Example 14. A machine manufactures bearings, which are considered suitable if the deviation $X$ from the design size in absolute value does not exceed 0.77 mm. What is the most probable number of suitable bearings out of 100, if the random variable $X$ is normally distributed with the parameter $\sigma=0.4$ mm?
Solution. First, we find the probability of deviation using formula (3.5.4) with $\delta=0.77$ and $\sigma=0.4$:

$$
P(|X-a|<0.77)=2 \Phi\left(\frac{0.77}{0.4}\right) \approx 2 \Phi(1.93)=2 \cdot 0.473197=0.946394
$$

Approximating $p=0.95$ and $q=0.05$, according to formula (3.1.5), i.e., $n p-q \leq k_{0} \leq n p+p$, we find for $n=100$:

$$
100 \cdot 0.95-0.05 \leq k_{0} \leq 100 \cdot 0.95+0.95
$$

from which $k_{0}=95$.