Proof of $\operatorname{Var}(\bar{x}) = \frac{\sigma^2}{n}$

Question

I'm trying to understand why we divide by $n - 1$ when calculating the variance of a sample in statistics.

Following Wikipedia's third alternate proof, I get stuck on

$$\operatorname{Var}(\bar{x}) = \frac{\sigma^2}{n}$$

How would one prove of this equation?

Answer 1

\begin{align} \operatorname{var}\left( \frac {X_1+\cdots+X_n} n \right) & = \frac 1 {n^2} \operatorname{var} (X_1+\cdots+X_n) \\[10pt] & = \frac 1 {n^2} \big( \operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n) \big) \quad \text{because of independence} \\[10pt] & = \frac 1 {n^2}(\sigma^2 + \cdots+\sigma^2) \\[10pt] & = \frac 1 {n^2}\cdot n\sigma^2 \\[10pt] & = \frac{\sigma^2} n. \end{align}