Question:1
Make a scatter diagram from the following data and interpret the result.
X
4
5
6
7
8
9
10
11
12
Y
78
72
66
60
54
48
42
36
30
Solution:
Thus, there is perfect negative correlation between X and Y.
Question:2
Represent correlation between the following figures through scatter diagram.
X
8
16
24
31
42
50
Y
70
58
50
32
26
12
Solution:
Thus, there is very high degree of negative correlation between X and Y series.
Question:3
Given the following pairs of values of the variables X and Y:
X
1
2
3
4
5
6
7
8
Y
11
12
15
20
24
18
26
29
Make a scatter diagram. Comment on the nature of relationship between variables X and Y.
Solution:
Thus, there is a high degree of positive correlation between X and Y.
Question:4
Given the following pair of values of the variables X and Y:
X
8
10
12
11
9
7
13
14
15
Y
5
7
9
8
6
4
10
11
12
Also describe relationship between X and Y.
Solution:
Thus, there is a perfect positive correlation between X and Y.
Question:5
Find the coefficient of correlation between X and Y series from the data:
X
10
12
8
15
20
25
40
Y
15
10
6
25
16
12
8
Solution:
X
Y
X2
Y2
XY
10
12
8
15
20
25
40
15
10
6
25
16
12
8
100
144
64
225
400
625
1600
225
100
36
625
256
144
64
150
120
48
375
320
300
320
ΣX
=130
ΣY
=92
ΣX 2
=3158
ΣY 2
=1450
ΣXY
=1633
N=7
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
7×1633-130×92
7×3158-(130) 2 7×1450-(92) 2
√
or, r = √
11431-11960
22106-16900 √ 10150-8464
-529
529
or, r = 72.15×41.06 or, r = - 2962.47 ⇒ r = -0. 178
Thus, coefficient of correlation is 0.178.
Question:6
The data on price and quantity purchased relating to a commodity for 10 months are given below:Calculate coefficient of correlation between price and quantity.
Price
Quantity
kg.
10
14
12
11
9
7
15
16
18
20
25
20
30
32
35
40
19
16
12
10
Solution:
Price
(X)
Quantity
(Y)
X2
Y2
XY
10
14
12
11
9
7
15
16
18
20
25
20
30
32
35
40
19
16
12
10
100
196
144
121
81
49
225
256
324
400
625
400
900
1024
1225
1600
361
256
144
100
250
280
360
352
315
280
285
256
216
200
ΣX
=132
ΣY
=239
ΣX 2
=1896
ΣY 2
=6635
ΣXY
=2794
N = 10
r =√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×2794-132×239
10×1896-(132) 2 10×6635-(239) 2
√
-3608
or, r = 39.19×96.06 ⇒ r = -0. 958
Thus, the coefficient of correlation between price and quantity is -0.958.
Question:7
From the following data, calculate coefficient correlation between the variables X and Y using Karl Pearson's method:
X
10
6
9
10
12
13
11
9
Y
9
4
6
9
11
13
8
4
Solution:
X
Y
X2
Y2
XY
10
6
9
10
12
13
11
9
9
4
6
9
11
13
8
4
100
36
81
100
144
169
121
81
81
16
36
81
121
169
64
16
90
24
54
90
132
169
88
36
∑ X = 80
∑ Y = 64
∑ X 2 = 832
∑ Y 2 = 584
∑ XY = 683
N=8
NΣXY-ΣXΣY
2
2
2
2
NΣX -(ΣX) NΣY -(ΣY)
√
r= √
or,
r =√
8×683-80×64
8×832-(80) 2 8×584-(64) 2
√
344
or, r = 16×24 ⇒ r = 0. 895
Thus, the coefficient of correlation between the variables X and Y is 0.895.
Question:8
Calculate coefficient of correlation for the ages of husband and wife.
Age of husband
24
25
22
30
34
37
Age of wife
20
21
18
26
28
30
Solution:
Age of Husband
(X)
Age of Wife
(Y)
X2
Y2
XY
24
25
22
30
34
37
20
21
18
26
28
30
576
625
484
900
1156
1369
400
441
324
676
784
900
480
525
396
780
952
1110
ΣX
=172
ΣY
=143
ΣX 2
=5110
ΣY 2
=3525
ΣXY
=4243
N=6
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
6×4243-172×143
6×5110-(172) 2 6×3525-(143) 2
√
862
or, r = 32.80×26.47 ⇒ r = 0. 992
Thus, the coefficient of correlation for the ages of husband and wife is 0.992.
