Example 9. The probability of an event occurring in each of 900 independent trials is 0.5. Find the probability that the relative frequency of the event's occurrence deviates from its probability in absolute value by no more than 0.02.
Solution. We will use formula (4.3.9). According to the condition, $n=900, p=0.5, q=0.5, \varepsilon=0.02$. In accordance with the specified formula, we obtain

$$
P\left(\left|\frac{m}{900}-0.5\right| \leq 0.02\right)=2 \Phi\left(0.02 \sqrt{\frac{900}{0.5 \cdot 0.5}}\right)=2 \Phi(1.2)
$$

From the table, we find $\Phi(1.2)=0.3849$. Therefore, $2 \Phi(1.2)=2 \cdot 0.3849=0.7698$. Thus, the required probability is approximately equal to 0.7698.