# The Sampling Distribution of the Mean

> Averages of samples behave better than single values. The mean and standard error of a sample mean, the central limit theorem, and sample proportions.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/sampling-distribution-of-the-mean/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The bakery inspector
A bakery says its loaves weigh 800 g on average. Single loaves vary a lot: some are 770 g, some are 830 g. An inspector cannot weigh every loaf, so she picks 25 at random and finds their average.

She notices something useful. Averages of 25 loaves hardly vary at all. They almost always land close to 800 g. If her sample average is far below 800 g, that is strong evidence the bakery is cheating. This chapter explains exactly how much sample averages wobble, and why their shape is nearly always a bell curve.

**In short:** The sample mean $\bar{X}$ changes from sample to sample, so it has its own distribution, with mean $\mu$ and standard error $\dfrac{\sigma}{\sqrt{n}}$. The central limit theorem says that for large samples, $\bar{X}$ is close to normal, whatever the population looks like.

## Starting point: what you already know

- **Mean and standard deviation** measure the centre and spread of data. Variance is the standard deviation squared.
- **Population and sample:** the population is every item. A sample is the part you actually measure. Greek letters $\mu$ (mean) and $\sigma$ (standard deviation) describe the population.
- **The normal distribution** (Mathematics Advanced): bell shaped, with about 68%, 95% and 99.7% of values within 1, 2 and 3 standard deviations of the mean.
- **z-scores:** $z = \dfrac{x - \mu}{\sigma}$ says how many standard deviations a value is from the mean.
- **The binomial distribution** (previous chapter): mean $np$ and variance $np(1 - p)$.

## The sample mean is a random variable

Take a random sample of size $n$ and work out its mean, $\bar{X}$. Take another sample and you get a slightly different $\bar{X}$. So $\bar{X}$ changes from sample to sample: it is a random variable with its own distribution, called the **sampling distribution of the mean**.

### Key idea: Centre and spread of the sample mean
For samples of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$:
$$E(\bar{X}) = \mu \qquad \mathrm{SD}(\bar{X}) = \frac{\sigma}{\sqrt{n}}$$
$\dfrac{\sigma}{\sqrt{n}}$ is called the **standard error**.

- The sample mean is aimed at the right place, $\mu$.
- Bigger samples wobble less. To halve the standard error, you need **four times** the sample size.

*Diagram: Two bell curves with the same centre. The wide, low curve shows single values. The narrow, tall curve shows sample means, which are much less spread out.*
*Sample means (the tall green curve) are centred in the same place as single values (the low wide curve) but are far less spread out.*

## The central limit theorem

Here is the surprising part. Even if single values are **not** bell shaped (lopsided, flat, anything), the sample mean is close to normal once $n$ is large enough. A common guide is $n \ge 30$.

### Key idea: Central limit theorem
For large $n$, $\bar{X}$ is approximately normal with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$. If the population is already normal, this is true for any $n$.

### Worked example: The bakery
Loaves have $\mu = 800$ g and $\sigma = 20$ g. The inspector weighs 25 loaves.

Standard error: $\dfrac{20}{\sqrt{25}} = 4$ g.

How likely is a sample mean below 792 g? $z = \dfrac{792 - 800}{4} = -2$. About 95% of sample means are within 2 standard errors, so $P(\bar{X} < 792) \approx 2.5\%$.

A single loaf below 792 g is common, with $z = -0.4$. A sample average that low is rare.

### Watch out: Use the right spread
For a single value, divide by $\sigma$. For a sample mean, divide by $\dfrac{\sigma}{\sqrt{n}}$. Mixing them up is the most common error in this topic.

### Calculator tip: Square roots with brackets
Type the whole fraction under the root: $\sqrt{\dfrac{0.4 \times 0.6}{100}}$ is `√( 0.4 × 0.6 ÷ 100 )` = 0.049. For a standard error $\dfrac{\sigma}{\sqrt{n}}$, type `15 ÷ √(25)`.

