1766. To determine the average percentage of raw protein in wheat grains, 626 grains were selected, and the examination showed that the sample mean is 16.8, and the sample variance is 4. What is the probability that the average percentage of raw protein differs from 16.8 by less than $0.2 \%$ in absolute value?
Solution. The volume of the general population is unknown to us. Therefore, we find the sampling error using the formula for repeated sampling $\mu \approx \frac{\sigma}{\sqrt{n-1}}$. In our case $n=626, \sigma^{2}=4$ and, consequently, $\mu \approx \frac{2}{\sqrt{625}}=\frac{2}{25}=0.08$. Since $P(|\tilde{X}-\bar{X}|<\varepsilon) \approx$ $\approx \Phi\left(\frac{\varepsilon}{\mu}\right)$, the required probability is

$$
P(|\bar{X}-16.8|<0.2) \approx \Phi\left(\frac{0.2}{0.08}\right)=\Phi(2.5)=0.98758
$$