# The variance of the arithmetic average

Why is the standard deviation of the arithmetic average equal to $\sigma/\sqrt{n}$ ?

For $X \mathrm{\;i.i.d.}\; \sim N(\mu,\sigma^2)$ (“i.i.d.” means identically and independently distributed):

$\begin{array}{l l}Var(\overline{X}) &= Var((1/n)\sum_{i=1}^n{x_i})\\&=(1/n^2) Var(\sum_{i=1}^n{x_i}), Var(aX)=a^2Var(X); a\in \mathbb{R} \mathrm{\,const.}\\&= (1/n^2) \sum_{i=1}^nVar({x_i}) \quad \mathrm{for\;} X \mathrm{\;i.i.d.}\\&=(1/n^2) \sum_{i=1}^n{x_i}{\sigma^2}\\&=(1/n^2)*n*{\sigma^2}\\&=\sigma^2/n\end{array}$

Hence, the standard deviation $\sigma=\sqrt{\sigma^2}$ of $\overline{X}$ is $\sigma/\sqrt{n}$.

Voilà. The world is safe again.