Explanation
$$Given\quad { b }_{ xy }=0.2\\ { b }_{ yx }=0.8\\ r=\sqrt { { b }_{ xy }\times { b }_{ yx } } \\ r=\sqrt { 0.36 } \\ r=0.6$$
Where r is coefficient of correlation.
Correlation coefficients are expressed as values between 1 and +1. A coefficient of +1 indicates a perfect positive correlation: A change in the value of one variable will predict a change in the same direction in the second variable. A coefficient of 1 indicates a perfect negative correlation: A change in the value of one variable predicts a change in the opposite direction in the second variable. Hence, correct answer is option C.
Physics$$(P)$$
Mathematics $$(M)$$
Rank $$(P)$$
$$|d|$$
$$d^2$$
98
60
72
62
56
40
39
52
30
78
65
70
38
54
32
31
6
3
1
2
4
7
8
5
9
0
$$n=9,\quad \sum d^2=40$$
$$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{40\times 6}{9(9^2-1)}=1-\cfrac{240}{720}=0.66$$
Since $$r>0$$ we can say that this indicate a moderate positive relationship between marks of physics and mathematics.
Marks of Physics $$(X)$$
Marks in Maths$$(Y)$$
$$XY$$
$$X^2$$
$$Y^2$$
35
23
47
17
10
43
28
33
45
49
12
1050
759
2115
391
80
2107
108
24
868
1225
529
2209
289
100
1849
81
784
900
1089
2025
64
2401
144
16
961
$$\sum X=218,\quad \sum Y=235,\quad \sum XY=7502,\quad \sum X^2=7102,\quad \sum Y^2=8129$$
$$N=09$$
Cov$$(x,y)=\cfrac{\sum XY}{N}-\cfrac{\sum X}{N}.\cfrac{\sum Y}{N}=\cfrac{7502}{09}-\cfrac{218}{9}.\cfrac{235}{9}=201.08$$
$$\sigma_x=\sqrt{\cfrac{\sum X^2}{N}-\left(\cfrac{\sum X^2}{N}\right)^2}=\sqrt{\cfrac{7102}{09}-\left(\cfrac{218}{09}\right)^2}=14.22$$
$$\sigma_y=\sqrt{\cfrac{\sum Y^2}{N}-\left(\cfrac{\sum Y^2}{N}\right)^2}=\sqrt{\cfrac{8129}{09}-\left(\cfrac{235}{9}\right)^2}=14.88$$
$$r=\cfrac{Cov(x,y)}{\sigma_x.\sigma_y}=\cfrac{201.08}{14.22\times 14.88}=0..95$$
Competition
$$1st$$ Judge
Rank
$$2nd$$ Judge
A
B
C
D
E
5.7
5.8
5.9
5.6
5.5
6.0
$$n=5,\quad \sum d^2=12$$
$$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 12}{5(5^2-1)}=1-\cfrac{72}{240}=0.44$$
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