Question

The two lines of regressions are $x+2y-5=0$ and $2x+3y-8=0$ and the variance of $x$ is $12$. Find the variance of $y$ and the coefficient of correlation.

Answer 1

The given equation of the lines of regression are
$x+2y-5=\mathrm{0.......}\left(i\right)$
and $2x+3y-8=\mathrm{0.....}\left(ii\right)$

Rewriting the equations $\left(i\right)$ and $\left(ii\right)$, we have
From equation $\left(i\right)$
$y=\frac{-x}{2}+\frac{5}{2}$
$y=-0.5x+2.5$ *regression line of $y$ on $x$)
$b_{yx}=r\frac{{\sigma }_y}{{\sigma }_x}=-\mathrm{0.5....}\left(iii\right)$

From eqution $\left(ii\right)$,
$x=\frac{-3}{2}y+\frac{8}{2}$
$x=-1.5y+4$( regression line of $x$ on $y$)
$b_{xy}=r\frac{{\sigma }_x}{{\sigma }_y}$
$\therefore r^2=b_{yx}\times b_{xy}=(-0.5)\times (-1.5)=0.75$
$\therefore$ $r=\sqrt{0.75}=\pm 0.866$

But $b_{xy}$ and $b_{yx}$ being both $-$ve therefore, $r$ is also $-$ve.

Correlation coefficient $\left(r\right)=-0.866$

Varianceof $x$ i.e., ${\sigma }_x^2=12$
$\therefore$ ${\sigma }_x=\sqrt{12}$

From equation $\left(iii\right)$
$r\frac{{\sigma }_y}{{\sigma }_x}=-0.5$
$-\mathrm{0.866.}\frac{{\sigma }_x}{\sqrt{12}}=-0.5$
${\sigma }_y=\frac{0.5\times \sqrt{12}}{0.866}=2$
$\therefore$ Variance of $y$ i.e., ${\sigma }_y^2=4$