1679. Let the random variable $X$ be distributed according to the normal law with the density

$$
p(x)=\frac{1}{\sigma \sqrt{2 \pi}} \cdot e^{-\frac{(x-a)^{2}}{2 \sigma^{2}}}
$$

It is required to find the probable deviation of this random variable, i.e., such a constant $E$ that

$$
P(|X-a|<E)=50 \%
$$
Solution. $\quad P(|X-a|<E)=\Phi\left(\frac{E}{\sigma}\right)$.

Here $\Phi\left(\frac{E}{\sigma}\right)$ is the value of the Laplace function at the point $\frac{E}{\sigma}$. According to the problem, $\Phi\left(\frac{E}{\sigma}\right)=0.5$. From the table (appendix 2), we find the value of the argument $t$ for which $\Phi(t)=0.5: t=0.675$. Therefore,

$$
\frac{E}{\sigma}=0.675 \text { and } E=0.675 \text { s. }
$$