Question 146
If the sum of the squares of deviations of ranks of two students in five subjects is 16, then the coefficient of rank correlation is
Solution
Given, Squares of deviation, D^2= 16
No. of subjects, N=5
Now, according to Spearmen's rank correlation
Coeffiecient of rank correlation ,R = 1 - (6 *Summation D^2)/[N(N^2 -1)]
= 1 - (6*16)/[5(25-1)]
= 1 - (6*16/(5*24)
= 1 - (4/5)
=1/5 Ans.
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