533. Find the skewness and kurtosis by the method of sums for the given sample distribution of size \( n=100 \):

\[
\begin{array}{rrrrrrrrrrr}
x_{i} & 48 & 52 & 56 & 60 & 64 & 68 & 72 & 76 & 80 & 84 \\
n_{i} & 2 & 4 & 6 & 8 & 12 & 30 & 18 & 8 & 7 & 5
\end{array}
\]
Solution. We will use the method of sums, for this purpose we will compose the calculation table 4. In § 2 of this chapter, when solving problem 529, it was already indicated how to fill in columns $1-4$ of the calculation table, so we will limit ourselves to brief explanations.

To fill in column 5, we will write zero in the cell of the row containing the false zero (68); above this zero and below it, we will place two more zeros.

In the cells above the zeros, we will write the accumulated frequencies, for which we will sum the frequencies of column 4 from top to bottom; in the end, we will have the following accumulated frequencies: $2 ; 2+8=10 ; 2+8+$ $+20=30$. By summing the accumulated frequencies, we will get the number $b_{3}=2+10+$ $+30=42$, which we will place in the top cell of the fifth column.

In the cells below the zeros, we will write the accumulated frequencies, for which we will sum the frequencies of column 4 from bottom to top; in the end, we will have the following accumulated frequencies: $5 ; 5+17=22$. By summing the accumulated frequencies, we will get the number $a_{3}=5+22=27$, which we will place in the bottom cell of the fifth column.

Similarly, we fill in column 6, and for this, we sum the frequencies of column 5. By summing the accumulated frequencies located above the zeros, we will get the number $b_{4}=2+12=14$, which we will write in the top cell of the sixth column. By summing the numbers located below the zeros (in our problem, there is only one addend), we will get the number $a_{4}=5$, which we will place in the bottom cell of the sixth column.

As a result, we will get the calculation table 4.

Control: the sum of the numbers located directly above the zero of the third column, to the left of it, and below it, should be equal to the sample size ( $32+30+38=100$ ); the sum of two numbers located above two steps of the stepped line (encircled by bold segments) should be equal to the numbers $b_{i}$, standing above the preceding step (when moving up the "ladder"): $32+40=72=b_{1} ; 40+30=70=b_{2} ; 30+12=42=b_{3}$. Similarly, the coincidence of the sums of two numbers located below the "steps of the ladder" leading downward is checked: $38+37=75=a_{1} ; 37+22=59=a_{2}$; $22+5=27=a_{3}$. In case of non-coincidence of at least one of the specified sums, it is necessary to look for an error in the calculation.

Let's find $d_{i}(i=1,2,3)$ and $s_{i}(i=1,2,3,4)$:

$$
d_{1}=a_{1}-b_{1}=75-72=3, \quad d_{2}=a_{2}-b_{2}=59-70=-11
$$

Table 4

| 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $x_{i}$ | $n_{i}$ | $b_{1}=72$ | $b_{2}=70$ | $b_{3}=42$ | $b_{4}=14$ |
| 48 | 2 | 2 | 2 | 2 | 2 |
| 52 | 4 | 6 | 8 | 10 | 12 |
| 56 | 6 | 12 | 20 | 30 | 0 |
| 60 | 8 | 20 | 40 | 0 | 0 |
| 64 | 12 | 32 | 0 | 0 | 0 |
| 68 | 30 | 0 | 0 | 0 | 0 |
| 72 | 18 | 38 | 0 | 0 | 0 |
| 76 | 8 | 20 | 37 | 0 | 0 |
| 80 | 7 | 12 | 17 | 22 | 0 |
| 84 | 5 | 5 | 5 | 5 | 5 |

Let's find the conditional moments of the first, second, third, and fourth orders:

$$
\begin{gathered}
M_{1}^{*}=\frac{d_{1}}{n}=\frac{3}{100}=0.03, M_{2}^{*}=\frac{s_{1}+2 s_{2}}{n}=\frac{147+2 \cdot 129}{100}=4.05 \\
M_{3}^{*}=\frac{d_{1}+6 d_{2}+6 d_{3}}{n}=\frac{3+6(-11)+6 \cdot(-15)}{100}=-1.53 ; \\
M_{4}^{*}=\frac{s_{1}+14 s_{2}+36 s_{3}+24 s_{1}}{n}=\frac{147+14 \cdot 129+36 \cdot 69+24 \cdot 19}{100}=48.93
\end{gathered}
$$

Let's find the central empirical moments of the third and fourth orders:

$$
\begin{gathered}
m_{2}=\left[M_{3}^{*}-3 M_{1}^{*} M_{2}^{*}+2\left(M_{1}^{*}\right)^{3}\right] \cdot h^{3}= \\
=\left[-1.53-3 \cdot 0.03 \cdot 4.05+2 \cdot(0.03)^{4}\right] 4^{3}=-121.245, \\
m_{4}=\left[M_{4}^{*}-4 M_{1}^{*} M_{3}^{*}+6\left(M_{1}^{*}\right)^{2} M_{2}^{*}-3\left(M_{1}^{*}\right)^{4}\right] h^{4}= \\
=[48.93-4 \cdot 0.03 \cdot(-1.53)+ \\
\left.+6 \cdot(0.03)^{2} \cdot 4.05-3(0.03)^{4}\right] \cdot 4^{4}=12578.679 .
\end{gathered}
$$

Let's find the required skewness and kurtosis, considering that $\sigma_{B}=$ $=\sqrt{D_{B}}=\sqrt{64.78}$ (variance $D_{B}$ was found earlier, see problem 529):

$$
\begin{gathered}
a_{s}=m_{3} / \sigma_{B}^{3}=-121.245 /(\sqrt{64.78})^{3}=-0.25 \\
e_{k}=m_{4} / \sigma_{B}^{4}-3=12578.679 /(\sqrt{64.78})^{4}-3=26.97
\end{gathered}
$$