The two regression lines between height (X) in inches and weight (Y) in kgs of girls are,

4y − 15x + 500 = 0

and 20x − 3y − 900 = 0

Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.

#### Solution

Given, X = Height (in inches), Y = weight (in Kg)

The equation of regression are

4y - 15x + 500 = 0

i.e., –15x + 4y = – 500 …(i)

and 20x – 3y – 900 = 0

i.e., 20x – 3y = 900 …(ii)

By 3 × (i) + 4 × (ii), we get

- 45x + 12y = - 1500

+ 80x - 12y = 3600

35x = 2100

∴ x = 60

Substituting x = 60 in (i), we get

–15(60) + 4y = –500

∴ 4y = 900 – 500

∴ y = 100

Since the point of intersection of two regression lines is `bar x, bar y`,

`bar x` = mean height of the group = 60 inches, and

`bar y` = mean weight of the group = 100 kg.

Let 4y – 15x + 500 = 0 be the regression equation of Y on X.

∴ The equation becomes 4y = 15x – 500

i.e., Y = `15/4"X" - 500/4` ...(i)

Comparing it with Y = b_{YX} X + a, we get

∴ `"b"_"YX" = 15/4`

∴ Now, other equation 20x – 3y – 900 = 0 be the regression equation of X on Y

∴The equation becomes 20x – 3y – 900 = 0

i.e., 20x = 3y + 900

X = `3/20"Y" + 900/20`

Comparing it with X = b_{XY} Y + a',

∴ `"b"_"XY" = 3/20`

Now, `"b"_"YX" * "b"_"XY" = 15/4 * 3/20 = 0.5625`

i.e., b_{XY} . b_{YX} < 1

∴ Assumption of regression equations is true.

Now, substituting x = 70 in (i) we get

y = `15/4 xx 70 - 500/4 = (1050 - 500)/4 = 550/4 = 137.5`

∴ Weight of girl having height 70 inches is 137.5 kg