1768. From a batch of 5000 electric lamps, 300 were selected using a simple random sampling without replacement. The average burning duration of the lamps in the sample was 1450 hours, and the variance was 40000. What is the probability that the average burning duration of the lamps in the entire batch is within the range of 1410 to 1490 hours?
Solution. In the problem, we are given $N=5000, n=300, \tilde{X}=1450$, $\sigma^{2}=40000$. We need to find $P(1410<\bar{X}<1490)=P(|\bar{X}-1450|<40)$.

First, let's calculate the sampling error

$$
\mu \approx \sqrt{\frac{\sigma^{2}}{n-1}\left(1-\frac{n}{N}\right)}=\sqrt{\frac{40000}{299}\left(1-\frac{300}{5000}\right)}=11.214
$$

Using the formula for the confidence probability, we find the required probability

$$
P(|\bar{X}-1450|<40) \approx \Phi\left(\frac{40}{11.214}\right)=\Phi(3.57)=0.99964
$$