Question:9
Find out the correlation between the marks in Statistics and marks in Accountancy:
No. of Students
1
Marks in Statistics
20 35 15 40 10 35 30 25 45 30
marks in
Accountancy
25 30 20 35 20 25 25 35 35 30
2
3
4
5
6
7
8
9
10
Solution:
Marks in
Statistics
(X)
Marks in
Accountancy
(Y)
X2
Y2
XY
20
35
15
40
10
35
30
25
45
30
25
30
20
35
20
25
25
35
35
30
400
1225
225
1600
100
1225
900
625
2025
900
625
900
400
1225
400
625
625
1225
1225
900
500
1050
300
1400
200
875
750
875
1575
900
ΣX
=285
ΣY
=280
ΣX 2
ΣY 2
=9225 =8150
ΣXY
=8425
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×8425-280×285
10×9225-(285) 2 10×8150-(280) 2
√
4450
or, r = 105×55.67 ⇒ r = 0. 761
Thus, there exists a high degree of positive correlation between the marks in Statistics and marks in Accountancy.
Question:10
Find Karl Pearson's coefficient of correlation between the values of X and Y given data:
X 128
129
130
140
132
135
125
130
132
135
Y
89
90
95
96
94
80
100
96
100
80
Solution:
X
Y
X2
Y2
XY
128
129
130
140
132
135
125
130
132
135
80
89
90
95
96
94
80
100
96
100
ΣX
ΣY
=1316 =920
16384
16641
16900
19600
17424
18225
15625
16900
17424
18225
6400
7921
8100
9025
9261
8836
6400
10000
9216
10000
10240
11481
11700
13300
12672
12690
10000
13000
12672
13500
ΣX 2
=173348
ΣY 2
=85159
ΣXY
=121255
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2
NΣY2-(ΣY) 2
√
or, r = √
10×121255-1316×920
10×173348-(1316) 2 10×85159-(920) 2
√
1830
or, r = 40.29×72.04 ⇒ r = 0. 63
Thus, the correlation coefficient between the values of X and Y is 0.63.
Question:11
Calculate coefficient of correlation from the following data:
Height of fathers
inches
66 68 69 72 65 59 62 67 61
71
Height of Sons
inches
65 64 67 69 64 60 59 68 60
64
Solution:
Height
Height of
of
Fathers
Sons
(X)
(Y)
66
68
69
72
65
59
62
67
61
71
65
64
67
69
64
60
59
68
60
64
ΣX
=660
ΣY
=640
X2
Y2
XY
4356
4624
4761
5184
4225
3481
3844
4489
3721
5041
4225
4096
4489
4761
4096
3600
3481
4624
3600
4096
4290
4352
4623
4968
4160
3540
3658
4556
3660
4544
ΣX 2
ΣY 2
=43726 =41068
ΣXY
=42351
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×42351-660×640
10×43726-(660) 2 10×41068-(640) 2
√
1110
or, r = 40.74×32.86 ⇒ r = 0. 829
Therefore, coefficient of correlation= 0.829
Question:12
Making use of the data given below, calculate the coefficient of correlation.
X
10
6
9
10
12
13
11
9
Y
9
4
6
9
11
13
8
4
X
Y
X2
Y2
XY
10
6
9
10
12
13
11
9
9
4
6
9
11
13
8
4
100
36
81
100
144
169
121
81
81
16
36
81
121
169
64
16
90
24
54
90
132
169
88
36
ΣX
=80
ΣY
=64
ΣX 2
=832
ΣY 2
=584
ΣXY
=683
Solution:
N=8
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
8×683-80×64
8×832-(80) 2 8×584-(64) 2
√
344
or, r = 16×24 ⇒ r = 0. 895
Therefore, coefficient of correlation= 0.895.
Question:13
The data on price and demand for a commodity is given below:
Price
Demand
kg.
