Example 3. Find the variance and standard deviation of a normal random variable.
Solution. Using formulas (2.5.15) and (3.5.11), we get

$$
D(X)=\int_{-\infty}^{+\infty}(x-a)^{2} p(x) d x=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty}(x-a)^{2} e^{-(x-a)^{2} / 2 \sigma^{2}} d x
$$

By transitioning to a new variable $t$ according to formula (3.5.9), we get

$$
D(X)=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty} t^{2} \sigma^{2} e^{-t^{2} / 2} \sigma d t=\frac{\sigma^{2}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t^{2} e^{-t^{2} / 2} d t
$$

We apply the method of integration by parts. Let $t=u$, $t e^{-t^{2} / 2} d t=d v$, then $d u=d t, v=-e^{-t^{2} / 2},\left(\right.$ because $d v=d\left(-e^{-t^{2} / 2}\right)$ ). Therefore,

$$
\begin{gathered}
D(X)=\frac{\sigma^{2}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t \cdot t e^{-t^{2} / 2} d t=-\frac{\sigma^{2}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} t d\left(e^{-t^{2} / 2}\right)= \\
=-\frac{\sigma^{2}}{\sqrt{2 \pi}}\left(\left.t e^{-t^{2} / 2}\right|_{-\infty} ^{+\infty}-\int_{-\infty}^{+\infty} e^{-t^{2} / 2} d t\right)=-\left.\frac{\sigma^{2}}{\sqrt{2 \pi}} t e^{-t^{2} / 2}\right|_{-\infty} ^{+\infty}+\frac{\sigma^{2}}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} e^{-t^{2} / 2} d t= \\
=0+\frac{\sigma^{2}}{\sqrt{2 \pi}} \cdot \sqrt{2 \pi}=\sigma^{2}
\end{gathered}
$$

(Here, equality (3.5.10) is also taken into account).

Thus,

$$
\begin{aligned}
& D(X)=\sigma^{2} \\
& \sigma=\sqrt{D(X)}
\end{aligned}
$$

Therefore, the parameter $\sigma$ in function (3.5.1) is the standard deviation of the normal random variable.