315. Find the variance and standard deviation of a random variable $X$ that is uniformly distributed in the interval $(a, b)$.
It is solved. We use the formula

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

Substituting $f(x)=1 /(b-a), M(X)=(a+b) / 2$ (see problem 313) and performing elementary calculations, we obtain the desired variance

$$
D(X)=(b-a)^{2} / 12
$$

The mean square deviation of the random variable $X$ is equal to the square root of its variance:

$$
\sigma(X)=(b-a) /(2 \sqrt{\overline{3}})
$$

In particular, the variance and mean square deviation of the random variable $R$, uniformly distributed in the interval $(0,1)$, are respectively: $D(R)=1 / 12, \sigma(R)=1 /(2 \sqrt{ } 3)$.