Problem 4. The operating time of a radio device element follows the law $f(x)=\lambda e^{-\lambda x}$. In the table

| $x_{2}$ | 2.5 | 7.5 | 12.5 | 17.5 | 22.5 | 27.5 | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $n_{2}$ | 133 | 45 | 15 | 4 | 2 | 1 | 200 |

the empirical distribution of the average operating time of $n=200$ elements is provided: $x_{i}$ - the average operating time of the element in hours, $n_{i}$ - the number of elements that operated for $x_{i}$ hours. Find $\lambda$.
Solution. As is known, for the exponential law, the mathematical expectation $M(X)$ is equal to $\frac{1}{\lambda}$. Therefore, the point estimate of $\lambda$ is $\frac{1}{\overline{\bar{x}}_{\mathrm{B}}}$. Let's calculate $\bar{x}_{\mathrm{B}}:$

$$
\bar{x}_{\mathrm{B}}=\frac{1}{200}(133 \cdot 2.5+45 \cdot 7.5+15 \cdot 12.5+4 \cdot 17.5+2 \cdot 22.5+1 \cdot 27.5)=
$$

$$
=4.8575
$$

Therefore, $\lambda=\frac{1}{4.8575}=0.2$.

Answer. $f(x)=0.2 e^{-0.2 x}$.