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

| variant | $x_{i}$ | 48 | 52 | 56 | 60 | 64 | 68 | 72 | 76 | 80 | 84 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 2 | 4 | 6 | 8 | 12 | 30 | 18 | 8 | 7 | 5 |
Solution. We will form Calculation Table 2 as follows:

1) write the variants in the first column;
2) write the frequencies in the second column; place the sum of the frequencies (100) in the bottom cell of the column;
3) as the false zero $C$, choose the variant (68), which has the highest frequency (as $C$, you can take any variant located approximately in the middle of the column); in the cells of the row containing the false zero, write zeros; in the fourth column above and below the already placed zero, write another zero each;
4) in the remaining unfilled cells above the zero in the third column (excluding the topmost one), write the accumulated frequencies sequentially: $2 ; 2+4=6 ; 6+6=12 ; 12+8=20 ; 20+12=32$; by summing all the accumulated frequencies, we get the number $b_{1}=72$, which we place in the top cell of the third column. In the remaining unfilled cells below the zero in the third column (excluding the bottommost one), write the accumulated frequencies sequentially: $5 ; 5+7=12$; $12+8=20 ; 20+18=38$; by summing all the accumulated frequencies, we get the number $a_{1}=75$, which we place in the bottom cell of the third column;
5) the fourth column is filled similarly, except that the frequencies of the third column are summed; by summing all the accumulated frequencies above the zero, we get the number $b_{2}=70$, which we place in the top cell of the fourth column; the sum of the accumulated frequencies below the zero is equal to the number $a_{2}$, which we place in the bottom cell of the fourth column.

In the end, we get Calculation Table 2.

Find $d_{1}, s_{1}, s_{2}$:

$$
\begin{gathered}
d_{1}=a_{1}-b_{1}=75-72=3 \\
s_{1}=a_{1}+b_{1}=75+72=147 \\
s_{2}=a_{2}+b_{2}=59+70=129
\end{gathered}
$$

Table 2

| 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: |
| $x_{i}$ | $n_{i}$ | $b_{1}=72$ | $b_{2}=70$ |
| 48 | 2 | 2 | 2 |
| 52 | 4 | 6 | 8 |
| 56 | 6 | 12 | 20 |
| 60 | 8 | 20 | 40 |
| 64 | 12 | 32 | 0 |
| 68 | 30 | 0 | 0 |
| 72 | 18 | 38 | 0 |
| 76 | 8 | 20 | 37 |
| 80 | 7 | 12 | 17 |
| 84 | 5 | 5 | 5 |
|  | $n=100$ | $a_{1}=75$ | $a_{2}=59$ |

Find the conditional moments of the first and second orders:

$$
\begin{gathered}
M_{1}^{*}=d_{1} / n=3 / 100=0.03 \\
M_{2}^{*}=\left(s_{1}+2 s_{2}\right) / n=(147+2 \cdot 129) / 100=4.05
\end{gathered}
$$

Calculate the required sample mean and sample variance, considering that the step (difference between two adjacent variants) $h=4$ and the false zero $C=68$:

$$
\begin{gathered}
\bar{x}_{\mathrm{B}}=M_{1}^{*} h+C=0.03 \cdot 4+68=68.12 \\
D_{\mathrm{B}}=\left[M_{2}^{*}-\left(M_{1}^{*}\right)^{2}\right] \cdot h^{2}=\left[4.05-0.03^{2}\right] \cdot 4^{2} \simeq 64.78
\end{gathered}
$$