Question

For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y – 5x + 180 = 0. The mean marks in accountancy is 44 and the variance of marks in statistics is `(9/16)^(th)` of the variance of marks in accountancy. Find the mean marks in statistics and the correlation coefficient between marks in the two subjects.

Answer 1

Given n = 50

Regression line of x on y is

3y – 5x + 180 = 0

⇒ 5x = `[3y + 180]/5`

∴ x = `3/5 y + 180/5`

∴ `b_(xy) = "coefficient of y" = 3/5`

Variance of marks in statistics = `9/16`

Variance of marks in accountancy.

i.e. V(x) = `9/16`V(y)

⇒ `V_(x)/V_(y) = 9/16`

Taking square roots,

`(σ_x)/(σ_y) = 3/4`

We have, `b_(xy) = r. (σ_x)/(σ_y)`

⇒ `3/5 = r xx 3/4`

⇒ `r = 4/5 = 0.8`

Given `bary` = 44

∴ Substituting y = 44 is 3y – 5x + 180 = 0

⇒ 3(44) – 5x + 18 = 0

⇒ 132 – 5x + 180 = 0

⇒ 5x = 132 + 180 = 312

⇒  x = `312/5 ` = 62.4

∴  `barx` = 62.4

∴  Mean marks in statistics `barx` = 62.4