139. The probability of an event occurring in each of 400 independent trials is 0.8. Find a positive number $\varepsilon$ such that with a probability of 0.99, the absolute value of the deviation of the relative frequency of the event's occurrence from its probability of 0.8 does not exceed $\varepsilon$.
Solution. According to the condition, $n=400, p=0.8, q=0.2$. Therefore,

$$
2 \Phi(\varepsilon \sqrt{400 /(0.8 \cdot 0.2)})=0.99 \quad \text { or } \quad \Phi(50 \varepsilon)=0.495 .
$$

From the table in Appendix 2, we find $\Phi(2.57)=0.495$, which means $50 \varepsilon=$ $=2.57$. From this, the required value $\varepsilon=0.05$.