Task 2. The quality control department checks 900 items for standard compliance. The probability that an item is standard is 0.9. Find, with a probability of 0.95, the boundaries within which the number $m$ of standard items among those checked will be contained.
Solution. Given: $n=900 ; p=0.9 ; q=0.1 ; P=2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)=$ $=0.95$. First, let's determine the deviation of the relative frequency from the probability of the event itself. The maximum value of this deviation $\varepsilon$ can be calculated from the relation $2 \Phi\left(\varepsilon \sqrt{\frac{900}{0.9 \cdot 0.1}}\right)=0.95$. From Table 1 in the appendix, we determine that the equality $\Phi(t)=0.475$ holds when $t=1.96$. Since $t=100 \cdot \varepsilon$, we have $\varepsilon=0.0196$.

It remains to solve the inequality $\left|\frac{m}{900}-0.9\right| \leqslant 0.0196$ for $m$. Sequentially, we have: $-0.0196 \leqslant \frac{m}{900}-0.9 \leqslant 0.0196 ; \quad 0.8804 \leqslant$ $\leqslant \frac{m}{900} \leqslant 0.9196 ; \quad 792.4 \leqslant m \leqslant 827.6$. Since $m$ must be an integer, after rounding we get $m \in[793 ; 827]$.

Answer. From 793 to 827.