127. The fleece weight of the Ovsiy breed sheep is a random variable $X$ distributed according to the normal law. Practically all possible values of this variable belong to the interval $(7 ; 10.6)$ kg. Find the interval, symmetric with respect to the mathematical expectation, in which with a probability of 0.95 the possible values of the fleece weight of these sheep are contained.
S o l u t i o n. The random variable $X$ - the wool clip in Aksaniyska breed sheep. Practically all possible values of any random variable distributed according to the normal law are contained in the interval $a \pm 3 \sigma$, where $a$ and $\sigma$ are its mathematical expectation and standard deviation (the three-sigma rule). According to the three-sigma rule and the condition of the problem, we have:

$$
\left\{\begin{array}{l}
a-3 \sigma=7 \\
a+3 \sigma=10.6
\end{array}\right.
$$

Solving the system of equations, we get that for the random variable $X$ under consideration, the mathematical expectation $a=8.8$, and the standard deviation $\sigma=0.6$. To answer the question of the problem, we will use formula (63). In this problem, the sought quantity is $\boldsymbol{\varepsilon}$.

Substituting the known quantities into formula (63), we get

$$
P(|X-8.8| \leq \varepsilon)=2 \Phi\left(\frac{\varepsilon}{0.6}\right)=0.95
$$

From this equality, we find $\Phi\left(\frac{\varepsilon}{0.6}\right)=0.475$. Using the table of values of the function $\Phi(x)$ (see table 3 in the appendix), we get that if $\Phi(x)=0.475$, then $x=1.96$. Thus, $\frac{\varepsilon}{0.6}=1.96$, from which $\varepsilon=1.96 \cdot 0.6=1.176$. With a probability of 0.95, we can assert that the possible values of the wool clip in Aksaniyska breed sheep are in the interval ( $8.8 \pm 1.176$ ) kg, i.e., the wool clip can be from 7.624 to 9.976 kg.