Question

The rank correlation coefficient between the marks in two subjects of a class is 0.8. The sum of the squares of the differences between the ranks is 33. Then the number of students is:

• A. 46
• B. 22
• C. 60
• D. 10

Hint:

Spearman’s rank correlation is useful when data is in rank form. Use \( r = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)} \) to relate rank differences and correlation.

Answer 1

The Correct Option is D

Spearman’s rank correlation formula: \[ r = 1 - \dfrac{6 \sum d^2}{n(n^2 - 1)} \] Given: \( r = 0.8 \), \( \sum d^2 = 33 \) Substitute into the formula: \[ 0.8 = 1 - \dfrac{6 \cdot 33}{n(n^2 - 1)} \Rightarrow 0.2 = \dfrac{198}{n(n^2 - 1)} \Rightarrow n(n^2 - 1) = 990 \Rightarrow n^3 - n = 990 \Rightarrow n = 10 \]