Sample means and the central-limit idea | Year 12

01 / LEARN

The idea, step by step

A mean varies between samples

Before selection, $\bar{X}$ is random. For independent observations from the same population, $E (\bar{X}) = μ$ and $Var (\bar{X}) = σ^{2} / n$ . Averaging keeps the centre and reduces the spread. Distinguish a distribution of individual observations from a distribution of means of samples of one fixed size.

In repeated-sampling experiments, each sample mean must use the same sample size. The standard deviation of those means is a standard error.

E (\bar{X}) = μ, SD (\bar{X}) = σ / \sqrt{n}

When a bell-shaped model is justified

A sample mean is exactly normal when independent observations come from a normal population. For a non-normal population with finite variance, the central limit theorem gives an increasingly useful normal approximation as sample size grows.

There is no universal sample size guaranteeing a good approximation for every population. Severe skewness and heavy tails may require larger samples. Independence also matters: repeated readings affected by the same drift need not behave as independent observations.

A common trap

A normal sample-mean distribution proves a normal population.

Instead: Large independent averages can be approximately normal even when individuals are not.

02 / WORKED EXAMPLES

See the thinking

Example 1

Exact sampling model

Fill volumes are normal with mean 500 mL and standard deviation 12 mL. Model the mean of 36 independent fills.

$E (\bar{X}) = 500$ .
$SD (\bar{X}) = 12 / \sqrt{36} = 2$ mL.
Normality is exact under the stated population model.

\bar{X} \sim N (500, 2^{2})

Example 2

Probability for an average

For that sample mean, find $P (498 < \bar{X} < 502)$ .

The standardised bounds are $- 1$ and $1$ .
Use $Φ (1) - Φ (- 1) = 0.682689 \dots$ .

Approximately 0.6827.

Check your understanding

For population mean 40, standard deviation 15 and n = 25, find the sample-mean centre and standard deviation.

Show answer

40 and 3.

$15 / \sqrt{25} = 3$ ; the centre stays 40.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $μ = 72, σ = 18, n = 36$ , find the mean and variance of $\bar{X}$ .
2 marks
Normal observations have mean 100 and standard deviation 20. Model a mean of 16 independent observations.
2 marks
For the previous model, find $P (\bar{X} > 110)$ .
2 marks
Why need a mean of four highly right-skewed observations not be approximately normal?
2 marks
What changes as independent sample size grows for a non-normal finite-variance population?
2 marks
Why can 100 consecutive readings from one drifting sensor fail independence?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $μ = 72, σ = 18, n = 36$ , find the mean and variance of $\bar{X}$ .
2 marks
$E (\bar{X}) = 72$ ; $Var (\bar{X}) = 18^{2} / 36 = 9$ .
72 and 9.
Normal observations have mean 100 and standard deviation 20. Model a mean of 16 independent observations.
2 marks
The standard error is $20 / \sqrt{16} = 5$ .
$\bar{X} \sim N (100, 5^{2})$ .
For the previous model, find $P (\bar{X} > 110)$ .
2 marks
$z = (110 - 100) / 5 = 2$ .
$1 - Φ (2) = 0.022750 \dots$ .
Approximately 0.0228.
Why need a mean of four highly right-skewed observations not be approximately normal?
2 marks
The central limit theorem is a limiting result, not a guarantee for every small sample.
A sample of four may be too small to average away strong skewness.
What changes as independent sample size grows for a non-normal finite-variance population?
2 marks
Its standard deviation is $σ / \sqrt{n}$ ; its expectation is $μ$ .
The mean distribution tends towards normality and becomes narrower; its centre remains the population mean.
Why can 100 consecutive readings from one drifting sensor fail independence?
2 marks
A large reading count is not evidence that the observations are independent.
Adjacent readings can share drift and be correlated.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: sampling distributions and confidence intervals for means and proportions. Mean tasks distinguish known population spread from sufficiently large-sample estimated-spread normal intervals. Repeated-interval computational simulation, approximation conditions, design and interpretation are included.

Official curriculum source ↗

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