75. The probability of infection by powdery smut for a certain variety of millet is 0.3. A random sample of 90 millet panicles of this variety is selected. Find the limit of the absolute value of the deviation of the relative frequency of infected panicles from the probability $p=0.3$, if this limit must be guaranteed with a probability of 0.9836.
Solution. According to the condition $p=0.3 ; q=0.7 ; n=90$; $P\left(\left|\frac{m}{n}-0.3\right| \leq \varepsilon\right)=0.9836$. We need to find $\varepsilon$.

According to formula (23)

$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right) \approx 2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)=0.9836
$$

From the last equality, we find $\Phi(x)=0.4918$, where $x=\varepsilon \sqrt{\frac{n}{p q}}$.

Using the table of values of the Laplace function (see table 3 in the appendix), we find $x=2.40$. Since

$$
\varepsilon \sqrt{\frac{n}{p q}}=\varepsilon \sqrt{\frac{90}{0.3 \cdot 0.7}} \approx 20.702 \varepsilon
$$

then $20.702 \varepsilon \approx 2.40$, from which $\varepsilon \approx 0.1159$.