**Mind map**

```text
Sampling distribution of the mean
  Sample mean
    Changes sample to sample
    A random variable
  Centre
    Mean of sample means is μ
  Spread
    Standard error σ/√n
    4 times n halves it
  Shape
    Central limit theorem
    Near normal for large n
  Using it
    z = (x̄ - μ)/(σ/√n)
    68, 95, 99.7 rule
  Sample proportion
    Mean p
    SD √(p(1 - p)/n)
```

### Year 11: Averages of dice
One fair die has mean 3.5 and standard deviation about 1.71. Its distribution is flat, not a bell.

Roll 36 dice and average them. The standard error is $\dfrac{1.71}{6} \approx 0.285$, and by the central limit theorem the average is close to normal. So about 95% of the time, the average of 36 dice lands between $3.5 - 0.57$ and $3.5 + 0.57$, that is from about 2.93 to 4.07.

### Year 12: Sample proportions
In a survey of $n$ people, let $\hat{p}$ be the proportion who say yes. Each answer is a Bernoulli trial (1 for yes, 0 for no), so $\hat{p}$ is a sample mean of 0s and 1s. The central limit theorem applies:
$$E(\hat{p}) = p \qquad \mathrm{SD}(\hat{p}) = \sqrt{\frac{p(1 - p)}{n}}$$
The approximation is good when $np$ and $n(1 - p)$ are both at least about 10.

**Example.** 40% of voters back a policy. In a random sample of 100, how likely is a majority, $\hat{p} > 0.5$?

$\mathrm{SD} = \sqrt{\dfrac{0.4 \times 0.6}{100}} \approx 0.049$, so $z = \dfrac{0.5 - 0.4}{0.049} \approx 2.04$. That is just past 2 standard deviations, so the chance is about 2%.

**Choosing a sample size.** To make $\mathrm{SD}(\hat{p}) = 0.03$ when $p = 0.4$: $\dfrac{0.24}{n} = 0.0009$, so $n \approx 267$. For a 95% chance of being within $0.05$ of $p$ (two standard deviations), with $p$ near $0.5$: $2\sqrt{\dfrac{0.25}{n}} = 0.05$ gives $n = 400$.

**The largest spread.** $p(1 - p)$ is largest at $p = 0.5$, so for samples of 64 the largest possible $\mathrm{SD}(\hat{p})$ is $\sqrt{\frac{0.25}{64}} = 0.0625$.

**Testing a claim.** If a claim says $p = 0.5$ and a sample of 100 gives $\hat{p} = 0.38$, then $z = \dfrac{0.38 - 0.5}{0.05} = -2.4$. Results that far out happen less than 2.5% of the time if the claim is true, so the claim looks doubtful. About 16% of sample means lie more than one standard error above the mean, so a result just past one standard error is not unusual.

## Quick summary

- The sample mean $\bar{X}$ is a random variable.
- $E(\bar{X}) = \mu$ and the standard error is $\dfrac{\sigma}{\sqrt{n}}$.
- Central limit theorem: for large $n$, $\bar{X}$ is close to normal, whatever the population looks like.
- Use $z = \dfrac{\bar{x} - \mu}{\sigma / \sqrt{n}}$ with the 68, 95, 99.7 rule or a normal table.
- Sample proportions: mean $p$, standard deviation $\sqrt{\dfrac{p(1 - p)}{n}}$.

## Quiz

Try these questions from the chapter quiz.

1. A population has mean 50. What is the mean of the sampling distribution of $\bar{X}$?
   - A. It depends on $n$
   - B. 50
   - C. $\dfrac{50}{\sqrt{n}}$
   - D. 0
   - Answer: B. 50. $E(\bar{X}) = \mu$.

2. $\sigma = 12$ and $n = 36$. The standard error is:
   - A. 12
   - B. 2
   - C. $\dfrac{1}{3}$
   - D. 6
   - Answer: B. 2. $\dfrac{12}{\sqrt{36}} = 2$.

3. To halve the standard error, the sample size must be:
   - A. doubled
   - B. multiplied by 4
   - C. halved
   - D. multiplied by 2.5
   - Answer: B. multiplied by 4. Standard error has $\sqrt{n}$ underneath.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/sampling-distribution-of-the-mean/#quiz