531. Find the skewness and 9-excess by the method of products for the given sample distribution of size $n=100$:

$$
\begin{array}{lrrrrrrr}
\text { variant } & x_{i} & 12 & 14 & 16 & 18 & 20 & 22 \\
\text { frequency } & n_{i} & 5 & 15 & 50 & 16 & 10 & 4
\end{array}
$$
Solution. We will use the method of products. We will compile the calculation table 3. In § 1 of this chapter, when solving problem 523, it was already indicated how to fill in columns 1-5 of the calculation table, so we will limit ourselves to brief explanations.

To fill in column 6, it is convenient to multiply the numbers of each row of columns 3 and 5. To fill in column 7, it is convenient to multiply the numbers of each row of columns 3 and 6. Column 8 serves to control calculations using the identity

$$
\sum n_{i}\left(u_{i}+1\right)^{4}=\sum n_{i} u_{i}^{4}+4 \sum n_{i} u_{i}^{3}+6 \sum n_{i} u_{i}^{2}+4 \sum n_{i} u_{i}+n
$$

| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | $n_{i}$ | $u_{i}$ | $n_{i} u_{i}$ | $n i u_{i}^{2}$ | $n i u_{i}^{\mathbf{3}}$ | $n_{i} u_{i}^{4}$ | $n_{i}\left(u_{i}+1\right)^{4}$ |
| 12 | 5 | -2 | -10 | 20 | -40 | 80 | 5 |
| 14 | 15 | -1 | -15 | 15 | -15 | 15 | - |
| 16 | 50 | 0 | -25 | - | -55 | - | 50 |
| 18 | 16 | 1 | 16 | 16 | 16 | 16 | 256 |
| 20 | 10 | 2 | 20 | 40 | 80 | 160 | 810 |
| 22 | 4 | 3 | 12 | 36 | 108 | 324 | 1024 |
|  |  |  | 48 |  | 204 |  |  |
|  | $n=100$ |  | $n_{i} u_{i}=$  $=23$ | $\sum n_{i} u_{i}^{2}$  $=127$ | $\underset{=149}{\sum n_{i} u_{i}^{3}}=$ | $\sum_{=595} n_{i} u_{i}^{4}=$ | $\sum n_{i}\left(u_{i}+\right.$  $+1)^{\prime}=$  $=2145$ |

We will present the calculation table 3.

Table 8

Verification:

$$
\begin{aligned}
& \sum n_{i}\left(u_{i}+1\right)^{4}=2145 \\
& \sum n_{i} u_{i}^{4}+4 \sum n_{i} u_{i}^{3}+6 \sum n_{i} u_{i}^{2}+4 \sum n_{i} u_{i}+n= \\
& =595+4 \cdot 149+6 \cdot 127+4 \cdot 23+100=2145 .
\end{aligned}
$$

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

We will find the conditional moments of the third and fourth orders (the conditional moments of the first and second orders were calculated in problem 523: $\left.M_{1}^{*}=0.23, M_{2}^{*}=1.27\right)$ :

$$
M_{3}^{*}=\frac{\sum n_{i} u_{i}^{3}}{n}=\frac{149}{100}=1.49 ; \quad M_{4}^{*}=\frac{\sum n_{i} u_{i}^{4}}{n}=\frac{595}{100}=5.95
$$

We will find the central empirical moments of the third and fourth orders:

$$
\begin{gathered}
m_{3}=\left[M_{3}^{*}-3 M_{1}^{*} M_{2}^{*}+2\left(M_{1}^{*}\right)^{3}\right] h^{3} \\
m_{4}=\left[M_{4}^{*}-4 M_{1}^{*} M_{3}^{*}+6\left(M_{1}^{*}\right)^{2} M_{2}^{*}-3\left(M_{1}^{*}\right)^{4}\right] h^{4}
\end{gathered}
$$

Substituting $h=2$ and $M_{1}^{*}=0.23, M_{2}^{*}=1.27, M_{3}^{*}=1.49, M_{4}^{*}=5.95$, we get $m_{3}=5.104, m_{4}=79.582$.

We will find the required skewness and kurtosis, taking into account that $D_{B}=4.87$ (see problem 523):

$$
\begin{gathered}
a_{s}=m_{3} / \sigma_{B}^{3}=5.124 /(\sqrt{4.87})^{3}=0.47 \\
e_{k}=m_{4} / \sigma_{B}^{4}-3=79.582 /(\sqrt{4.87})^{4}-3=0.36
\end{gathered}
$$