Example 6.35. Find the confidence interval for estimating the mathematical expectation $a$ of a normal distribution with a reliability of 0.95, knowing the sample mean $\bar{X}_{\mathrm{B}}=10.43$, the sample size $n=100$, and the standard deviation $\sigma=5$.
Solution. We will use the formula

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

Let's find $t$. From the relation $2 \Phi(t)=0.95$ we get: $\Phi(t)=0.475$. By the table, we find $t=1.96$. Let's find the accuracy of the estimate

$$
\delta=\frac{t \sigma}{\sqrt{n}}=\frac{1.96 \cdot 5}{\sqrt{100}}=0.98
$$

The desired confidence interval is ( $10.43-0.98<a<10.43+0.98$ ) or $(9.45<a<11.41)$.

The meaning of the obtained result is as follows: if a large number of samples are taken, then $95 \%$ of them will determine such confidence intervals in which the mathematical expectation will be included.

## 6.5. Methods for Calculating Sample Characteristics

Let's consider rational methods for determining the sample characteristics $\bar{x}_{\mathrm{B}}$ and $d_{\mathrm{B}}$.