Question

From a random sample space, \[5\] students have been selected. Their marks in Maths and Statistics are given below. The rank correlation coefficient is

Roll number	\[1\]	\[2\]	\[3\]	\[4\]	\[5\]
Marks in Maths	\[85\]	\[60\]	\[73\]	\[40\]	\[90\]
Marks in Statistics	\[93\]	\[75\]	\[65\]	\[50\]	\[80\]

1). \[0.8\]
2). \[0.6\]
3). \[0.5\]
4). \[ - 0.8\]

Answer 1

Hint: Here we are asked to find the rank correlation coefficient of the given data of five students. To find the correlation coefficient we first need to set up the rank of the given data. The rank will be given according to the marks from highest to lowest like one, two, three, etc. Then we will find the difference in the ranks for each student and find their sum. After that, we will apply the formula to find the rank correlation coefficient.
Formula Used: The rank correlation coefficient \[ = 1 - \dfrac{{6\sum {d_i^2} }}{{n\left( {{n^2} - 1} \right)}}\] where \[d\] is the difference in ranks of each observation and \[n\] is the total number of observations. It is denoted by \[\rho \].

Complete step-by-step solution:
It is given the marks of five students in two subjects, Math and Statistics.
We aim to find the rank correlation coefficient of the given data. For that, we will first set up the rank for each subject for each student.

Roll number	Math	Statistic	\[x\](Rank of math)	\[y\](Rank of statistic)
\[1\]	\[85\]	\[93\]	\[2\]	\[1\]
\[2\]	\[60\]	\[75\]	\[4\]	\[3\]
\[3\]	\[73\]	\[65\]	\[3\]	\[4\]
\[4\]	\[40\]	\[50\]	\[5\]	\[5\]
\[5\]	\[90\]	\[80\]	\[1\]	\[2\]

Here \[x\] is the rank of the subject math and \[y\] is the rank of the subject statistics.
Now we will find the difference in rank and its square.

Roll number	Math	Statistic	\[x\](Rank of math)	\[y\](Rank of statistic)	\[d\](Difference in rank:\[x \sim y\])	\[{d^2}\](Square of the difference)
\[1\]	\[85\]	\[93\]	\[2\]	\[1\]	\[1\]	\[1\]
\[2\]	\[60\]	\[75\]	\[4\]	\[3\]	\[1\]	\[1\]
\[3\]	\[73\]	\[65\]	\[3\]	\[4\]	\[1\]	\[1\]
\[4\]	\[40\]	\[50\]	\[5\]	\[5\]	\[0\]	\[0\]
\[5\]	\[90\]	\[80\]	\[1\]	\[2\]	\[1\]	\[1\]

Now we will find the sum of the square of the difference in the rank \[(x \sim y)\].
\[\sum {{d^2} = 1 + 1 + 1 + 0 + 1 = 4} \]
Thus, we have found the necessary thing we need to find the rank correlation coefficient.
We know that the rank correlation coefficient \[\rho = 1 - \dfrac{{6\sum {{d^2}} }}{{n\left( {{n^2} - 1} \right)}}\] where \[d\] is the difference in ranks of each observation and \[n\]is the total number of observations.
Now let us substitute the values in the formula.
\[\rho = 1 - \dfrac{{6\sum {{d^2}} }}{{n\left( {{n^2} - 1} \right)}}\]\[ = 1 - \dfrac{{6(4)}}{{5({5^2} - 1)}}\]
Let us simplify the above expression.
\[ = 1 - \dfrac{{24}}{{5(25 - 1)}}\]
\[ = 1 - \dfrac{{24}}{{5(24)}}\]
\[ = 1 - \dfrac{1}{5}\]
On further simplification, we get
\[\rho = \dfrac{{5 - 1}}{5} = \dfrac{4}{5} = 0.8\]
Thus, the rank correlation coefficient of the given observation is \[0.8\]. Now let us see the options.
Option (1) \[0.8\] is the correct answer as we got the same value in our calculation above.
Option (2) \[0.6\] is an incorrect option as we got that \[0.8\] is the value of the rank correlation coefficient.
Option (3) \[0.5\] is an incorrect option as we got that \[0.8\] is the value of the rank correlation coefficient.
Option (4) \[ - 0.8\] is an incorrect option as we got that \[0.8\] is the value of the rank correlation coefficient.

Note: The rank correlation coefficient is also known as the Spearman Rank correlation coefficient. The formula that we used above is only for the case when all the ranks are distinct. If any one of the ranks is repeated then the formula that we used above cannot be used to find the rank correlation coefficient.