13.1.2. Calculate the sample mean $\bar{X}$, sample variance $D_{x}$, and sample standard deviation $\sigma_{x}$.
Solution. To calculate $\bar{X}, D_{x}$, and $\sigma_{x}$, we will use the method of products. We introduce conditional variants

$$
u_{1}=\frac{\left(x_{i}-c_{x}\right)}{h_{x}}
$$

where $c_{x}$ is the value of $x_{i}$ corresponding to the highest frequency, $c_{x}=3.8$, and $h_{x}$ is the step of the sample, $h_{x}=1.2$.

Then, by calculating $u_{i}$, we obtain the conditional series:

| $u_{i}$ | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m_{i}$ | 5 | 13 | 25 | 25 | 19 | 10 | 3 |

For this series, we will construct a calculation table:

| $i$ | $u_{i}$ | $m_{i}$ | $m_{i} u_{i}$ | $m_{i} u_{i}^{2}$ | $m_{i}\left(u_{i}+1\right)^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | -3 | 5 | -15 | 45 | 20 |
| 2 | -2 | 13 | -26 | 52 | 13 |
| 3 | -1 | 25 | -25 | 25 | 0 |
| 4 | 0 | 25 | 0 | 0 | 25 |
| 5 | 1 | 19 | 19 | 19 | 76 |
| 6 | 2 | 10 | 20 | 40 | 90 |
| 7 | 3 | 3 | 9 | 27 | 48 |
| $\Sigma$ |  | 100 | -18 | 208 | 272 |

Verification:

$$
\sum m_{i}\left(u_{i}+1\right)^{2}=\sum m_{i} u_{i}^{2}+2 \sum m_{i} u_{i}+\sum m_{i}, \quad 272=272
$$

Now, let's find the conditional characteristics:

$$
\bar{U}=\frac{\sum m_{i} u_{i}}{N}=-0.18 ; D_{u}=\frac{\sum m_{i} u_{i}^{2}}{N}-(\bar{U})^{2}=2.048 ; \sigma_{u}=D_{u}^{1 / 2}=1.43
$$

Returning to the original variational series, using the equality $x_{i}=h_{x} u_{i}+c_{x}$, we get:

$$
\bar{X}=h_{x} \bar{U}+c_{x}=3.58 ; \quad D_{x}=h_{x}^{2} D_{u}=2.949 ; \quad \sigma_{x}=h_{x} \sigma_{u}=1.72
$$