Variance of Sample Means

Suppose $X_1, X_2, \ldots, X_n$ is a random sample from a distribution with mean $\mu$ and variance $\sigma^2$. By definition, the random variables $X_i$ are independent and identically distributed, so each one has variance $\sigma^2$.

The sample mean is the random variable

$$\bar{X} = \dfrac{1}{n}\sum_{i=1}^n X_i.$$

We want to know how variable $\bar{X}$ is. Pulling the constant $\dfrac{1}{n}$ outside (which squares when we take variance) and using independence to split the variance of the sum into a sum of variances, we get

$$\begin{align*}\operatorname{Var}(\bar{X}) &= \operatorname{Var}\!\left(\dfrac{1}{n}\sum_{i=1}^n X_i\right) \\[3pt] &= \dfrac{1}{n^2}\sum_{i=1}^n \operatorname{Var}(X_i) \\[3pt] &= \dfrac{1}{n^2}\cdot n\sigma^2 \\[3pt] &= \dfrac{\sigma^2}{n}.\end{align*}$$

This is the variance of the sample mean:

$$\boxed{\operatorname{Var}(\bar{X}) = \dfrac{\sigma^2}{n}}.$$

Notice that as $n$ grows, $\operatorname{Var}(\bar{X})$ shrinks. Larger samples produce sample means that cluster more tightly around the true mean $\mu$.

For a quick illustration: if a population has $\sigma^2 = 20$ and we take a sample of size $n = 5$, then $\operatorname{Var}(\bar{X}) = \dfrac{20}{5} = 4.$