1. **Stating the problem:** We are given two regression lines: $$y = 20 - 9x$$ and $$y = 4 + 5x$$ with the variance of $x$ as 9. We need to find: a) The mean values of $x$ and $y$. b) The standard deviation of $y$. 2. **Understanding regression lines:** The regression lines are usually given as: $$y - \bar{y} = b_{yx}(x - \bar{x})$$ and $$x - \bar{x} = b_{xy}(y - \bar{y})$$ where $\bar{x}$ and $\bar{y}$ are the means of $x$ and $y$, and $b_{yx}$ and $b_{xy}$ are regression coefficients. 3. **Rewrite the given lines in regression form:** Given lines are: $$y = 20 - 9x$$ $$y = 4 + 5x$$ We interpret these as two regression lines of $y$ on $x$ and $x$ on $y$ respectively. 4. **Identify regression coefficients and means:** From the first line: $$y = 20 - 9x \implies y = -9x + 20$$ So, slope $b_{yx} = -9$ and intercept $a_{yx} = 20$. From the second line: $$y = 4 + 5x \implies x = \frac{y - 4}{5}$$ Rearranged as: $$x = 0.2y - 0.8$$ So, slope $b_{xy} = 0.2$ and intercept $a_{xy} = -0.8$. 5. **Find means $\bar{x}$ and $\bar{y}$:** Using the regression line formulas: $$y - \bar{y} = b_{yx}(x - \bar{x})$$ At $x=\bar{x}$, $y=\bar{y}$, so: $$\bar{y} = b_{yx}\bar{x} + a_{yx}$$ Similarly, for the second line: $$x - \bar{x} = b_{xy}(y - \bar{y})$$ At $y=\bar{y}$, $x=\bar{x}$, so: $$\bar{x} = b_{xy}\bar{y} + a_{xy}$$ 6. **Set up equations:** From first line: $$\bar{y} = -9\bar{x} + 20$$ From second line: $$\bar{x} = 0.2\bar{y} - 0.8$$ 7. **Solve the system:** Substitute $\bar{x}$ from second into first: $$\bar{y} = -9(0.2\bar{y} - 0.8) + 20 = -1.8\bar{y} + 7.2 + 20 = -1.8\bar{y} + 27.2$$ Bring terms together: $$\bar{y} + 1.8\bar{y} = 27.2$$ $$2.8\bar{y} = 27.2$$ $$\bar{y} = \frac{27.2}{2.8} = 9.7143$$ Now find $\bar{x}$: $$\bar{x} = 0.2 \times 9.7143 - 0.8 = 1.9429 - 0.8 = 1.1429$$ 8. **Find standard deviation of $y$:** Recall: $$b_{yx} = r \frac{\sigma_y}{\sigma_x}$$ $$b_{xy} = r \frac{\sigma_x}{\sigma_y}$$ where $r$ is the correlation coefficient, $\sigma_x$ and $\sigma_y$ are standard deviations. Multiply $b_{yx}$ and $b_{xy}$: $$b_{yx} b_{xy} = r^2$$ Calculate: $$-9 \times 0.2 = -1.8$$ Since $r^2$ must be positive and less than or equal to 1, this is inconsistent. 9. **Re-examine the problem:** The given lines are: $$20 - 9x = y$$ and $$4 + 5x = y$$ This suggests the first line is: $$y = 20 - 9x$$ and the second line is: $$y = 4 + 5x$$ But two regression lines cannot both be $y$ on $x$. Assuming the first is regression of $x$ on $y$: $$x = 20 - 9y$$ and the second is regression of $y$ on $x$: $$y = 4 + 5x$$ 10. **Use this assumption:** From regression of $y$ on $x$: $$y = 4 + 5x$$ So $b_{yx} = 5$, $a_{yx} = 4$. From regression of $x$ on $y$: $$x = 20 - 9y = -9y + 20$$ So $b_{xy} = -9$, $a_{xy} = 20$. 11. **Find means:** At means: $$\bar{y} = 4 + 5\bar{x}$$ $$\bar{x} = -9\bar{y} + 20$$ Substitute $\bar{y}$ into second: $$\bar{x} = -9(4 + 5\bar{x}) + 20 = -36 - 45\bar{x} + 20 = -16 - 45\bar{x}$$ Bring terms together: $$\bar{x} + 45\bar{x} = -16$$ $$46\bar{x} = -16$$ $$\bar{x} = -\frac{16}{46} = -0.3478$$ Find $\bar{y}$: $$\bar{y} = 4 + 5(-0.3478) = 4 - 1.739 = 2.261$$ 12. **Find standard deviation of $y$:** Given variance of $x$ is 9, so $\sigma_x = 3$. Use: $$b_{yx} = r \frac{\sigma_y}{\sigma_x} = 5$$ $$b_{xy} = r \frac{\sigma_x}{\sigma_y} = -9$$ Multiply: $$b_{yx} b_{xy} = r^2 = 5 \times (-9) = -45$$ This is impossible since $r^2$ must be between 0 and 1. 13. **Conclusion:** The problem as stated has inconsistent regression lines for $x$ and $y$. Assuming the first line is regression of $x$ on $y$: $$x = 20 - 9y$$ and the second line is regression of $y$ on $x$: $$y = 4 + 5x$$ Given variance of $x$ is 9, we find: $$\bar{x} = -0.3478$$ $$\bar{y} = 2.261$$ Standard deviation of $y$ cannot be found due to inconsistency in regression coefficients. **Final answers:** $$\bar{x} = -0.3478$$ $$\bar{y} = 2.261$$ Standard deviation of $y$ cannot be determined from given data due to inconsistency.