14
16
17
18
19
20
21
22
23
84
78
70
75
66
67
62
58
60
Calculate the coefficient of correlation between price and demand and comment on its sign and magnitude.
Solution:
Price demand
(X)
(Y)
X2
Y2
XY
14
16
17
18
19
20
21
22
23
84
78
70
75
66
67
62
58
60
196
256
289
324
361
400
441
484
529
7056
6084
4900
5625
4356
4489
3844
3364
3600
1176
1248
1190
1350
1254
1340
1302
1276
1380
ΣX
=170
ΣY
=620
ΣX 2
=3280
ΣY 2
=43318
ΣXY
=11516
N=9
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
9×11516-170×620
9×3280-(170) 2 9×43318-(620) 2
√
-1756
or, r = 24.89×73.90 ⇒ r = -0. 954
Hence, there is a high degree of negative correlation between price and demand.
Question:14
Find coefficient of correlation from the following figures:
X
0.5
Y
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000
Solution:
X
Y
X2
Y2
XY
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
1000
2000
3000
4000
5000
6000
7000
8000
9000
0.25
1.00
2.25
4.00
6.25
9.0
12.25
16.0
20.5
1000000
4000000
9000000
16000000
25000000
36000000
49000000
64000000
81000000
500
2000
4500
8000
12500
18000
24500
32000
40500
ΣY 2
=285000000
ΣXY
=142500
ΣX
ΣY
ΣX 2
=22.5 =45000 =71.25
N=9
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
9×142500-22.5×45000
9×71.25-(22.5) 2 9×285000000-(45000) 2
√
270000
or, r = 11.61×23237.90 ⇒ r = 9. 99 = 1
Thus, the coefficient of correlation is 1.
Question:15
Calculate product moment correlation between the values of X and Y.
X
2
3
1
5
6
4
Y
4
5
3
4
6
2
Solution:
X
Y
X2
Y2
XY
2
3
1
5
6
4
4
5
3
4
6
2
4
9
1
25
36
16
16
25
9
16
36
4
8
15
3
20
36
8
ΣX
=21
ΣY
=24
ΣX 2
=91
ΣY 2
=106
ΣXY
=90
N=6
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
6×90-24×21
6×91-(21) 2 6×106-(24) 2
√
36
or, r = 10.24×7.74 ⇒ r = 0. 454
Hence, product moment correlation = 0.454
Question:16
Following are the heights and weights of 10 students of a class:
Height
inches
60
64
68
62
67
69
70
72
65
61
Weight
kg
50
48
56
65
49
52
57
60
59
47
Calculate the coefficient of correlation by Karl Pearson's method:
Solution:
Height
(X)
Weight
(Y)
X2
Y2
XY
60
64
68
62
67
69
70
72
65
61
50
48
56
65
49
52
57
60
59
47
3600
4096
4624
3844
4489
4761
4900
5184
4225
3721
2500
2304
3136
4225
2401
2704
3249
3600
3481
2209
3000
3072
3808
4030
3283
3588
3990
4320
3835
2867
ΣX
=658
ΣY
=543
ΣX 2
=43444
ΣY 2
=29809
ΣXY
=35793
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×35793-658×543
10×43444-(658) 2 10×29809-(543) 2
√
636
or, r = 38.41×56.92 ⇒ r = 0. 29
Hence, coefficient of correlation is 0.29.
