Question

If sum of the squares of rank differences in the marks of 10 students in 44,then coefficient of Rank correlation is

Answer 1

Answer:

P=1−6n(n2−1)∑di2

0.6=1−n(n2−1)6×66

n(n2−1)6×66=0.4

n(n2−1)=23×66×10=990

n=10

10(102−1)=10(99)=990

Answer 2

The coefficient of Rank correlation is 0.73

Given:

• If sum of the squares of rank differences in the marks of 10 students in 44.

To find:

• Find the coefficient of Rank correlation.

Solution:

Concept/Formula to be used:

$\bf \rho=1-6\frac{\Sigma di^2}{n(n^2-1)}$

Step 1:

Write the given values.

$\Sigma di^2 = 44$

Number of students (n)= 10

Step 2:

Find the coefficient of rank correlation.

$\rho=1-6 \left(\frac{44}{10(100-1)} \right)$

$\rho= 1 - \frac{6 \times 44}{10 \times 99}$

Cancel common factors from numerator and denominator.

$\rho = 1 - \frac{ 4}{5 \times 3}$

$\rho = 1 - \frac{ 4}{15}$

$\rho = \frac{15 - 4}{15 }$

$\rho = \frac{ 11}{15 }$

$\bf \rho = 0.73$

Thus,

The coefficient of Rank correlation is 0.73

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