Example 6.36. Given the sample values $x_{i}$ and corresponding frequencies $n_{i}$ of a quantitative characteristic $X$:

| $x_{i}$ | 10 | 15 | 20 | 25 | 30 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $n_{i}$ | 6 | 16 | 50 | 24 | 4 |

Find the sample mean, variance, and standard deviation using the method of products.
Solution. We will form a calculation table for which:

1) we will write the variants $x_{i}$ in the first column;
2) we will write the frequencies $n_{i}$ in the second column;
3) as the false zero $C$, we will choose the variant 20 (this variant has the highest frequency); in the cell of the third column, which belongs to the row containing the variant 20, we write 0; above the zero, we sequentially record the conditional variants $-1, -2$, and below the zero - sequentially 1, 2;
4) the products of frequencies by conditional variants $u_{i}$ will be written in the fourth column; we find the sum of these products and place it in the lower cell of the column;
5) the products of frequencies by the squares of conditional variants will be written in the fifth column; the sum of the numbers in the column (80) will be placed in the lower cell of the column;
6) the products of frequencies by the squares of conditional variants, increased by one, will be written in the sixth (control) column; the sum of the numbers in the column (188) will be placed in the lower cell of the column.

In the end, we will get the following calculation table:

| $x_{i}$ | $n_{i}$ | $u_{i}$ | $n_{i} \cdot u_{i}$ | $n_{i} \cdot u_{i}^{2}$ | $n_{i}\left(u_{i}+1\right)^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 10 | 6 | -2 | -12 | 24 | 6 |
| 15 | 16 | -1 | -16 | 16 | 0 |
| 20 | 50 | 0 | 0 | 0 | 50 |
| 25 | 24 | 1 | 24 | 24 | 96 |
| 30 | 4 | 2 | 8 | 16 | 36 |
|  | $n=100$ |  | $\sum n_{i} u_{i}=4$ | $\sum n_{i} u_{i}^{2}=80$ | $\sum n_{i}\left(u_{i}+1\right)^{2}=188$ |

Control

$$
\begin{gathered}
\sum n_{i} \cdot u_{i}^{2}+2 \sum n_{i} \cdot u_{i}+n=80+2 \cdot 4+100=188 \\
\sum n_{i}\left(u_{i}+1\right)^{2}=188
\end{gathered}
$$

The coincidence of the found sums indicates that the calculations have been performed correctly.

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

$$
M_{1}^{*}=\frac{\sum n_{i} u_{i}}{n}=\frac{4}{100}=0.04 ; \quad M_{2}^{*}=\frac{\sum n_{i} u_{i}^{2}}{n}=\frac{80}{100}=0.8
$$

We will find the step (difference between two adjacent variants): $h=15-10=5$.

We will find the required sample mean:

$$
\bar{x}_{\mathrm{B}}=M_{1}^{*} \cdot h+C=0.04 \cdot 5+20=20.2
$$

We will find the required sample variance:

$$
d_{\mathrm{B}}=\left[M_{2}^{*}-\left(M_{1}^{*}\right)^{2}\right] \cdot h=\left[0.8-(0.04)^{2}\right] \cdot 5^{2}=19.96
$$

We will find the required standard deviation:

$$
\sigma_{\mathrm{B}}=\sqrt{d_{\mathrm{B}}}=\sqrt{19.96}=4.47
$$