WAEC Further Mathematics 2018 Paper

Concept:Spearman's rank correlation coefficient $\rho$ measures how closely two sets of rankings agree with each other.Explanation:Here, $n = 4$ students are ranked in Mathematics and Physics.For each student, calculate the rank difference $d = |r_x - r_y|$ and then square it to obtain $d^2$.The marks, ranks, and computed values are given in the table below.

Maths mark	Rank $r_x$	Physics mark	Rank $r_y$	$d=|r_x-r_y|$	$d^2$
3	2	4	1	1	1
4	1	3	2	1	1
2	3	1	4	1	1
1	4	2	3	1	1
Total					4

Adding the $d^2$ column gives $\sum d^2 = 4$.Use the formula for Spearman's rank correlation coefficient.$\rho = 1 - \frac{6\sum d^2}{n(n^2 - 1)}$Substitute $n = 4$ and $\sum d^2 = 4$.$\rho = 1 - \frac{6 \times 4}{4(4^2 - 1)} = 1 - \frac{24}{60}$Since $\frac{24}{60} = 0.4$, we get $\rho = 1 - 0.4 = 0.6$.Answer:The rank correlation coefficient is $0.6$, which is option C.