1777. It is planned to conduct a selective survey of 5000 light bulbs to determine the average duration of their burning. What should be the volume of the non-repeated sample to ensure with a probability of 0.9802 that the general average differs from the sample average by less than 15 hours in absolute value, if the general variance is less than 10000?
Solution. According to the table (Appendix 2), we find that the argument $t=2.33$ corresponds to the given confidence probability of 0.9802. For sampling without replacement, the sample size is determined by the formula

$$
n \approx \frac{N t^{2} \sigma_{0}^{2}}{N \varepsilon^{2}+t^{2} \sigma_{0}^{2}}
$$

In our case $\sigma_{0}^{2}<10000$, therefore

$$
\begin{gathered}
n \approx \frac{5000 \cdot(2.33)^{2} \sigma_{0}^{2}}{5000 \cdot(15)^{2}+(2.33)^{2} \sigma_{0}^{2}}<\frac{5000 \cdot 5.4289 \cdot 10000}{5000 \cdot 225+5.4289 \cdot 10000}= \\
=230.2 ; n_{\max }=230
\end{gathered}
$$