1776. What should be the sample size to ensure that when determining the percentage of non-standard items, the maximum error $\varepsilon=5\%$ is achieved with a confidence probability $P=0.99896$.
Solution. Given the confidence probability $P=$ $=0.99896$, we find the argument $t$ of the Laplace function $\Phi(t)$, for which $\Phi(t)=0.99896 ; t=3.28$. Therefore, the sample size, at which with a probability of 0.99896 it can be stated that the sample proportion differs from the population proportion in absolute value by less than $5 \%$, is determined by the formula

$$
n \approx \frac{t^{2} p(1-p)}{\varepsilon^{2}}=\frac{(3.28)^{2} p(1-p)}{(0.05)^{2}}
$$

If the proportion of the characteristic $p$ were known at least approximately, then the approximate value of $n$ would follow directly from this formula. Since there is no information about $p$ in the problem statement, let's consider $n$ as a function of $p$ and find its maximum value. The derivative of $n$ with respect to $p n^{\prime}=$ $=\frac{(3.28)^{2}}{(0.05)^{2}}[p(1-p)]^{\prime}=\frac{(3.28)^{2}}{(0.05)^{2}}(1-2 p)$ is zero at $p=0.5$, and changes sign from positive to negative as $p$ passes through this value. Thus, $n$ attains its maximum value at $p=0.5$:

$$
n_{\max }=\frac{(3.28)^{2}}{(0.05)^{2}} \cdot \frac{1}{4} \approx 1076
$$