Consider the following data of a bivariate distribution: - The mean of the variables and are 25 and…

Given $\bar{x} = 25$, $\bar{y} = 30$, $b_{xy} = 0\cdot 4$, $b_{yx} = 1\cdot 6$. Line of regression of $x$ on $y$: $x - \bar{x} = b_{xy}(y - \bar{y}) \implies x - 25 = 0\cdot 4(y - 30) \implies x = 0\cdot 4y + 13$ (or $5x - 2y - 65 = 0$). Line of regression of $y$ on $x$: $y - \bar{y} = b_{yx}(x - \bar{x}) \implies y - 30 = 1\cdot 6(x - 25) \implies y = 1\cdot 6x - 10$ (or $8x - 5y - 50 = 0$).

To estimate $y$ when $x = 60$, use the regression line of $y$ on $x$: $y = 1\cdot 6(60) - 10 = 96 - 10 = 86$.

Coefficient of correlation $r = \pm\sqrt{b_{xy} \cdot b_{yx}}$. Since both regression coefficients are positive, $r$ is positive: $r = \sqrt{(0\cdot 4)(1\cdot 6)} = \sqrt{0\cdot 64} = 0\cdot 8$.