1615. On average, $85 \%$ of the items coming off the conveyor are of the first grade. How many items need to be taken so that with a probability of 0.997, the deviation of the frequency of first-grade items from 0.85 in absolute value does not exceed 0.01?
Solution. Here $p=0.85, q=1-0.85=0.15, \varepsilon=0.01$, $P=0.997, n$-? Since in the equality $P\left(\left|\frac{m}{n}-p\right|<\varepsilon\right) \approx$ $\approx \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)$ the probability $P$ on the left is known, we first solve the equation $\Phi(t)=P$. Let $t_{P}$ be the root of this equation. Then $\varepsilon \sqrt{\frac{n}{p q}} \approx t_{p}, \varepsilon^{2} \frac{n}{p q} \approx t_{P}^{2}$ and, consequently, $n \approx$
$\approx \frac{p q t_{P}^{2}}{\varepsilon^{2}}$. For our case $t_{0.997}=3$, therefore

$$
n \approx \frac{0.85 \cdot 0.15 \cdot 3^{2}}{(0.01)^{2}}=\frac{1.1475}{0.0001}=11475
$$