The rank correlation coefficient between marks obtained by 10 students in English and Statistics was found to be 0.5. Find the sum of squares of different of ranks.

Explanation

Given : r=0.5,N=10
Now, `r=1-(6 sumD^(2))/(N^(3)-N)`
`(6sumD^(2))/(N^(3)-N)=1-r`
`(6sumD^(2))/(10^(3)-10)=1-0.5`
`(6sumD^(2))/(1,000-10)=0.5`
`(6sumD^(2))/(990)=0.5`
`implies6sumD^(2)=0.5xx990`
`implies 6sumD^(2)=495`
`implies sumD^(2)=(495)/(6)`
`impliessumD^(2)=82.5`
Sum of squares of Difference of Ranks=82.5.