Regression line of X on Y is 3Y-5X+180=0.

Find:

(i) Mean of Y

(ii) Correlation r between X and Y.

(iii) Regression line of Y on X.

Solution:

Given:

Mean of $X=\bar{X}=62.4$ --------(1)

Regression line of X on Y is

3Y-5X+180=0 --------(2)

Substituting (1) in (2) we get,

3Y-5(62.4)+180=0

Y=44

$\therefore \ \bar{Y}=44$ --------(3)

Given, 16 Var(X)=9 Var(Y)

$16 \sigma_{x}^{2}=9 \sigma_{y}^{2}$

$4 \sigma_{x} = 3 \sigma_{y}$

$\cfrac{\sigma_{x}}{\sigma_{y}}=\cfrac{3}{4}$ --------(4)

From (2), 3Y+180=5X

$X=\cfrac{3}{5}Y-\cfrac{180}{5}$

$G_{xy}=\cfrac{3}{5}$ --------(5)

But $G_{xy}=r \cfrac{\sigma_{x}}{\sigma_{y}}$

$\cfrac{3}{5}=r\cfrac{3}{4}$ from (4) and (5)

$\therefore \ r=\cfrac{4}{5}=0.8$ --------(6)

Also, $G_{xy}=\cfrac{16}{15}$ --------(7)

Regression line of Y on X-axis,

$Y-\bar{Y}=G_{xy} (X- \bar{X})$

$Y-44=\cfrac{16}{15}X-66.56+44$

$Y=1.0667X-22.56$