Example 6.38. A statistical distribution of a sample is given:

| $x_{i}$ | 1.6 | 3.0 | 4.4 | 5.8 | 7.2 | 6.6 | 10.0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $n_{i}$ | 3 | 7 | 15 | 35 | 22 | 13 | 5 |
## Solution.

1. We will find the sample mean, variance, and standard deviation using the method of products. To do this, we will create Table 1.

Table 1

| $x_{i}$ | $n_{i}$ | $u_{i}$ | $n_{i} u_{i}$ | $n_{i} u_{i}^{2}$ | $n_{i}\left(u_{i}+1\right)^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1.6 | 3 | -3 | -9 | 27 | 12 |
| 3.0 | 7 | -2 | -14 | 28 | 7 |
| 4.4 | 15 | -1 | -15 | 16 | 0 |
| 5.8 | 35 | 0 | 0 | 0 | 35 |
| 7.2 | 22 | 1 | 22 | 22 | 88 |
| 8.6 | 13 | 2 | 26 | 52 | 117 |
| 10.0 | 5 | 3 | 15 | 45 | 80 |
|  | $n=\sum n_{i}=100$ | $\sum n_{i} \cdot u_{i}=25$ | $\sum n_{i} \cdot u_{i}^{2}=189$ | $\sum n_{i}\left(u_{i}+1\right)^{2}=339$ |  |

We take $C=5.8$ - the variant with the highest frequency 35 as the false zero. The step of the sample $h=x_{2}-x_{1}=3.0-1.6=1.4$. Then the conditional variants are determined by the formula

$$
u_{i}=\frac{x_{i}-C}{h}=\frac{x_{i}-5.8}{1.4} .
$$

We calculate the conditional variants $u_{i}$ and fill in all the columns.

The last column serves as a control for the calculations by the identity:

$$
\sum n_{i}\left(u_{i}+1\right)^{2}=\sum n_{i} u_{i}^{2}+2 \sum n_{i} u_{i}+n
$$

Control: $339=189+2 \cdot 25+100$.

The calculations are correct. Let's find the conditional moments.

$$
M_{1}^{*}=\frac{\sum n_{i} u_{i}}{n}=\frac{25}{100}=0.25 ; \quad M_{2}^{*}=\frac{\sum n_{i} u_{i}^{2}}{n}=\frac{189}{100}=1.89
$$

We calculate the sample mean:

$$
\bar{x}_{\mathrm{B}}=M_{1}^{*} \cdot h+C=0.25 \cdot 1.4+5.8=6.15
$$

We find the sample variance:

$$
d_{\mathrm{B}}=\left[M_{2}^{*}-\left(M_{1}^{*}\right)^{2}\right] \cdot h^{2}=\left[1.89-(0.25)^{2}\right] \cdot 1.4^{2}=3.58
$$

We determine the sample standard deviation:

$$
\sigma_{\mathrm{B}}=\sqrt{d_{\mathrm{B}}}=\sqrt{3.58}=1.89
$$

## 2. We construct the normal curve.

To simplify the calculations, we summarize all the calculations in Table 2.

Table 2

| $x_{i}$ | $n_{i}$ | $x_{i}-\bar{x}_{\mathrm{B}}=$  $=x_{i}-6.15$ | $u_{i}=\frac{x_{i}-x_{\mathrm{B}}}{\sigma_{\mathrm{B}}}=\frac{x_{i}-6.15}{1.89}$ | $\phi\left(u_{i}\right)$ | $n_{i}^{\prime}=74.07 \cdot \phi\left(u_{i}\right)$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1.6 | 3 | $-4.55$ | $-2.41$ | 0.0219 | 2 |
| 3.0 | 7 | $-3.15$ | $-1.67$ | 0.0989 | 7 |
| 4.4 | 15 | $-1.75$ | $-0.92$ | 0.2613 | 19 |
| 5.8 | 35 | $-0.35$ | $-0.18$ | 0.3925 | 30 |
| 7.2 | 22 | 1.05 | 0.56 | 0.3410 | 25 |
| 8