523. Find the sample mean and sample variance using the product method for the given sample distribution of size $n=100$:

| variant | $x_{i}$ | 12 | 14 | 16 | 18 | 20 | 22 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 5 | 15 | 50 | 16 | 10 | 4 |
Solution. We will form Calculation Table 1; for this:

1) we will write the variants in the first column;
2) we will write the frequencies in the second column; the sum of the frequencies (100) will be placed in the bottom cell of the column;
3) as the false zero $C$, we will choose the variant (16), which has the highest frequency (as $C$, we can take any variant located approximately in the middle of the column); in the cell of the third column, which belongs to the row containing the false zero, we write 0; above the zero, we sequentially record $-1, -2$, and below the zero, $1, 2, 3$;
4) the products of the frequencies $n_{i}$ by the conditional variants $u_{i}$ will be written in the fourth column; separately, we find the sum (-25) of the negative numbers and the sum (48) of the positive numbers; adding these numbers, their sum (23) is placed in the bottom cell of the fourth column;
5) the products of the frequencies by the squares of the conditional variants, i.e., $n_{i} u_{i}^{2}$, will be written in the fifth column (it is more convenient to multiply the numbers of each row of the third and fourth columns: $u_{i} \cdot n_{i} u_{i}=n_{i} u_{i}^{2}$); the sum of the numbers in the column (127) is placed in the bottom cell of the fifth column;
6) the products of the frequencies by the squares of the conditional variants, increased by one, i.e., $n_{i}\left(u_{i}+1\right)^{2}$, will be written in the sixth control column; the sum of the numbers in the column (273) is placed in the bottom cell of the sixth column.

In the end, we will get Calculation Table 1.

For the control of calculations, we use the identity

$$
\sum n_{i}\left(u_{i}+1\right)^{2}=\sum n_{i} u_{i}^{2}+2 \sum n_{i} u_{i}+n
$$

Control:

$$
\begin{gathered}
\sum n_{i}\left(u_{i}+1\right)^{2} \stackrel{f}{=} 273, \sum n_{i} u_{i}^{2}+2 \sum n_{i} u_{i}+n= \\
=127+2 \cdot 23+100=273 .
\end{gathered}
$$

The coincidence of the control sums testifies to the correctness of the calculations.

We will calculate the conditional moments of the first and second orders:

$$
M_{1}^{*}=\left(\sum n_{i} u_{i}\right) / n=23 / 100=0.23 ; \quad M_{2}^{*}=\left(\sum n_{i} u_{i}^{2}\right) / n=127 / 100=1.27
$$

We will find the step (the difference between any two adjacent variants): $h=14-12=2$.

We will calculate the required sample mean and variance, taking into account that the false zero (the variant with the highest frequency) $C=16:$

$$
\begin{gathered}
\bar{x}_{\mathrm{B}}=M_{1}^{*} h+C=0.23 \cdot 2+16=16.46 \\
D_{\mathrm{B}}=\left[M_{2}^{*}-\left(M_{1}^{*}\right)^{2}\right] h^{2}=\left[1.27-0.23^{2}\right] \cdot 2^{2}=4.87
\end{gathered}
$$

Table 1

| 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | $n_{i}$ | $u_{i}$ | $n_{i} u_{i}$ | $n_{i} u_{i}^{2}$ | $n_{i}\left(u_{i}+1\right)^{2}$ |
| 12 | 5 | -2 | -10 | 20 | 5 |
| 14 | 15 | -1 | -15 | 15 | - |
| 16 | 50 | 0 | -25 | - | 50 |
| 18 | 16 | 1 | 16 | 16 | 64 |
| 20 | 10 | 2 | 20 | 40 | 90 |
| 22 | 4 | 3 | 12 | 36 | 64 |
|  |  |  | 48 |  |  |
|  | $n=100$ |  | $\sum n_{i} u_{i}=23$ | $\sum n_{i} u_{i}^{2}=127$ | $\sum n_{i}\left(u_{i}+1\right)^{2}=273$ |