13.2.1. Find $\bar{Y}$ and $\sigma_{y}$ for the sample

| $y_{j}$ | 0.5 | 1.3 | 2.1 | 2.9 | 3.7 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $m_{y_{j}}$ | 5 | 20 | 45 | 27 | 3 |
Solution. To calculate $\bar{Y}, D_{y}$, and $\sigma_{y}$, we will use the method of products. Introduce conditional variants $v_{j}=\left(y_{j}-c_{y}\right) / h_{y}$, where $c_{y}$ is the value of $y_{j}$ corresponding to the highest frequency, $c_{y}=2.1$, and $h_{y}$ is the step of the sample, $h_{y}=0.8$.

Then, calculating $v_{j}$, we obtain the conditional series:

| $v_{j}$ | -2 | -1 | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $m_{j}$ | 5 | 20 | 45 | 27 | 3 |

For this series, we will construct a calculation table:

| $j$ | $v_{j}$ | $m_{j}$ | $m_{j} v_{j}$ | $m_{j} v_{j}^{2}$ | $m_{j}\left(v_{j}+1\right)^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | -2 | 5 | -10 | 20 | 5 |
| 2 | -1 | 20 | -20 | 20 | 0 |
| 3 | 0 | 45 | 0 | 0 | 45 |
| 4 | 1 | 27 | 27 | 27 | 108 |
| 5 | 2 | 3 | 6 | 12 | 27 |
| $\sum$ |  | 100 | 3 | 79 | 185 |

Verification:

$$
m_{j}\left(v_{j}+1\right)^{2}=\sum m_{j} v_{j}^{2}+2 \sum m_{j} v_{j}+\sum m_{j}, 185=185
$$

Conditional characteristics:

$$
\begin{gathered}
\bar{V}=\frac{\sum m_{j} v_{j}}{N}=0.03 ; \quad D_{v}=\frac{\sum m_{j} u_{j}^{2}}{N}-(\bar{V})^{2}=0.789 ; \\
\sigma_{v}=D_{v}^{1 / 2}=0.89
\end{gathered}
$$

Returning to the original variational series, using the equality $y_{j}=h_{y} v_{j}+c_{y}$, we obtain

$$
\begin{gathered}
\bar{Y}=h_{y} \bar{V}+c_{y}=2.12 ; \quad D_{y}=h_{y}^{2} D_{v}=0.505 ; \\
\sigma_{y}=h_{y} \sigma_{v}=0.71
\end{gathered}
$$