130. The diameters of balls for bearings are controlled as follows: if a ball does not pass through a hole of diameter $d_{1}$, but does pass through a hole of diameter $d_{2}>d_{1}$, then its size is considered acceptable. If either of these conditions is not met, the ball is rejected. It is known that the diameter of the ball is a random variable $X$, distributed according to the normal law, the mathematical expectation of this variable is $a=\left(d_{1}+d_{2}\right) / 2$, and the standard deviation is $\sigma=\left(d_{2}-d_{1}\right) / 4$. Find the probability that a checked ball will be rejected.
S o l u t i o n. From the formula $a=\left(d_{1}+d_{2}\right) / 2$, it follows that the mathematical expectation $a$ of the considered random variable $X$ is the midpoint of the interval ( $d_{1}, d_{2}$ ), i.e., the sizes $d_{1}$ and $d_{2}$ are symmetric with respect to $a$. Considering that $\varepsilon=\left(d_{2}-d_{1}\right) / 2$, and $\sigma=\left(d_{2}-d_{1}\right) / 4$, we find the probability that the ball will not be rejected using formula (63):

$$
P(X-a \mid \leq \varepsilon)=P\left(|X-a| \leq \frac{d_{2}-d_{1}}{2}\right)=2 \Phi\left(\frac{d_{2}-d_{1}}{2 \sigma}\right)=2 \Phi(2)
$$

The probability that the ball will be rejected is

$$
P(|X-a|>\varepsilon)=1-P(|X-a| \leq \varepsilon)=1-2 \Phi(2)=1-2 \cdot 0.4772=0.0456
$$