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For a certain bivariate data of a group of 10 students, the following information gives the internal marks obtained in English (X) and Hindi (Y):
X  Y  
Mean  13  17 
Standard Deviation  3  2 
If r = 0.6, Estimate x when y = 16 and y when x = 10
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Solution
Given, `barx` = 13, `bary` = 17, `sigma_x` = 3, `sigma_y` = 2, r = 0.6
b_{yx} = `"r" sigma_y/sigma_x = 0.6 xx 2/3` = 0.4
b_{xy} = `"r" sigma_x/sigma_y = 0.6 xx 3/2` = 0.9
The regression equation of X on Y is given by `("X"  barx) = "b"_(xy) ("Y"  bary)`
(X – 13) = 0.9(Y – 17)
X – 13 = 0.9Y – 15.3
X = 0.9Y – 15.3 + 13
X = – 2.3 + 0.9Y ......(i)
For Y = 16, from equation (i) we get
X = – 2.3 + (0.9)(16)
= – 2.3 + 14.4
= 12.1
The regression equation of Y on X is given by `("Y"  bary) = "b"_(yx) ("X"  barx)`
(Y – 17) = 0.4(X – 13)
Y – 17 = 0.4X – 5.2
Y = 0.4X – 5.2 + 17
Y = 11.8 + 0.4X .....(ii)
For X = 10, from equation (ii) we get
Y = 11.8 + 0.4(10)
= 11.8 + 4
= 15.8
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