Q.The two lines of regression are $x-2y+16=0$ and $5x-6y+32=0$. Find the means $\bar x,\bar y$ and the coefficient of correlation.

Solve the two lines for the means, then read off the coefficients and take their geometric mean.

$r=+\sqrt{0.5\times1.2}=\sqrt{0.6}$.

Means = intersection. Solve

From (1), $x=2y-16$. Substitute in (2): $5(2y-16)-6y+32=0\Rightarrow10y-80-6y+32=0\Rightarrow4y-48=0\Rightarrow y=12.$ Then $x=2(12)-16=8.$ So $\bar x=8,\ \bar y=12.$

Coefficients (identify the lines with the product test). Assume the first line is $Y$ on $X$: solve $x-2y+16=0$ for $y$: $2y=x+16\Rightarrow y=0.5x+8\Rightarrow b_{yx}=0.5.$ Then the second line is $X$ on $Y$: solve $5x-6y+32=0$ for $x$: $5x=6y-32\Rightarrow x=1.2y-6.4\Rightarrow b_{xy}=1.2.$ Product $=0.5\times1.2=0.6\le1$, so the assignment is valid.

Correlation. Both coefficients positive, so

You can confirm the line identification by checking the alternative assignment (first line as $X$ on $Y$): it gives $b_{xy}=2$ and $b_{yx}=5/6\approx0.83$, product $1.67>1$ — impossible, so the assignment used above is the correct one.

Reading a regression coefficient straight from the $x$- and $y$-coefficients of the line without first solving for the correct dependent variable. Always rearrange the line into $y=\ldots$ (for $b_{yx}$) or $x=\ldots$ (for $b_{xy}$).