501. Find the confidence interval for estimating the unknown expected value $a$ of a normally distributed feature $X$ of the population with a reliability of 0.95, if the population standard deviation $\sigma=5$, the sample mean $\bar{x}_{\text {v }}=14$ and the sample size $n=25$.
Solution. It is required to find the confidence interval

$$
\bar{x}_{\mathrm{B}}-t \frac{\sigma}{\sqrt{n}}<a<\bar{x}_{\mathrm{B}}+t \frac{\sigma}{\sqrt{n}}
$$

All quantities except $t$ are known. We will find $t$ from the relation $\Phi(t)=0.95 / 2=0.475$. From the table in Appendix 2, we find $t=1.96$. Substituting $t=1.96, \bar{x}_{\mathrm{B}}=14, \sigma=5, n=25$ into (*), we will finally obtain the required confidence interval $12.04<a<15.96$.