Distribution of the sample mean (normal)
Study the distribution of the sample mean for a normal population in Year 12 Specialist Mathematics in Queensland (QCAA). When the parent is normal, the sample mean is exactly normal, keeping the same centre but a smaller spread that depends on the sample size.
You will learn that the sample mean has standard deviation sigma over root n, how to standardise it, and how to find probabilities for the sample mean — the foundation for statistical inference, confidence intervals and estimating a population mean later in the course.
Theory
The distribution of the sample mean is the pattern of values that \(\bar X\) takes over repeated samples. In Year 12 Specialist Mathematics (QCAA, Queensland), if the parent \(X\sim N(\mu,\sigma^2)\) is normal, then \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\) exactly. Its centre stays at \(\mu\) but its standard deviation is \(\dfrac{\sigma}{\sqrt n}\), so the spread narrows as \(n\) grows. This page shows how to standardise and find probabilities for \(\bar X\).
Take a random sample of size \(n\) from a population and record its mean. The sample mean \(\bar X\) is itself a random variable: its value changes from sample to sample. The pattern of those values across all possible samples is the sampling distribution of the sample mean.
When the parent variable is normal, \(X\sim N(\mu,\sigma^2)\), the sample mean is exactly normal for every sample size: \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\). The centre is unchanged — the mean of \(\bar X\) equals \(\mu\).
What changes is the spread. The standard deviation of \(\bar X\) is \(\dfrac{\sigma}{\sqrt n}\) (its variance is \(\dfrac{\sigma^2}{n}\)), which is smaller than \(\sigma\). Averaging cancels out extremes, so sample means cluster more tightly around \(\mu\) than single observations do.
Because \(\dfrac{\sigma}{\sqrt n}\) shrinks as \(n\) increases, larger samples give a narrower distribution: the sample mean becomes a more reliable estimate of \(\mu\). To find a probability for \(\bar X\), standardise with \(Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\).
For a normal parent \(X\sim N(\mu,\sigma^2)\), the sample mean of a random sample of size \(n\) is exactly normal:
Its centre and spread are:
Standardise a value of the sample mean with:
How to find a probability for the sample mean
- State the sampling distribution: for a normal parent write \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\).
- Find the standard deviation of \(\bar X\): compute \(\dfrac{\sigma}{\sqrt n}\) (divide \(\sigma\) by \(\sqrt n\), never use \(\sigma\) alone).
- Standardise each boundary value with \(Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\).
- Read the probability from the standard normal distribution, using tails or the between-region as required.
Keep the same centre \(\mu\); the variance divides by \(n\):
| \(\bar X\) | \(\sim\) | \(N\!\left(\mu,\ \dfrac{\sigma^2}{n}\right)\) |
| \(\sim\) | \(N\!\left(30,\ \dfrac{6^2}{9}\right)\) | |
| \(\sim\) | \(N\!\left(30,\ \dfrac{36}{9}\right)\) | |
| \(\sim\) | \(N(30,\ 4)\) |
\(\bar X\sim N(30,\ 4)\); the standard deviation of \(\bar X\) is \(\dfrac{6}{\sqrt 9}=2\) cm.
First the standard deviation of \(\bar X\), then standardise the boundary:
| \(\text{sd}(\bar X)\) | \(=\) | \(\dfrac{\sigma}{\sqrt n} = \dfrac{8}{\sqrt{16}} = 2\) |
| \(Z\) | \(=\) | \(\dfrac{224-220}{2}\) |
| \(=\) | \(2\) |
Read the upper tail from the standard normal:
| \(P(\bar X > 224)\) | \(=\) | \(P(Z > 2)\) |
| \(=\) | \(1 - 0.9772\) | |
| \(=\) | \(0.0228\) |
\(P(\bar X > 224)=0.0228\).
Find the standard deviation of \(\bar X\), then standardise each endpoint:
| \(\text{sd}(\bar X)\) | \(=\) | \(\dfrac{4}{\sqrt{16}} = 1\) |
| \(Z_1\) | \(=\) | \(\dfrac{99-100}{1} = -1\) |
| \(Z_2\) | \(=\) | \(\dfrac{101.5-100}{1} = 1.5\) |
Subtract the two standard-normal areas:
| \(P(99 < \bar X < 101.5)\) | \(=\) | \(P(-1 < Z < 1.5)\) |
| \(=\) | \(0.9332 - 0.1587\) | |
| \(=\) | \(0.7745\) |
\(P(99 < \bar X < 101.5)=0.7745\).
Set \(\dfrac{\sigma}{\sqrt n}\) equal to the target and solve for \(n\):
| \(\dfrac{\sigma}{\sqrt n}\) | \(=\) | \(3\) |
| \(\dfrac{15}{\sqrt n}\) | \(=\) | \(3\) |
| \(\sqrt n\) | \(=\) | \(\dfrac{15}{3} = 5\) |
| \(n\) | \(=\) | \(5^2\) |
| \(=\) | \(25\) |
\(n=25\); a larger \(n\) is what shrinks \(\dfrac{\sigma}{\sqrt n}\) and narrows the distribution of \(\bar X\).
Common pitfalls
Frequently asked questions
What is the distribution of the sample mean for a normal population?
If \(X\sim N(\mu,\sigma^2)\) then the sample mean is exactly normal: \(\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)\), for every sample size \(n\).
What is the standard deviation of the sample mean?
It is \(\dfrac{\sigma}{\sqrt n}\), the population standard deviation divided by \(\sqrt n\). This is smaller than \(\sigma\), and its square is the variance \(\dfrac{\sigma^2}{n}\).
How do you find a probability for the sample mean?
Standardise with \(Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\), then read the required tail or between-region from the standard normal distribution.
Why does the sample mean have a smaller spread than a single value?
Averaging cancels out high and low observations, so sample means vary less than individual values. The spread \(\dfrac{\sigma}{\sqrt n}\) shrinks as \(n\) grows.
What happens to the distribution of the sample mean as \(n\) increases?
The centre stays at \(\mu\) but the standard deviation \(\dfrac{\sigma}{\sqrt n}\) gets smaller, so the curve becomes taller and narrower and \(\bar X\) estimates \(\mu\) more precisely.