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}.

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