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

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