Question:17
Calculate coefficient of correlation from the following data:
X
30
36
42
48
54
60
72
82
Y
50
50
54
54
62
66
70
82
Solution:
X
Y
X2
Y2
XY
30
36
42
48
54
60
72
82
50
50
54
54
62
66
70
82
900
1296
1764
2304
2916
3600
5184
6724
2500
2500
2916
2916
3844
4356
4900
6724
1500
1800
2268
2592
3348
3960
5040
6724
ΣX ΣY
ΣX 2
ΣY 2
ΣXY
=424 =488 =24688 =30656 =27232
N=8
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
8×27232-424×488
8×24688-(424) 2 8×30656-(488) 2
√
10944
or, r = 133.14×84.28 ⇒ r = 0. 9753
Hence, coefficient of correlation= 0.9753
Question:18
Calculate Karl Pearson's coefficient of correlation between income and expenditure of 10 families from the following data:
Income
in′000
59
55
58
60
65
63
54
56
66
64
Expenditure
in′000
52
53
55
58
57
66
59
54
52
54
Solution:
Income Expenditure
(X)
(Y)
59
55
58
60
65
63
54
56
66
64
52
53
55
58
57
66
59
54
52
54
ΣX
=600
ΣY
=560
X2
Y2
XY
3481
3025
3364
3600
4225
3969
2916
3136
4356
4096
2704
2809
3025
3364
3249
4356
3481
2916
2704
2916
3068
2915
3190
3480
3705
4158
3186
3024
3432
3456
ΣX 2
ΣY 2
ΣXY
=36168 =31524 =33614
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×33614-600×560
10×36168-(600) 2 10×31524-(560) 2
√
Thus, coefficient of correlation = 0.0843
Question:19
Calculate Karl Pearson's coefficient between the marks outof30
in English and Hindi obtained by 10 students.
Marks in English
10 25 13 25 22 11 12 25 21 20
Marks in Hindi
12 22 16 15 18 18 17 23 24 17
Solution:
140
or, r = 40.98×40.49 ⇒ r = 0. 0843
Marks in
English
(X)
Marks in
Hindi
(Y)
X2
Y2
XY
10
25
13
25
22
11
12
25
21
20
12
22
16
15
18
18
17
23
24
17
100
625
169
625
484
121
144
625
441
400
144
484
256
225
324
324
289
529
576
289
120
550
208
375
396
198
204
575
504
340
ΣX
=184
ΣY
=182
ΣX 2
ΣY 2
=3734 =3440
ΣXY
=3470
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×3470-184×182
10×3734-(184) 2 10×3440-(182) 2
√
1212
or, r = 59.02×35.72 ⇒ r = 0. 574
Thus, coefficient of correlation = 0.574
Question:20
Calculate coefficient of correlation from the following data:
i
Sum of deviation of X values = −6
ii
Sum of deviation of Y values = 1
iii
Sum of squares of deviations of X values = 196
iv
Sum of squares of deviation of Y values = 87
v
Sum of the product of deviations of X and Y values =124
vi
No of pairs of observations = 6
Solution:
Given:
(∑ d )× (∑ d )
x
∑ ∑
y
∑ dx = -6∑ dy = 1∑ d2x = 196∑ d2y = 87∑ dx dy = 124N = 6r =
√
(∑ d )
x
2
∑ d x-
N
(-6×1)
N
d x d y2
×
√
(∑ d )
y
2
∑ d y-
124- 6
2
N
or, r =
√
(-6)
2
196- 6 ×
√
(1)
87- 6
2
124+1
Hence, coefficient of correlation=0.974
Question:21
Two judges in a beauty competition rank the 12 entries as follows:
X
1
2
3
4
5
6
7
8
9
10
11
12
Y
12
9
6
10
3
5
4
7
8
2
11
1
Calculate the Spearman's rank coefficient of correlation.
Solution:
X
Y
R1
R2
1
2
3
4
5
6
7
8
9
10
11
12
12
9
6
10
3
5
4
7
8
2
11
1
1
2
3
4
5
6
7
8
9
10
11
12
12
9
6
10
3
5
4
7
8
2
11
1
D=R1−R2
−11
−7
−3
−6
2
1
3
1
1
8
0
11
D2
121
49
9
36
4
1
9
1
1
64
0
121
ΣD2
=416
N = 12
[
1
(
1
) (
)
]
6 ΣD 2- 12 m3-m +12 m3-m + ...
N 3-N
rk = 1 Here, m = 0
6ΣD 2
3
-N
rk = 1 - N
6×416
125
125
or, r = 13.78×9.31 or, r = 13.78×9.31 or, r = 128.29 ⇒ r = 0. 974
2496
or, rk = 1 - 1728-12 or, rk = 1 - 1716 ⇒ rk = -0. 454
Hence, spearmen's rank coefficient of correlation = -0.454
Question:22
A group of ten workers of a factory is ranked according to their efficiency by two different judges as follows:
Name of Worker
A
B
C D E
F G H
Rank by Judge A
4
8
6
7
1
3
2
5
10
I
J
9
Rank by Judge B
3
9
6
5
1
2
4
7
8
10
Calculate the coefficient of rank correlation.
