525. Find the sample mean and sample variance using the method of products for the given sample distribution of size \( n=100 \):

| \( x_{i} \) | 2 | 3 | 7 | 9 | 11 | 12.5 | 16 | 18 | 23 | 25 | 26 |
| ---: | ---: | ---: | ---: | ---: | ---: | :---: | ---: | ---: | ---: | ---: | ---: |
| \( n_{i} \) | 3 | 5 | 10 | 6 | 10 | 4 | 12 | 13 | 8 | 20 | 9 |
Solution. Let's divide the interval $2-26$ into four sub-intervals of length $h=6: 2-8 ; 8-14 ; 14-20 ; 20-26$. Taking the midpoints of the sub-intervals as new variants $y_{i}$, we obtain equidistant variants: $y_{1}=5, y_{2}=11, y_{3}=17, y_{4}=23$.

As the frequency $n_{1}$ of the variant $y_{1}=5$, we take the sum of the frequencies of the variants that fall into the first interval: $n_{1}=3+5+10=18$.

Calculating the frequencies of the other variants in the same way, we obtain the distribution of equidistant variants:

| $y_{i}$ | 5 | 11 | 17 | 23 |
| ---: | ---: | ---: | ---: | ---: |
| $n_{i}$ | 18 | 20 | 25 | 37 |

Using the method of products, we find $\bar{y}_{\mathrm{B}}=15.86, \quad D_{\mathrm{B}}=45.14$.

Considering that the number of sub-intervals (4) is small, we will apply Sheppard's correction:

$$
D_{\mathrm{B}}^{\prime}=D_{\mathrm{B}}-(1 / 12) h^{2}=45.14-6^{2} / 12=42.14
$$