1701. To determine the average burning duration of light bulbs in a batch of 200 identical boxes, one bulb was taken from each box as a sample. Estimate from below the probability that the average burning duration of the 200 selected light bulbs differs from the average burning duration of the entire batch by less than 5 hours in absolute value, given that the standard deviation of the burning duration of the bulbs in each box is less than 7 hours.
Solution. Let $X_{i}$ be the burning duration of a light bulb taken from the $i$-th box. It is given in the problem that $D X_{i}<7^{2}=$

=49. Obviously, the average burning duration of a bulb in the sample is

$$
\frac{X_{1}+X_{2}+\ldots+X_{200}}{200}
$$

and the average burning duration of a bulb in the entire batch is

$$
\frac{E X_{1}+E X_{2}+\ldots+E X_{200}}{200}
$$

We need to estimate from below the probability

$$
P\left(\left|\frac{X_{1}+X_{2}+\ldots+X_{200}}{200}-\frac{E X_{1}+E X_{2}+\ldots+E X_{200}}{200}\right|<5\right)
$$

This is the left side of inequality (1). Since $X_{1}, X_{2}, \ldots$, $X_{200}$ are independent random variables, this probability is estimated from below by the right side of inequality (1), where we should set $C=49, \varepsilon=5, n=200$. Therefore, the required probability is

$$
P \geqslant 1-\frac{49}{200 \cdot 25}=1-\frac{49}{5000}=1-0.0098=0.9902
$$