131. The probability of an event occurring in each of 625 independent trials is 0.8. Find the probability that the relative frequency of the event's occurrence deviates from its probability in absolute value by no more than 0.04.
Solution. According to the condition, $n=625 ; p=0.8 ; q=0.2 ; \varepsilon=0.04$. It is required to find the probability $P(|m / 625-0.8|<0.04)$. We will use the formula

$$
P\left(\left|\frac{m}{n}-p\right|<\varepsilon\right)=2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$

We have

$$
P\left(\left|\frac{m}{625}-0.8\right|<0.04\right)=2 \Phi\left(0.04 \sqrt{\frac{625}{0.8 \cdot 0.2}}\right)=2 \Phi(2.5)
$$

From the table in Appendix 2, we find $\Phi(2.5)=0.4938$. Therefore, $2 \Phi(2.5)=2 \cdot 0.4938=0.9876$. Thus, the required probability is approximately 0.9876.