Question

Karl Pearson's coefficient of rank correlation between the ranks obtained by ten students in Mathematics and Chemistry in a class test as given below is: Rank in Mathematics $12345678910$. Rank in Chemistry $31051294876$

• A. $0.224$
• B. $0.204$
• C. $0.240$
• D. None of these

Answer 1

The correct option is A

$0.224$

Explanation for correct option:

Calculating the rank correlation:

Given the Rank in mathematics and chemistry are:

Mathematics $x_i$

$1$

$2$

$3$

$4$

$5$

$6$

$7$

$8$

$9$

$10$

Chemistry $y_i$

$3$

$10$

$5$

$1$

$2$

$9$

$4$

$8$

$7$

$6$

Now converting this data into tabular form we get

$x_i$	$y_i$	$x_i{y}_i$	${x_i}^2$	${y_i}^2$
$1$	$3$	$3$	$1$	$9$
$2$	$10$	$20$	$4$	$100$
$3$	$5$	$15$	$9$	$25$
$4$	$1$	$4$	$16$	$1$
$5$	$2$	$10$	$25$	$4$
$6$	$9$	$54$	$36$	$81$
$7$	$4$	$28$	$49$	$16$
$8$	$8$	$64$	$64$	$64$
$9$	$7$	$63$	$81$	$49$
$10$	$6$	$60$	$100$	$36$

Now,

$\sum x_i=55$,

$\sum y_i=55$,

$\sum {x_i}^2=385$

$\sum {y_i}^2=385$

$\sum x_i{y}_i=321$

Number of terms, $N=10$

Now coefficient of correlation is given by,

$\begin{array}{l}r(x,y)=\frac{N\sum x_i{y}_i-\sum x_i\sum y_i}{\left(\sqrt{N\sum x_i^2-{\left(\sum x_i\right)}^2}\right)\left(\sqrt{N\sum y_i^2-{\left(\sum y_i\right)}^2}\right)}\\ =\frac{(10\times 321)-(55\times 55)}{\left(\sqrt{10\times 385-(55{)}^2}\right)\left(\sqrt{10\times 385-(55{)}^2}\right)}\\ =\frac{3210-3025}{\left(\sqrt{825}\right)\left(\sqrt{825}\right)}\\ =\frac{185}{825}\\ =0.224\end{array}$

Therefore, Karl Pearson's coefficient of rank correlation between the ranks obtained by ten students in Mathematics and Chemistry in a class test is equal to $0.224$

Hence, option (A) is the correct answer