1774. To determine the percentage of second-grade items in a batch, a random repeated sample of 100 units is taken. Determine the confidence limits for the percentage of second-grade items in the entire batch, which can be guaranteed with a confidence probability of \( P = 0.997 \), if 25 second-grade items were found in the sample.
Solution. The frequency of the feature in the sample (sample proportion) $w=\frac{25}{100}=0.25$. Therefore, the sampling error

$$
\mu_{\text {d }} \approx \sqrt{\frac{w(1-w)}{n-1}}=\sqrt{\frac{0.25 \cdot 0.75}{100-1}}=0.0435
$$

We need to find such $\varepsilon$ that $P(|w-p|<\varepsilon)=0.997$. Since $P(|w-p|<\varepsilon) \approx \Phi\left(\frac{\varepsilon}{\mu_{\AA}}\right)$, then approximately $\varepsilon$ can be found
from the equation $\Phi\left(\frac{\varepsilon}{\mu_{\AA}}\right)'=0.997$. From the table (Appendix 2) we find $\Phi(t)=0.997$ at $t=3$. Then $\frac{\varepsilon}{\mu_{\text {d }}}=3, \varepsilon=3 \mu_{\text {d }}=3 \cdot 0.0435=$ $=0.1305=13.05\%$ and, therefore, the population proportion with a probability of 0.997 is contained between

and

$$
\begin{aligned}
& w+\varepsilon=25 \%+13.05 \%=38.05 \% \\
& w-\varepsilon=25 \%-13.05 \%=11.95 \%
\end{aligned}
$$