Example 13. The probability of an event occurring in each of 900 independent trials is 0.5. Find such a positive number $\varepsilon$, so that with a probability of 0.77, the absolute deviation of the frequency of the event's occurrence from its probability 0.5 does not exceed $\varepsilon$.
Solution. We will use formula (4.3.9). In this case, $n=900, p=0.5, q=1-0.5=0.5$. Therefore,

$$
\begin{gathered}
2 \Phi\left(\varepsilon \sqrt{\frac{900}{0.5 \cdot 0.5}}\right)=0.77, \quad \text { or } \\
2 \Phi(60 \varepsilon)=0.77, \Phi(60 \varepsilon)=0.385
\end{gathered}
$$

From the table of values of the Laplace function, we find $\Phi(1.2)=0.385$; hence $60 \varepsilon=1.2$, from which $\varepsilon=0.02$.