The equations of two regression lines are x − 4y = 5 and 16y − x = 64. Find means of X and Y. Also, find correlation coefficient between X and Y.

#### Solution

Given equations of regression lines are

x - 4y = 5 …(i)

16y - x = 64

i.e., - x + 16y = 64 …(ii)

Adding (i) and (ii), we get

x - 4y = 5

- x + 16y = 64

12y = 69

∴ y = `69/12 = 5.75`

Substituting y = 5.75 in (i), we get

x - 4(5.75) = 5

∴ x - 23 = 5

∴ x = 5 + 23 = 28

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

∴ `bar x = 28 and bar y = 5.75`

Let, x - 4y = 5 be the regression equation of X on Y

∴ The equation becomes X = 4Y + 5

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

b_{XY} = 4

Now, the other equation i.e. 16y - x = 64 is regression equation of Y on X

∴ The equation becomes 16Y = X + 64

i.e., Y = `1/16 "X" + 64/16`

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

`"b"_"YX" = 1/16`

r = `+-sqrt("b"_"XY" * "b"_"YX")`

`= +- sqrt(4 xx 1/16) = +- sqrt(1/4) = +- 1/2 = +- 0.5`

Since b_{XY} and b_{YX} both are positive,

r is also positive.

∴ r = 0.5