508. A sample of size $n=10$ is drawn from the general population:

$$
\begin{array}{lllllllll}
\text { value } & x_{i} & -2 & 1 & 2 & 3 & 4 & 5 \\
\text { frequency } & n_{i} & 2 & 2 & 2 & 2 & 2 & 1
\end{array}
$$

Estimate the mathematical expectation $a$ of the normally distributed characteristic of the general population with a reliability of 0.95 using the confidence interval based on the sample mean.
Solution. The sample mean and the "corrected" standard deviation will be found, respectively, by the formulas:

$$
\bar{x}_{\mathrm{B}}=\frac{\sum n_{i} x_{i}}{n}, s=\sqrt{\frac{\sum n_{i}\left(x_{i}-\bar{x}_{\mathrm{B}}\right)^{2}}{n-1}}
$$

Substituting the data from the problem into these formulas, we get $\overline{x_{B}}=2, s=2.4$.

Let's find $t_{\gamma}$. Using the table in Appendix 3, for $\gamma=0.95$ and $n=10$, we find $t_{\gamma}=2.26$.

Let's find the required confidence interval:

$$
\bar{x}_{\mathrm{B}}-t_{\gamma} s / \sqrt{ } \bar{n}<a<\bar{x}_{\mathrm{B}}+t_{\gamma} s / \sqrt{n}
$$

Substituting $\bar{x}_{\mathrm{B}}=2, t_{\gamma}=2.26, s=2.4, n=10$, we get the required confidence interval $0.3<a<7$, covering the unknown expected value $a$ with a reliability of 0.95.