In a partially destroyed record, the following data are available: variance of X = 25, Regression equation of Y on X is 5y − x = 22 and regression equation of X on Y is 64x − 45y = 22 Find

- Mean values of X and Y
- Standard deviation of Y
- Coefficient of correlation between X and Y.

#### Solution

Given, `sigma_"X"^2 = 25`

∴ `sigma_"X"` = 5

Regression equation of Y on X is

5y – x = 22

Regression equation of X on Y is

64x - 45y = 22

**(i) **Consider, the two regression equation

- x + 5y = 22 ....(i)

64x - 45y = 22 ....(ii)

By (i) × + (ii), we get

- 9x + 45y = 198

+ 64x - 45y = 22

55x = 220

∴ x = 4

Substituting x = 4 in (i), we get

- 4 + 5y = 22

∴ 5y = 22 + 4

∴ y = `26/5 = 5.2`

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

`bar x` = mean value of X = 4 and

`bar y` = mean value of Y = 5.2

**(ii)** To find standard deviation of Y we should first find the coefficient of correlation between X and Y.

Regression equation of Y on X is

5y - x = 22

i.e., 5Y = X + 22

i.e., Y = `"X"/5 + 22/5`

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

`"b"_"YX" = 1/5`

Now, regression equation of X on Y is

64x - 45y = 22

i.e., 64X - 45Y = 22

i.e., 64X = 45Y + 22

i.e., X = `"45Y"/64 + 22/64`

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

`"b"_"XY" = 45/64`

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

`= +- sqrt((1/5)(45/64)) = +- sqrt(9/64) = +- 3/8`

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

r is also positive.

∴ r = `3/8`

Now, `"b"_"YX" = "r" sigma_"Y"/sigma_"X"`

∴ `1/5 = 3/8 xx sigma_"Y"/5`

∴ `sigma_"Y" = 1/5 xx 8/3 xx 5`

∴ `sigma_"Y"`= Standard deviation of Y = `8/3`

**(iii) **The correlation coefficient of X and Y is

r = `8/3 = 0.375`