# Following data are given for marks in subject A and B in a certain examination : SUBJECT A SUBJECT B MEAN MARKS 36 85 STANDARD DEVIATION 11 8 Coefficient of correlation between A and B = ±0.66 i) Determine the two equations of regression ii) Calculate the expected marks in A corresponding to 75 marks obtained in B.

1
by sachinBorra118

2015-04-06T22:47:34+05:30

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Theory:
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Covariance  Cov(X, Y) = σ_X * σ_Y * Corr(X, Y)  = E [(X - X_bar) (Y - Y_bar) ]

Slope of the Linear regression line:  beta β = Covariance (X, Y) / variance(X)
β = σ_X * σ_Y *  Corr(X, Y) / σ_X²  =  Corr(X, Y) * σ_Y / σ_X
α = alpha = Y_bar  -  β * X_bar  = Y intercept of the line.

equation of linear regression:     Y = β * X +  α

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The given problem:
Here X is the variable denoting the marks in subject A and  Y is the variable denoting marks in subject B.
Given data:    X_bar = 36 ,     Y_bar = 85,     σ_X = 11     ,   σ_Y = 8
and    Corr(X, Y) = +0.66    or  -0.66

So     β = 0.66 * 8 / 11  = 0.48
α = alpha = 85 -  0.48 * 36 = 67.72
=>   Equation:   Y = 0.48 X + 67.72  ,  OR,    B = 0.48 A + 67.72    ---- (1)
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I am not sure of the following.  I am taking X as variable for the marks in subject B  and Y as the variable for the marks in  subject A.  But correlation coefficient remains the same as:    Corr(X,Y) = Corr(Y, X).
β = 0.66 * 11 / 8 = 0.9075
α = 36 - 0.9075 * 85 = - 41.1375
=>  equation is:     Y = 0.9075 X - 41.1375.
writing in terms of A and B,       A = 0.9075 B - 41.1375.        --- (2)
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marks obtained in subject B = 75.
As per (1),     75 = 0.48 A + 67.72
A = 7.28 /0.48 = 15.17 marks
as per (2) ,     A = 0.9075 * 75 - 41.1375 = 26.925 marks
I am not really too sure.  Please verify.