356. Find: a) variance; b) standard deviation of the exponential distribution given by the probability density function: $f(x)=\lambda \mathrm{e}^{-\lambda x}$ for $x \geqslant 0 ; f(x)=0$ for $x<0$.
Solution. a) Using the formula

$$
D(X)=\int_{-\infty}^{\infty} x^{2} f(x) \mathrm{d} x-[M(X)]^{2}
$$

Considering that $f(x)=0$ for $x<0, M(X)=1 / \lambda$ (see problem 358), we get

$$
D(X)=\lambda \int_{0}^{\infty} x^{2} e^{-\lambda x} \mathrm{~d} x-(1 / \lambda)^{2}
$$

Integrating by parts twice, we find

$$
\lambda \int_{0}^{\infty} x^{2} \mathrm{e}^{-\lambda x} \mathrm{~d} x=\frac{2}{\lambda^{2}}
$$

Therefore, the sought variance is

$$
D(X)=2 / \lambda^{2}-1 / \lambda^{2}=1 / \lambda^{2}
$$

i.e., the variance of the exponential distribution is equal to the value inverse to $\lambda^{2}$.

b) Let's find the standard deviation:

$$
\sigma(X)=\sqrt{D(X)}=\sqrt{1 / \lambda^{2}}=1 / \lambda
$$

i.e., the standard deviation of the exponential distribution is equal to the value inverse to $\lambda$.