Example 15. An automatic machine produces rollers, controlling their diameters $X$. Assuming that the random variable $X$ is normally distributed, with parameters $a=10$ mm, $\sigma=0.1$ mm, find the interval in which the diameters of the produced rollers will be contained with a probability of 0.9973.
Solution. We will use formula (3.5.4). In this case, it is known that $a=10, \sigma=0.1, 2 \Phi(\delta / \sigma)=0.9973$; it is required to determine $\delta$ and the interval $(a-\delta, a+\delta)$.

From the table of values of the Laplace function, we find that $\delta / \sigma=3$. This follows from the equality $\Phi(\delta / \sigma)=0.9973 / 2=0.4987$. Therefore, $\delta=3 \sigma=3 \cdot 0.1=0.3$. From the inequality $|X-10|<0.3$ we get $-0.3<X-10<0.3 ; 10-0.3<X<10+0.3, 9.7<X<10.3$. Thus, the required interval is $(9.7 ; 10.3)$.

Example 16. Find $P(|X-a|<\sigma)$ for a random variable distributed according to the normal law with parameters $a$ and $\sigma$.

Solution. According to formula (3.5.4) for $\delta=\sigma$, we find

$$
P(|X-a|<\sigma)=2 \Phi\left(\frac{\delta}{\sigma}\right)=2 \Phi\left(\frac{\sigma}{\sigma}\right)=2 \Phi(1)=2 \cdot 0.341345=0.682690
$$