Year 12 Maths Extension 1 (2027) Statistical analysis

Sampling distribution of the mean & Central Limit Theorem

Q: What is the sampling distribution of the mean?

The distribution of the sample means of all samples of a given size n.

Q: What are its mean and variance?

The mean is mu and the variance is sigma squared over n.

Q: What is the standard error?

The standard deviation of the sample mean, sigma over the square root of n.

Q: What does the Central Limit Theorem say?

For n at least thirty, the sample mean is approximately normal with mean mu and variance sigma squared over n, whatever the population shape.

Q: How do you find a probability for the mean?

Standardise using z equals xbar minus mu over sigma over root n, then use the standard normal.

20 practice questions 2 video lessons Theory + worked examples

Understand the sampling distribution of the mean in NSW Year 12 Mathematics Extension 1. Sample means vary from one sample to the next, yet their distribution follows a predictable pattern as the sample size grows.

You will learn how the mean and variance of the sampling distribution relate to the population, and how the Central Limit Theorem makes it approximately normal for large samples — the basis for statistical prediction in politics, finance and science in Extension 1.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Central Limit Theorem - Sampling Distribution of Sample Means - Stats & Probability Watch
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Theory

The sample mean $\bar{X}$ has $E (\bar{X}) = μ$ and $SD (\bar{X}) = \frac{σ}{\sqrt{n}}$ . The Central Limit Theorem says that for $n \geq 30$ , $\bar{X} \approx N (μ, \frac{σ^{2}}{n})$ , whatever the population shape. This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-06.

The sampling distribution of the mean describes how the sample mean $\bar{X} = \frac{1}{n} \sum X_{i}$ varies from sample to sample. For samples of size $n$ from a population with mean $μ$ and variance $σ^{2}$ :

E (\bar{X}) = μ, Var (\bar{X}) = \frac{σ^{2}}{n}, SD (\bar{X}) = \frac{σ}{\sqrt{n}} .

The Central Limit Theorem (CLT): provided $n$ is large $(n \geq 30)$ , the sampling distribution is approximately normal — whatever the population's shape: $\bar{X} \approx N (μ, \frac{σ^{2}}{n})$ .

NESA link. Part of the Year 12 The binomial distribution and sampling distribution of the mean focus area, outcome ME1-12-06 ("solves problems involving binomial distributions, sampling distribution of the mean and the central limit theorem") with MAO-WM-01.

Larger

n

gives a smaller standard error, so

\bar{X}

clusters near

μ

Any population shape gives an approximately normal

\bar{X}

E (\bar{X}) = μ, Var (\bar{X}) = \frac{σ^{2}}{n}, SD (\bar{X}) = \frac{σ}{\sqrt{n}} (standard error) .

E(Xbar) = mu, Var(Xbar) = sigma^2/n, SD(Xbar) = sigma/sqrt(n)

To find a probability, standardise and use the standard normal:

z = \frac{\bar{x} - μ}{σ / \sqrt{n}} .

z = (xbar - mu) / (sigma / sqrt(n))

Divide by $\sqrt{n}$ . The standard error is $\frac{σ}{\sqrt{n}}$ , not $\frac{σ}{n}$ . If the population is already normal, $\bar{X}$ is normal for any $n$ .

How to solve a sampling problem

Mean and standard error: $E (\bar{X}) = μ$ , $SD (\bar{X}) = \frac{σ}{\sqrt{n}}$ .
Check the CLT applies ( $n \geq 30$ , or the population is normal).
Standardise: $z = \frac{\bar{x} - μ}{σ / \sqrt{n}}$ .
Use the standard normal to find the probability.

Example 1 — Standard error

A population has mean

80

and SD

20

. For samples of size

n = 100

, find

E (\bar{X})

and the standard error.

Solution

$E (\bar{X})$	$=$	$80$
$SD (\bar{X})$	$=$	$\frac{20}{\sqrt{100}} = 2$

E(Xbar) = 80, standard error 2

$E (\bar{X}) = 80$ , standard error $2$ .

Example 2 — Find the sample size

A population has variance

σ^{2} = 144

. What sample size gives

Var (\bar{X}) = 9

Solution

\frac{144}{n}

=

9 \Rightarrow n = 16

n = 16

A sample of $n = 16$ .

Example 3 — A tail probability

A population has mean

40

and SD

15

. A sample of

n = 25

is taken. Find

P (\bar{X} > 46)

Solution

Standard error $= \frac{15}{\sqrt{25}} = 3$ .

$z$	$=$	$\frac{46 - 40}{3} = 2$
$P (\bar{X} > 46)$	$=$	$P (Z > 2) \approx 0.023$

P(Xbar > 46) approx 0.023

$P (\bar{X} > 46) \approx 0.023$ .

Example 4 — Within bounds

A population has mean

200

and SD

24

. For a sample of

n = 36

, find

P (196 < \bar{X} < 204)

Solution

Standard error $= \frac{24}{\sqrt{36}} = 4$ .

$z$	$=$	$\frac{\pm 4}{4} = \pm 1$
$P (196 < \bar{X} < 204)$	$=$	\(P(-1

P(196 < Xbar < 204) approx 0.683

About $0.683$ (the $68 %$ rule).

Common pitfalls

Standard error.

Var (\bar{X}) = \frac{σ^{2}}{n}

, so

SD (\bar{X}) = \frac{σ}{\sqrt{n}}

— divide by

\sqrt{n}

, not

n

When you need $n \geq 30$ . Only for a non-normal population; a normal population gives a normal

\bar{X}

for any

n

Standardising. Use

z = \frac{\bar{x} - μ}{σ / \sqrt{n}}

, with the standard error in the denominator.

Bigger $n$ . Larger samples give a smaller standard error and a tighter distribution.

Frequently asked questions

What is the sampling distribution of the mean?

The distribution of the sample means of all samples of a given size $n$ .

What are its mean and variance?

$E (\bar{X}) = μ$ and $Var (\bar{X}) = \frac{σ^{2}}{n}$ .

What is the standard error?

$SD (\bar{X}) = \frac{σ}{\sqrt{n}}$ .

What does the Central Limit Theorem say?

For $n \geq 30$ , $\bar{X}$ is approximately $N (μ, \frac{σ^{2}}{n})$ , whatever the population shape.

How do you find a probability for the mean?

Standardise with $z = \frac{\bar{x} - μ}{σ / \sqrt{n}}$ and use the standard normal.

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📖 Theory

How to solve a sampling problem

Common pitfalls

Frequently asked questions

Theory