Solution:
Name of
Worker
Rank by
Judge A
(RA)
Rank by
Judge B
(RB)
A
B
C
D
E
F
G
H
I
J
4
8
6
7
1
3
2
5
10
9
3
9
6
5
1
2
4
7
8
10
D= RA−RB
D2
1
-1
0
2
0
1
-2
-2
2
-1
1
1
0
4
0
1
4
4
4
1
ΣD2
=20
N = 10
6ΣD 2
3
-N or,
6×20
rk = 1 - N
rk = 1 - 1000-10 or, rk = 1 -0. 121 ⇒ rk = 0. 88
Hence, coefficient of rank correlation=0.88
Question:23
Ten competitors in a debate contest are ranked by three judges in the following order.
Competitors
A
B
C
D
E F G H
I
J
Ranks By 1st Judge
7
4
10
5
9 8
6
2
1
3
Judge
4
1
9
10
7 3
2
5
6
8
Ranks By 3rd Judge
10
2
8
5
7 6
9
1
4
3
Ranks By 2
nd
Using the ranking correlation method and state which pair of judges have the nearest approach.
Solution:
Competitors
R1
R2
R3
A
B
C
D
E
F
G
H
I
J
7
4
10
5
9
8
6
2
1
3
4
1
9
10
7
3
2
5
6
8
10
2
8
5
7
6
9
1
4
3
D1 = R1 R2
D21
3
3
1
−5
2
5
4
−3
−5
−5
9
9
1
25
4
25
16
9
25
25
D2 = R1
D22
- R3
−3
2
2
0
2
2
−3
1
−3
0
9
4
4
0
4
4
9
1
9
0
ΣD1 2
=148
ΣD2 2
=44
D3 = R2 R3
−6
−1
1
5
0
−3
−7
4
2
5
D23
36
1
1
25
0
9
49
16
4
25
ΣD3 2
=166
N = 10
Rank Correlation between Judge 1 and Judge 2
2
6∑ D 1
rk 1,2 = 1 -
N 3-N
6×148
= 1 - 1000-10 or, rk 1,2 = 1 -0. 896 ⇒ rk 1,2 = 0. 103
Rank Correlation between Judge 1 and Judge 3
2
6∑ D 2
rk 1,3 = 1 -
N 3-N
6×44
= 1 - 1000-10 or, rk 1,3 = 1 -0. 266 ⇒ rk 1,3 = 0. 73
Rank Correlation between Judge 2 and Judge 3
2
6∑ D 3
rk 2,3 = 1 -
N 3-N
6×166
= 1 - 1000-10 or, rk 2,3 = 1 -1. 006 ⇒ rk 2,3 = 0. 006
As the rank correlation coefficient between Judge 1 and Judge 3 is highest, thus Judge 1 and Judge 3 has the nearest approach.
Question:24
A group of 8 students got the following marks in a test in Maths and Accountancy.
Marks in Maths
50 60 65 70
75
40
80
85
Marks in Accountancy
80 71 60 75
90
82
70
50
Compute the coefficient of rank correlation.
