Example 19. Find the probability that a normal random variable with a mean of 3 and a variance of 4 will take values: 1) in the interval $(-1,5)$; 2) no more than $8$; 3) no less than 5; 4) in the interval $(-3,9)$.
Solution. We will use formula (3.5.2).

1) According to the condition, $\alpha=-1, \beta=5, a=3, \sigma=\sqrt{4}=2$, therefore

$$
\begin{gathered}
P(-1<X<5)=\Phi\left(\frac{5-3}{2}\right)-\Phi\left(\frac{-1-3}{2}\right)=\Phi(1)-\Phi(-2)=\Phi(1)+\Phi(2)= \\
=0.3413+0.4772=0.8185
\end{gathered}
$$

2) $P(X \leq 8)=P(-\infty<x \leq 8)=\Phi\left(\frac{8-3}{2}\right)-\Phi\left(\frac{-\infty-3}{2}\right)=\Phi(2.5)+\Phi(+\infty)=$

$$
=0.4938+0.5=0.9938
$$

3) $P(X \geq 5)=P(5 \leq X<+\infty)=\Phi\left(\frac{+\infty-3}{2}\right)-\Phi\left(\frac{5-3}{2}\right)=\Phi(+\infty)-\Phi(1)=$

$$
=0.5-0.3643=0.1587
$$

4) 

$$
\begin{gathered}
P(-3<X<9)=\Phi\left(\frac{9-3}{2}\right)-\Phi\left(\frac{-3-3}{2}\right)=\Phi(3)-\Phi(-3)=2 \Phi(3)= \\
=2 \cdot 0.4986=0.9972 \\
P(-3<X<9)=0.9972
\end{gathered}
$$

Thus, the random variable $X$ taking values in the interval $(-3,9)$ is almost certain.