1614. 600 corn seeds are planted with a probability of 0.9 for each seed to germinate. Find the boundary of the absolute value of the deviation of the frequency of germinated seeds from the probability $p=0.9$, if this boundary is to be guaranteed with a probability $P=0.995$.
Solution. We know that if $n$ is the number of independent trials and $p$ is the probability of the event occurring in a single trial, then for any $\varepsilon>0$ the approximate equality holds:

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

where $q=1-p$. In our case, $n=600, p=0.9, q=1-0.9=$ $=0.1, P=0.995, \varepsilon-$ ? According to formula (4),

$P\left(\left|\frac{m}{n}-0.9\right|<\varepsilon\right) \approx \Phi\left(\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}\right)$ or $\Phi\left(\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}\right)=0.995$.

Using the table (Appendix 2), we solve the equation $\Phi(t)=$ $=0.995: t=2.81$. Hence, $\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}=2.81$ and, therefore,

$$
\varepsilon=2.81 \sqrt{\frac{0.09}{6 \cdot 100}}=\frac{2.81 \cdot 0.3}{10 \sqrt{6}}=0.034
$$