Solution:
Marks in
Marks in
R1
Maths
Accountancy
50
60
65
70
75
40
80
85
80
71
60
75
90
82
70
50
2
3
4
5
6
1
7
8
D= R1−R2
R2
6
4
2
5
8
7
3
1
D2
−4
−1
2
0
−2
−6
4
7
16
1
4
0
4
36
16
49
ΣD2
=126
N=8
6ΣD 2
3
-N or,
rk = 1 - N
6×126
rk
= 1 - 512-8 ⇒ rk = -0. 5
Hence, coefficient of rank correlation = -0.5
Question:25
Calculate rank correlation between advertisement cost and sales as per the data given below:
Cost
in′000
78
36
98
25
75
82
90
62
65
39
Sales
inlakh
84
51
91
60
68
62
86
58
53
47
Solution:
Advertising
Sales
Cost
(Y)
(X)
78
36
98
25
75
82
90
62
65
39
84
51
91
60
68
62
86
58
53
47
R1
R2
7
2
10
1
6
8
9
4
5
3
8
2
10
5
7
6
9
4
3
1
D= R1−R2
D2
−1
0
0
−4
−1
2
0
0
2
2
1
0
0
16
1
4
0
0
4
4
ΣD2
=30
N= 1
6ΣD 2
3
-N or,
rk = 1 - N
6×30
rk = 1 - 1000-10 or, rk = 1 - 0. 1818 = -0. 818 ⇒ rk = -0. 82
Hence, coefficient of rank correlation = -0.82
Question:26
Calculate the coefficient of rank correlation from the following data:
X
48
33
40
9
16
16
65
24
16
57
Y
13
13
24
6
15
4
20
9
6
19
Solution:
X
Y
R1
R2
D = R1−R2
D2
48
33
40
9
16
16
65
24
16
57
13
13
24
6
15
4
20
9
6
19
8
6
7
1
3
3
10
5
3
9
5.5
5.5
10
2.5
7
1
9
4
2.5
8
2.5
0.5
−3
−1.5
−4
2
1
1
0.5
1
6.25
0.25
9.0
2.25
16
4
1
1
0.25
1.0
ΣD2
=41.00
N = 10
[
1
(
1
) (
1
) (
6 ΣD 2+12 m13-m1 +12 m23-m2 +12 m33-m3
rk = 1 -
N 3-N
)]
or, rk = 1 -
[
1
( ) ( ) ( )]
1
1
6 41+12 3 3-3 +12 2 3-2 +12 2 3-2
1000-10
or, rk = 1 -
6[41+2+0.5+0.5]
990
or,
rk = 1 -0. 266 ⇒ rk = 0. 733
Hence, coefficient of rank correlation = 0.733
Question:27
The coefficient of rank correlation of the marks obtained by 10 students in two particular subjects was found to be 0.5. It was later discovered that the difference in ranks in two subjects
obtained by one of the students was wrongly taken as 3 instead of 7. What should be the correct value of coefficient of rank correlation.
Solution:
Given, rk = 0.5
N = 10
Now,
6ΣD 2
3-N
Substituting
rk = 1 - N
6ΣD 2
the given values in the formula we get, 0. 5 = 1 - 1000-10 or, 495 = 990 -6ΣD2 or, ΣD2 =
495
6 or,
ΣD2 = 82. 5Wrong ΣD2 = 82. 5Correct ΣD2 = 82. 5 -(3)2 + (7)2
Hence, correct coefficient of rank correlation is 0.257.
Question:28
Find the standard deviation of X series, if coefficient of correlation between two series X and Y is 0.35 and their covariance in 10.5 and variance of Y series is 56.25.
Solution:
Given, r = 0.35
Cov (x,y) = 10.5
σ2y = 56. 25 ⇒ σ = 7. 5
y
Substituting the given values in the formula we get,
r=
Cov (x , y )
σx σy
10.5
10.5
or, 0. 35 = 7.5 σxσx = 7.5×0.35 ⇒ σx = 4
Hence, standard deviation of X series is 4.
Question:29
The ranking of ten students in two subjects A and B as follows:
Subject A
3
5
8
4
7
10
2
1
6
9
Subject B
6
4
9
8
1
2
3
10
5
7
What is the coefficient of rank correlation?
Solution:
RA
RB
D = RA − RB
D2
3
5
8
4
7
10
2
1
6
9
6
4
9
8
1
2
3
10
5
7
−3
1
−1
−4
6
8
−1
−9
1
2
9
1
1
16
36
64
1
81
1
4
ΣD2
=214
N = 10
6ΣD 2
3
-N or,
rk = 1 - N
6×214
rk = 1 - 1000-10 ⇒ rk = -0. 2969
Hence, coefficient of rank correlation is -0.297.
