Example 8. The average length of a part is $50 \mathrm{~cm}$, and the variance is 0.1. Estimate the probability that the length of the manufactured part will be no less than 49.5 cm and no more than 50.5 cm.
Solution. Since here $a=50$, the condition $49.5<X<50.5$, in which the random variable $X$ denotes the possible length of the part, is transformed by subtracting the number $a=50$ from each term to the form $|X-a| \leq 0.5$.

Applying inequality (4.1.4.) in the case $\varepsilon=0.5$ and $D(X)=0.1$, we get:

$$
P(|X-a| \leq 0.5) \geq 1-\frac{0.1}{(0.5)^{2}}=1-\frac{0.1}{0.25}=0.6
$$