Sample Means From Normal Populations

Let $X_1, X_2, \ldots, X_n$ be i.i.d. random variables drawn from a normal population with mean $\mu$ and variance $\sigma^2$. The sample mean is

$$\bar{X} = \dfrac{X_1 + X_2 + \cdots + X_n}{n}.$$

Sums of independent normal variables are normal, and scaling a normal variable keeps it normal, so $\bar{X}$ is also normally distributed. Its mean is $\mu$ and its variance is $\sigma^2/n$:

$$\bar{X} \sim N\!\left(\mu,\; \dfrac{\sigma^2}{n}\right).$$

The center is unchanged, but the variance shrinks by a factor of $n$. The standard deviation of $\bar{X}$, namely $\sigma/\sqrt{n}$, is called the standard error of the sample mean.

For example, if $X_i \sim N(50, 16)$ and we average $n = 4$ of them, then

$$\bar{X} \sim N\!\left(50,\; \dfrac{16}{4}\right) = N(50, 4),$$

with standard error $\sigma/\sqrt{n} = 4/2 = 2$.