Variance of the normal distribution

Statement

Let $\mu \in \mathbb{R}$, let $\sigma^2 > 0$ and let $X$ have the normal distribution $\mathcal{N}(\mu, \sigma^2)$. Then $$\operatorname{Var}(X) = \sigma^2.$$

Remarks

Proof: $Z = (X - \mu)/\sigma$ has the law $\mathcal{N}(0, 1)$, where $\sigma = \sqrt{\sigma^2}$. Integration by parts gives $E[Z^2] = \int z^2 \varphi(z) \, dz = \int \varphi(z) \, dz = 1$. Since $E[Z] = 0$ by the mean of the normal distribution, $\operatorname{Var}(Z) = 1$. The scaling and translation rules then give $\operatorname{Var}(X) = \sigma^2 \operatorname{Var}(Z) = \sigma^2$. So the two parameters of $\mathcal{N}(\mu, \sigma^2)$ are its mean and its variance.

Explanations

L1 human

The second parameter of the normal distribution is exactly its variance. So $\sigma$ is the standard deviation, the typical distance of a value from the center.