Question

If the lines of regression are $3x+12y=19$ and $3y+9x=46$, then $r_{xy}$ will be

• A. $0.289$
• B. $-0.289$
• C. $0.209$
• D. None of these

Answer 1

The correct option is B

$-0.289$

Explanation for the correct option:

Step 1: Solve for the regression line of $y$ on $x$

The given lines of regression are

$3x+12y=19$ $...\left(i\right)$

$3y+9x=46$ $...\left(ii\right)$

From $\left(i\right)$, regression line of $y$ on $x$ is

$12y=-3x+19$

$\Rightarrow$ $y=\frac{-3}{12}x+\frac{19}{12}$

$\Rightarrow$ $y=\frac{-1}{4}x+\frac{19}{12}$

$\Rightarrow$ $b_{yx}=\frac{-1}{4}$

Step 2: Solve for the regression line of $x$ on $y$

From $\left(ii\right)$,regression line of $x$ on $y$ is

$9x=-3y+46$

$\Rightarrow$ $x=\frac{-3}{9}y+\frac{46}{9}$

$\Rightarrow$ $x=\frac{-1}{3}y+\frac{46}{9}$

$\Rightarrow$$b_{xy}=\frac{-1}{3}$

Step 3: Solve for the required value

$r_{xy}=\sqrt{b_{yx}\times b_{xy}}$

$=-\sqrt{\frac{-1}{4}\times \frac{-1}{3}}$ [Since both $b_{yx}$ and $b_{xy}$ are negative, $r_{xy}$ is also negative]

$=-\sqrt{\frac{1}{12}}$

$\Rightarrow$$r_{xy}=-0.289$

Thus the value of $r_{xy}$ is $-0.289$

Hence option(B) is the correct answer.