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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 variance of marks in statistics is `(9/16)^"th"` of the variance of marks in accountancy. Find the correlation coefficient between marks in two subjects.
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Solution
Given, n = 50, X = marks in Statistics,
Y = marks in Accountancy,
Regression equation of X on Y is
3y – 5x + 180 = 0,
`bar y = 44, sigma_X^2 = 9/16 sigma_Y^2`
Now, 3y – 5x + 180 = 0 is the regression equation of X on Y.
∴ The equation becomes 5X = 3Y + 180
i.e., X = `3/5` Y + `180/5`
Comparing it with X = b_{XY} Y + a', we get
`b_(XY) = 3/5, a = 180/5` = 36
a = `barx  b_(XY) bary`
∴ 36 = `bar x  3/5 xx 44`
∴ 36 = `bar x` – 26.4
∴ `bar x` = 36 + 26.4 = 62.4
Also, `sigma_X^2 = 9/16 sigma_Y^2`
∴ `sigma_X^2/sigma_Y^2 = 9/16`
∴ `sigma_X/sigma_Y = 3/4`
`b_(XY) = r xx sigma_X/sigma_Y`
∴ `3/5 = r xx 3/4`
∴ `3/5 xx 4/3` = r
∴ r = `4/5` = 0.8
∴ Mean marks in statistics `(barx)` are 62.4 and correlation coefficient (r) between marks in the two subjects is 0.8.
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