The solution is as follows. Before finding the desired probability, we need to calculate the sampling error $\mu_{\text {d }}: \mu_{\pi}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.15 \cdot 0.85}{1000}}=$ $=\sqrt{0.0001275}=0.0113$, since $n=1000, p=15 \%=0.15$. It is known that $P(|w-p|<\varepsilon) \approx \Phi\left(\frac{\varepsilon}{\mu_{\mathrm{I}}}\right)$. In our case, $\varepsilon=2 \%$, so $P(|w-0.15|<0.02) \approx \Phi\left(\frac{0.02}{0.0113}\right)=\Phi(1.77)=0.92327$.

Note that, as in the previous problem, the size of the population is unknown. Therefore, despite the fact that the sampling is without replacement, we calculate the sampling error using the formula for sampling with replacement. As a result, the obtained probability is slightly less than the desired one.
1767. Among the standard products of one factory, on average $15 \%$ belong to the second grade. With what probability can it be stated that the percentage of second-grade products among 1000 standard products of this factory differs from $15 \%$ by less than $2 \%$ in absolute value?