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The ranks of same 15 students in two subjects mathematics and statistics are given below:

(1,10) (2,7) (3,2)(4,6) (5,4) (6,8) (7,3) (8,1) (9,11) (10,15) (11,9) (12,5) (13,14) (14,12) (15,13)

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1 Answer

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Let xi denotes the ranks of students in math & yi denotes the ranks of students in statistics.

xi yi xi - \(\bar {\text x}\) yi - \(\bar y\) (xi - \(\bar {\text x}\))2 (yi - \(\bar y\))2 (xi - \(\bar {\text x}\) )(yi - \(\bar{\text y}\))
1 10 -7 2 49 7 -14
2 7 -6 -1 36 1 6
3 2 -5 -6 25 36 30
4 6 -4 -2 16 4 8
5 4 -3 -4 9 16 12
6 8 -2 0 4 0 0
7 3 -1 -5 1 25 5
8 1 0 -7 0 49 0
9 11 1 3 1 9 3
10 15 2 7 4 49 14
11 9 3 1 9 1 3
12 5 4 -3 16 9 -12
13 14 6 4 36 16 24
14 12 6 4 36 16 24
15 13 7 5 49 25 35
Σxi = 120
 
Σyi = 120
 
Σ(xi - \(\bar{\text x}\))2
 = 280


 

Σ(yi - \(\bar{\text y}\))2

= 280

Σ(xi - \(\bar {\text x}\)) (yi - \(\bar {\text y}\))
= 144

Σxi = 1 + 2 + 3....+15 = \(\frac{15(15+1)}2\) = \(\frac{15\times16}2\) = 15 x 8 = 120

Σyi = 1 + 2 + 3 + ....15 = 120

\(\therefore\) \(\bar {\text x} = \frac{\sum\text x_i}{n}=\frac{120}{15}\) = 8

\(\bar y\) = \( \frac{\sum\text y_i}{n}=\frac{120}{15}\) = 8

\(\therefore\) Rank correlation = \(\frac{\sum(X_i-\bar x)(y_i-\bar y)}{\sqrt{\sum(x_i-\bar x)^2\sum(y_i-\bar y)^2}}\)

\(=\frac{144}{\sqrt{280\times280}}=\frac{144}{280}\) = \(\frac{18}{35}\) = 0.5143

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