Problem 3. The random variable $X$ - the number of weed seeds in a grain sample, is distributed according to the Poisson law. In the table

| $x_{i}$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m_{i}$ | 405 | 366 | 175 | 40 | 8 | 5 | 2 | 1000 |

the distribution of the number of weed seeds in $n=1000$ grain samples is provided. In the first row - the number $x_{i}$ of weed seeds in one sample, in the second row - the number of samples containing $x_{i}$ weed seeds. Find the point estimate of the unknown parameter of the Poisson distribution.
Solution. The Poisson law has the form $p_{m}\left(x_{\imath}\right)=\frac{\lambda^{x_{2}} e^{-\lambda}}{x_{\imath}!}$, where $m$ is the number of trials conducted in one experiment, $x_{\imath}$ is the number of occurrences of the event in the $i$-th experiment. For the Poisson law, the mathematical expectation coincides with the parameter $\lambda$. On the other hand, the unbiased estimate of $M(X)$ is taken as the sample mean $\bar{x}_{\text {v }}$. We calculate $\bar{x}_{\text {v }}$ using the formula from point $4^{\circ}:$

$\bar{x}_{\mathrm{B}}=\frac{1}{1000}(405 \cdot 0+366 \cdot 1+175 \cdot 2+40 \cdot 3+8 \cdot 4+4 \cdot 5+2 \cdot 6)=0.9$.

Answer. $\lambda=0.9 ; p_{m}\left(x_{i}\right)=\frac{0.9^{x_{2}} e^{-0.9}}{x_{\imath}!}$.