example 2: Unbiased estimates from a sample

A random sample gives the values 4, 6, 7, 9, 9. Find unbiased estimates of the population mean and variance.

x̄ = (4 + 6 + 7 + 9 + 9) / 5 = 7
Σ(x − x̄)² = 9 + 1 + 0 + 4 + 4 = 18
s² = 18 / (5 − 1) = 4.5

Answer: x̄ = 7, s² = 4.5

💡 Exam Traps & Secrets

⚠️ Divide by √n for a sample meanFor a single value use σ; for the mean of n values use σ/√n. Forgetting the √n is the standard error in this unit.

⚠️ n − 1 for the sample varianceThe unbiased estimate of the population variance divides by n − 1. Dividing by n gives the variance of the sample itself, which underestimates σ².

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Sample mean)A population has mean 120 and standard deviation 15. Find the probability that the mean of a sample of 25 is less than 114.

(Point estimates)Find unbiased estimates of the population mean and variance from the sample 12, 15, 11, 18, 14.

(Effect of n)Show how the distribution of the sample mean changes when n increases from 4 to 36, for a population N(50, 8²).