1811. From the general population, where the distribution of characteristics $X$ and $Y$ is normal, a sample of 530 units has been taken. The results of measuring characteristics $X$ and $Y$ of the sample members are presented in the following correlation table:

| $x$ | $15-$ | $25-3$ | $35-4$ | $5-5$ | $35-6$ | 5-7 | $5-85$ | $m_{x}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $200-300$ | 19 | 5 |  |  |  |  |  | 24 |
| 300-400 | 23 | 116 | 11 |  |  |  |  | 150 |
| 400-500 | 1 | 41 | 98 | 9 |  |  |  | 149 |
| $500-600$ |  | 4 | 32 | 65 | 7 |  |  | 108 |
| 600—700 |  | 1 | 4 | 21 | 36 | 3 |  | 65 |
| $700-800$ |  |  | 1 | 2 | 11 | 13 | 1 | 28 |
| $800-900$ |  |  |  |  | 1 | 3 | 2 | 6 |
| $m_{y}$ | 43 | 167 | 146 | 97 | 55 | 19 | 3 | 530 |

Find the sample coefficient of linear correlation and its standard deviation. Write the equations of the regression lines $y$ on $x$ and $x$ on $y$.
Solution. First, we replace the intervals of the values of features $X$ and $Y$ with their midpoints. As a result, we obtain a correlation table that has the form of the table placed at the beginning of this paragraph. Since the linear correlation coefficient does not change from changes in the origins of measurement and scales of the features (property 2), we will simplify the calculations by replacing the values of features $X$ and $Y$ with the values of features $U$ and $V$, which are related to the first by the following relationships:

$$
U=\alpha\left(X-x_{0}\right), V=\beta\left(Y-y_{0}\right)
$$

where we will set $x_{0}=550 ; y_{0}=50 ; \alpha=0.01 ; \beta=0.1$. The values of $\alpha$ and $\beta$ obtained will be written above the corresponding intervals. All further calculations will be performed on the values of features $U$ and $V$ (table on p. 403):

$$
\begin{gathered}
\overline{U V}=\frac{\sum_{i} \sum_{j} m_{i j} u_{i} v_{j}}{N}=\frac{1161}{530}=2.191 ; \\
\bar{V}=\frac{\sum_{i} m_{y_{j}} v_{j}}{N}=\frac{-507}{530}=-0.957 ; \bar{U}=\frac{\sum_{i} m_{x_{i}} u_{i}}{N}=\frac{-282}{530}=-0.721 ; \\
\overline{V^{2}}=\frac{\sum_{j} m_{y_{j}} v_{j}^{2}}{N}=\frac{1359}{530}=2.564 ; \quad \overline{U^{2}}=\frac{\sum_{i} m_{x_{i}} u_{i}^{2}}{N}=\frac{1196}{530}=2.257 \\
\sigma_{V}^{2}=\bar{V}^{2}-(\bar{V})^{2}=2.564-0.916=1.648 ; \sigma_{U}^{2}=\overline{U^{2}}-(\bar{U})^{2}= \\
=2.257-0.520=1.737 ; \sigma_{V}=1.28 ; \sigma_{U}=1.32 ;
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-403.jpg?height=1745&width=1014&top_left_y=52&top_left_x=117)

$$
\begin{aligned}
& r=\frac{\overline{U V}-\bar{U} \bar{V}}{{ }^{v^{G} V}}=\frac{2.191-0.690}{1.32 \cdot 1.28}=\frac{1.501}{1.690}=0.888 ; \bar{X}=x_{0}+ \\
& +\frac{\bar{U}}{\alpha}=550+\frac{-0.721}{0.01}=550-72.1=477.9 ; \bar{Y}=y_{0}+ \\
& +\frac{\bar{V}}{\beta}=50+\frac{-0.957}{0.1}=50-9.57=40.43 ; \rho_{y / x}=r{ }^{\sigma_{Y}}{ }^{\sigma_{X}}= \\
& =0.888 \cdot \frac{\frac{1.28}{0.1}}{\frac{1.32}{0.01}}=0.0888 \cdot \frac{128}{132}=0.0849 ; \rho_{x / y}=r \frac{\sigma_{X}}{{ }^{\sigma} Y}=9.16 ; \\
& y-40.43=0.085(x-477.9) ; x-477.9=9.16(y-40.43) ; \\
& \sigma_{r} \approx \frac{1-r^{2}}{\sqrt{N}}=0.009
\end{aligned}
$$

| ${ }_{X} \quad Y$ | 1 | 2 | 3 | 4 | 5 | $m_{x}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 2 |  |  |  |  | 2 |
| 2 | 1 | 1 | 2 |  |  | 4 |
| 3 |  | 2 | 2 | 1 |  | 5 |
| 4 |  |  |  | 2 | 2 | 4 |
| $m_{y}$ | 3 | 3 | 4 | 3 | 2 | 15 |