Problem 989

Solution. The calculation of correlation relationships will be carried out according to the following scheme (table on p. 410):

$$\begin{aligned} & \bar{X}=\frac{\Sigma_{x} m_{x} x}{\Sigma_{x} m_{x}}=\frac{8}{42}=0.190 ; \bar{X}^{2}=\frac{\Sigma_{x} m_{x} x^{2}}{\sum_{x} m_{x}}= \\ & =\frac{60}{42}=1.424 ; \sigma_{X}^{2}=\overline{X^{2}}-(\bar{X})^{2}=1.388 ; \end{aligned}$$

$$\begin{gathered} \bar{V}=\frac{\Sigma m_{y} v}{\sum_{y} m_{y}}=\frac{0}{42}=0 ; \overline{V^{2}}=\frac{\Sigma m_{y} v^{2}}{\Sigma m_{y}}=\frac{302}{42}=7.190 ; \sigma_{V}^{2}=7.190 ; \\ \overline{\left(\overline{V_{x}}\right)^{2}}=\frac{\sum_{x} \frac{1}{m_{x}}\left(\sum_{y}^{\Sigma} m_{x y} v\right)^{2}}{\Sigma_{x} m_{x}}=\frac{0}{42}=0 ; \sigma^{2}\left(\overline{V_{x}}\right)=\overline{\left(\bar{V}_{x}\right)^{2}}- \\ -(\bar{V})^{2}=0-0=0 ; \\ \overline{\left.(\bar{X})_{y}\right)^{2}}=\frac{\left.\sum_{y} \frac{1}{\bar{m}_{y}}\left(\sum_{x} m_{x y}\right)^{x}\right)^{2}}{\Sigma_{y}^{\Sigma m_{y}}}=\frac{52.542}{42}=1.251 ; \sigma^{2}\left(\overline{X_{v}}\right)=\overline{\left(\overline{\bar{X}_{v}}\right)^{2}}- \\ -(\bar{X})^{2}=1.215 . \end{gathered}$$

!

From this, considering property 5, $\eta_{g / x}=\eta_{o / x}=\sqrt{\frac{\sigma^{2}\left(\bar{V}_{\mathrm{x}}\right)}{\sigma_{V}^{2}}}=\sqrt{\frac{0}{7.190}}=0 ; \eta_{x / g}=\eta_{x / D}=\sqrt{\frac{\sigma^{2}\left(\bar{X}_{v}\right)}{\sigma_{X}^{2}}}=\sqrt{\frac{1.215}{1.388}}=\sqrt{0.875}=$ $=0.935$. Obviously, $\rho_{y / x}=\rho_{x / y}=0$ and $r(X, Y)=0$.