Example 9. It is known that the variance of each of the given independent random variables does not exceed 4. Determine the number of such variables for which the probability of the deviation of the arithmetic mean of the random variable from the arithmetic mean of their mathematical expectations by no more than 0.25 exceeds 0.99.
Solution. Inequality (4.2.1) for $C=4$ and $\varepsilon=0.25$ takes the form

$$
m\left(\left\lvert\, \frac{1}{n} \sum_{i=1}^{n} X_{i}-\frac{1}{n} \sum_{i=1}^{n} M\left(X_{i} \mid < 0.25\right) \geq 1-\frac{4}{n \cdot (0.25)^{2}}\right.\right.
$$

From the condition, it follows that

$$
\begin{aligned}
& 1-\frac{4}{n \cdot 0.0625} \geq 0.99, \text{ hence } 1-0.99 \geq \frac{4}{n \cdot 0.0625}, \text{ or } \\
& 0.01 \geq \frac{4}{n \cdot 0.0625}, n \geq \frac{4}{0.01 \cdot 0.0625}=\frac{4}{0.000625}=6400
\end{aligned}
$$

Thus, $n \geq 6400$.