1693. The distribution of the random variable $X$ is given by the following table:

| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p(x)$ | 0.05 | 0.10 | 0.25 | 0.30 | 0.20 | 0.10 |

What is the probability that $|X-E X|<2$? Estimate this probability using Chebyshev's inequality.
Solution. First, we find the mathematical expectation and variance of the random variable $X$:

$E X=1 \cdot 0.05+2 \cdot 0.10+3 \cdot 0.25+4 \cdot 0.30+5 \cdot 0.20+6 \cdot 0.10=3.80;$

$E X^{2}=1^{2} \cdot 0.05+2^{2} \cdot 0.10+3^{2} \cdot 0.25+4^{2} \cdot 0.30+5^{2} \cdot 0.20+$ $+6^{2} \cdot 0.10=16.10$

$$
D X=E X^{2}-(E X)^{2}=16.10-14.44=1.66
$$

To find $P(|X-E X|<2)$, we need to know the values of $X$ that satisfy the inequality $|X-E X|<2:|X-3.8|<$ $<2, \quad-2<X-3.8<2, \quad 1.8<X<5.8$. Only the values $X: 2,3,4$ and 5 satisfy these inequalities. Therefore,

$$
\begin{gathered}
P(|X-3.8|<2)=P(X=2)+P(X=3)+P(X=4)+ \\
+P(X=5)=0.10+0.25+0.30+0.20=0.85
\end{gathered}
$$

Using Chebyshev's inequality, we can obtain a lower bound for this probability. For this, we substitute the previously found values of $E X$, $D X$, and $\varepsilon=2$ into the inequality
$P(|X-E X|<\varepsilon) \geqslant 1-\frac{D X}{\varepsilon^{2}}$. Then

$$
P(|X-3.8|<2) \geqslant 1-\frac{1.66}{4}=0.585
$$