Question:30
Calculate the number of items, when
i
Standard deviation of series Y = 10
ii
Coefficient of correlation = 0.6
iii
Sum of the product of deviation of X and Y from actual means = 150
iv
Sum of squares of deviations of X from actual means = 125
Solution:
Given, σ y = 10 ⇒ σ 2y = 100r = 0. 6Σxy = 150Σx 2 = 125 σ 2y =
Σy 2
N 100
=
Σy 2
N
⇒ Σy 2 = 100 NNow, r = √
Σx y
Σx 2×Σy 2
0. 6 = √
150
125 √ 100 N
0. 6 = √
150
12500 N
Squaring both sides, we get
22500
0. 36 = 12500 N ⇒ N = 5
Hence, number of items = 5
Question:31
Find the coefficient of rank correlation between the marks obtained in Mathematics and those in Statistics by 10 students of a class.
Marks in
Mathematics
12 18 32 18 25 24 25 40 38 22
Marks in Statistics
16 15 28 16 24 22 25 36 34 19
Solution:
Marks in Marks in
Maths
Statistics
R1
R2
D = R1−R2
D2
= 122
12
18
32
18
25
24
25
40
38
22
16
15
28
16
24
22
25
36
34
19
1
2.5
8
2.5
6.5
5
6.5
10
9
4
2.5
1
8
2.5
6
5
7
10
9
4
-1.5
1.5
0
0
0.5
0
-0.5
0
0
0
2.25
2.25
0
0
0.25
0
0.25
0
0
0
ΣD2
=5
Here, N = 10
m 1= m 2 = m 3 = 2
[
1
1
(
1
) (
) (
6 ΣD 2+12 m13-m1 +12 m23-m2 +12 m33-m3
N 3-N
rk = 1 -
)]
[
1
1
1
6 5+12(8-2)+12(8-2)+12(8-2)
or, rk = 1 -
]
⇒ rk = 0. 960
990
Hence, the value of rank correlation coefficient is 0.960.
Question:32
From the following data, calculate coefficient of correlation.
X
57
59
62
63
64
65
58
66
70
72
Y
113
117
126
125
130
128
110
132
140
149
Solution:
X
Y
X2
Y2
XY
57
59
62
63
64
65
58
66
70
72
113
117
126
125
130
128
110
132
140
149
3249
3481
3844
3969
4096
4225
3364
4356
4900
5184
12769
13689
15876
15625
16900
16384
12100
17424
19600
22201
6441
6903
7812
7875
8320
8320
6380
8712
9800
10728
ΣX
ΣY
ΣX 2
ΣY 2
ΣXY
=636 =1270 =40668 =162568 =81291
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×81291-636×1270
10×40668-(636) 2 10×162568-(1270) 2
√
5190
or, r = 46.733×113.04 ⇒ r = 0. 982
Hence, coefficient of correlation is 0.982.
Question:33
Calculate the coefficient of correlation, if:
i
Covariance between X and Y = 9.2
ii
Variance of X = 11.5
iii
Variance of Y = 14.2
Solution:
Given, Cov(x,y) = 9.2
σ 2x = 11. 5 ⇒ σ = 3. 39σ 2y = 14. 2 ⇒ σ = 3. 76Now, r =
x
Cov (x , y )
σσ
x y
9.2
Substituting the given values in the formula we get, or, r = 3.39×3.76 ⇒ r = 0. 721
y
Hence, coefficient of correlation is 0.721.
Question:34
Find Karl Pearson's coefficient for the following data:
Marks in English
45 70 65 30 90 40 50 75 85 60
Marks in Maths
35 90 70 40 95 40 60 80 80 50
Solution:
Marks in
English
(X)
Marks in
Maths
(Y)
X2
Y2
XY
45
70
65
30
90
40
50
75
85
60
35
90
70
40
95
40
60
80
80
50
2025
4900
4225
900
8100
1600
2500
5625
7225
3600
1225
8100
4900
1600
9025
1600
3600
6400
6400
2500
1575
6300
4550
1200
8550
1600
3000
6000
6800
3000
ΣX
=610
ΣX 2
ΣY 2
=40700 =45350
ΣY
=640
ΣXY
=42575
N = 10
r=√
NΣXY-ΣXΣY
NΣX2-(ΣX) 2 NΣY2-(ΣY) 2
√
or, r = √
10×42575-610×640
10×40700-(610) 2 10×45350-(640) 2
√
Hence, Karl Pearson's coefficient is 0.9031.
35350
or, r = 186.81 × 209.52 ⇒ r = 0. 9031