Example 1. The random variable $\xi$ is distributed according to the normal law with an unknown expected value $a$ and variance 1. There is a sample $x_{1}, \ldots, x_{100}$ of values of $\xi, \bar{x}=1.3$. Find the confidence interval for the expected value $a$ at the confidence level $p=0.98$.
## Solution.

1. Find the function $f\left(x_{1}, \ldots, x_{100}, a\right)$, which has a known distribution function $F(x)$.

In this case,

$$
f\left(x_{1}, \ldots, x_{100}, a\right)=\frac{x_{1}+\ldots+x_{100}}{100}-a
$$

is distributed according to the normal law with mean 0 and variance $1 / 100$. Therefore,

$$
F(x)=\frac{10}{\sqrt{2 \pi}} \int_{-\infty}^{x} e^{-100 u^{2} / 2} d u=\Phi(10 x)
$$

2. Find the quantiles $q_{1}$ and $q_{2}$ from the conditions

$$
\mathrm{P}\left(\frac{x_{1}+\ldots+x_{100}}{100}-a \leqslant q_{1}\right)=\frac{1-p}{2}=0.01
$$

and

$$
\mathrm{P}\left(\frac{x_{1}+\ldots+x_{100}}{100}-a>q_{2}\right)=\frac{1-p}{2}=0.01
$$

By the definition of the distribution function $F(x)$,

$$
\mathrm{P}\left(\frac{x_{1}+\ldots+x_{100}}{100}-a \leqslant q_{1}\right)=F\left(q_{1}\right)
$$

and

$$
\mathrm{P}\left(\frac{x_{1}+\ldots+x_{100}}{100}-a \geqslant q_{2}\right)=1-\mathrm{P}\left(\frac{x_{1}+\ldots+x_{100}}{100}-a \leqslant q_{2}\right)=1-F\left(q_{2}\right)
$$

Therefore, the quantiles $q_{1}$ and $q_{2}$ are determined by the equations

$$
F\left(q_{1}\right)=\frac{1-p}{2}=0.01, \quad F\left(q_{2}\right)=\frac{1+p}{2}=0.99
$$

To solve these equations for $q_{1}$ and $q_{2}$, we use a table of values of the function $F(x)=\Phi(10 x)$ or the software package SOLVER.STAT.

We obtain $q_{1}=-0.233$ and $q_{2}=0.233$.

3. Solve the inequalities

$$
q_{1}<\frac{x_{1}+\ldots+x_{100}}{100}-a<q_{2}
$$

for $a$ and get

$$
\frac{x_{1}+\ldots+x_{100}}{100}-q_{2}<a<\frac{x_{1}+\ldots+x_{100}}{100}-q_{1}
$$

4. The random interval

$$
\left(\frac{x_{1}+\ldots+x_{100}}{100}-q_{2}, \frac{x_{1}+\ldots+x_{100}}{100}-q_{1}\right)
$$

contains $a$ with probability $p=0.98$. Substitute the specific sample values $x_{1}, \ldots, x_{100}$ of the random variable $\xi$ and the quantiles $q_{1}, q_{2}$ into (2) and get the specific realization $(1.3-0.233,1.3+0.233)$ of the random interval (2), i.e., the confidence interval for $a$ with confidence level $p=0.98$.

Answer. $a \in(1.067,1.533)_{0.98}$.

Remark. The point estimate $\left(x_{1}+\ldots+x_{100}\right) / 100$ of the mean $a$ coincides with the midpoint of the confidence interval. Therefore, the answer can be written as $a=1.3 \pm 0.233_{0.98}$.