1694. The average length of a part is 50 cm, and the variance is 0.1. Using Chebyshev's inequality, estimate the probability that a randomly selected part will have a length of no less than 49.5 cm and no more than 50.5 cm.
Solution. Let $X$ be the length of a randomly taken part. From the condition of the problem, it follows that $E X=50$, and $D X=0.1$. It is required to estimate from below the value of $P(49.5<X<50.5)$. Since the inequalities $49.5<X<50.5$ are equivalent to the inequality $|X-50|<0.5$, it is necessary to estimate $P(|X-50|<0.5)$. But this is the left-hand side of Chebyshev's inequality, since $E X=50$. Hence,

$$
P(|X-50|<0.5) \geqslant 1-\frac{D X}{\varepsilon^{2}}=1-\frac{0.1}{(0.5)^{2}}=0.6
$$