Example 10. 600 corn seeds are planted with a probability of 0.9 for each seed to germinate. Find the limit of the modulus of deviation of the frequency of germinated seeds from the probability $p=0.9$, if this limit is to be guaranteed with a probability $P=0.995$.
Solution. We will use formula (4.3.9), i.e.

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

In this case, $n=600, p=0.9, q=1-0.9=0.1, P=0.995$; we need to determine the value of $\varepsilon$. The specified formula takes the form:

$$
\begin{gathered}
P\left(\left|\frac{m}{n}-0.9\right| \leq \varepsilon\right)=2 \Phi\left(\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}\right) \\
2 \Phi\left(\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}\right)=0.995, \quad \Phi\left(\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}\right)=0.4975
\end{gathered}
$$

Using the table of values of the Laplace function (see appendix), we solve the equation $\Phi(t)=0.4975 ; t=2.81$.

Therefore,

$$
\varepsilon \sqrt{\frac{600}{0.9 \cdot 0.1}}=2.81, \varepsilon=2.81 \cdot \sqrt{\frac{0.09}{600}}=\frac{2.81 \cdot 0.3}{10 \sqrt{6}}=0.0034
$$