If the regression lines of a bivariate distribution are and , then Calculate the arithmetic mean of…

The regression lines intersect at the means $(\bar{x}, \bar{y})$: $4\bar{x} - 5\bar{y} + 33 = 0$ $20\bar{x} - 9\bar{y} - 107 = 0$ Multiplying the first equation by 5: $20\bar{x} - 25\bar{y} + 165 = 0$ Subtracting from the second equation: $16\bar{y} - 272 = 0 \implies 16\bar{y} = 272 \implies \bar{y} = 17$. Substituting into the first equation: $4\bar{x} - 5(17) + 33 = 0 \implies 4\bar{x} = 52 \implies \bar{x} = 13$. Arithmetic means: $\bar{x} = 13$ and $\bar{y} = 17$.

To estimate $x$ given $y$, we identify the regression line of $x$ on $y$. Taking $20x - 9y - 107 = 0$ as $x$ on $y$ gives $b_{xy} = \frac{9}{20} = 0\cdot 45$, and $4x - 5y + 33 = 0$ as $y$ on $x$ gives $b_{yx} = \frac{4}{5} = 0\cdot 8$. Since $b_{xy} \cdot b_{yx} = (0\cdot 45)(0\cdot 8) = 0\cdot 36 \le 1$, this assignment is valid. Using the regression line of $x$ on $y$: $20x = 9y + 107$ For $y = 7$: $20x = 9(7) + 107 = 63 + 107 = 170 \implies x = \frac{170}{20} = 8\cdot 5$.

From parts (a) and (b), $b_{yx} = 0\cdot 8$ and $b_{xy} = 0\cdot 45$. Correlation coefficient $r = \sqrt{(0\cdot 8)(0\cdot 45)} = \sqrt{0\cdot 36} = 0\cdot 6$. Using $b_{yx} = r \frac{\sigma_y}{\sigma_x}$ with $\sigma_x = 3$: $0\cdot 8 = 0\cdot 6 \left(\frac{\sigma_y}{3}\right) \implies 0\cdot 8 = 0\cdot 2 \sigma_y \implies \sigma_y = 4$. Variance of $y = \sigma_y^2 = 4^2 = 16$.