Example 11. The random variable $X$ is given by the distribution function

$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{3} & \text { if } 0 < x \leq 1 \\ 1 & \text { if } x > 1\end{cases}
$$

Find the numerical characteristics of the random variable $X: M(X), D(X), \sigma(X)$
Solution. First, we find the probability density function $p(x)$ using formula (2.3.5). Since $p(x)=F^{\prime}(x)$, we have

$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 3 x^{2} & \text { if } 0 < x \leq 1\end{cases}
$$

Using formula (2.4.7), we calculate the expected value:

$$
M(X)=\int_{0}^{1} x \cdot 3 x^{2} d x=3 \int_{0}^{1} x^{3} d x=3 \cdot \frac{x^{4}}{4}=\frac{3}{4}
$$

According to formulas (2.5.13) and (2.5.16), we find the variance and standard deviation:

$$
\begin{gathered}
D(X)=\int_{0}^{1}\left(x-\frac{3}{4}\right)^{2} \cdot 3 x^{2} d x=3 \int_{0}^{1}\left(x^{2}-\frac{3}{2} x+\frac{9}{16}\right) x^{2} d x= \\
=3 \int_{0}^{1}\left(x^{4}-\frac{3}{2} x^{3}+\frac{9}{16} x^{2}\right) d x=\left.3\left(\frac{x^{5}}{5}-\frac{3}{2} \cdot \frac{x^{4}}{4}+\frac{9}{16} \cdot \frac{x^{3}}{3}\right) \cdot\right|_{0} ^{1}= \\
=3\left(\frac{1}{5}-\frac{3}{8}+\frac{3}{16}\right)=\frac{3}{80} ; \quad D(X)=\frac{3}{80} \\
\sigma(X)=\sqrt{3 / 80} \approx 0.19
\end{gathered}
$$