Sandeep Garg 2018 Solutions for Class 11 Commerce Economics Chapter 8 Measures Of Correlation are provided here with simple step-by-step explanations. These solutions for Measures Of Correlation are extremely popular among class 11 Commerce students for Economics Measures Of Correlation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Sandeep Garg 2018 Book of class 11 Commerce Economics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Sandeep Garg 2018 Solutions. All Sandeep Garg 2018 Solutions for class 11 Commerce Economics are prepared by experts and are 100% accurate.

Page No 11.51:

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

Answer:



Thus, there is perfect negative correlation between X and Y.

Page No 11.51:

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

Answer:



Thus, there is very high degree of negative correlation between X and Y series.

Page No 11.51:

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.

Answer:



Thus, there is a high degree of positive correlation between X and Y.

Page No 11.51:

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.

Answer:

 

Thus, there is a perfect positive correlation between X and Y.



Page No 11.52:

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

Answer:

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 ΣX2=3158 ΣY2=1450 ΣXY=1633

N = 7

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r =7×1633-130×927×3158-13027×1450-922or, r=11431-1196022106-1690010150-8464or, r =-52972.15×41.06or, r =-5292962.47r =-0.178

Thus, coefficient of correlation is 0.178.

Page No 11.52:

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 (₹) 10 14 12 11 9 7 15 16 18 20
Quantity (kg.) 25 20 30 32 35 40 19 16 12 10

Answer:

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 ΣX2=1896 ΣY2=6635 ΣXY=2794


N = 10

r =NΣXY-ΣXΣYNΣX2-ΣX2NΣY2-ΣY2or, r =10×2794-132×23910×1896-132210×6635-2392or, r=-360839.19×96.06 r =-0.958

Thus, the coefficient of correlation between price and quantity is -0.958.

Page No 11.52:

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

Answer:

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 X2=832 Y2=584 XY=683

N = 8

r=NΣXY-ΣXΣYNΣX2-ΣX2NΣY2-ΣY2or, r =8×683-80×648×832-8028×584-642or, r =34416×24 r =0.895

Thus, the coefficient of correlation between the variables X and Y is 0.895.

Page No 11.52:

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

Answer:

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 ΣX2=5110 ΣY2=3525 ΣXY=4243

N = 6

r=NΣXY-ΣYNΣX2-ΣX2 NΣY2-ΣY2or, r =6×4243-172×1436×5110-17226×3525-1432or, r =86232.80×26.47r =0.992

Thus, the coefficient of correlation for the ages of husband and wife is 0.992.

Page No 11.52:

Question 9:

Find out the correlation between the marks in Statistics and marks in Accountancy:

No. of Students 1 2 3 4 5 6 7 8 9 10
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

Answer:

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 ΣX2=9225 ΣY2=8150 ΣXY=8425

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r =10×8425-280×28510×9225-285210×8150-2802or, r =4450105×55.67r =0.761

Thus, there exists a high degree of positive correlation between the marks in Statistics and marks in Accountancy.

Page No 11.52:

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 80 89 90 95 96 94 80 100 96 100

Answer:

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
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=1316 ΣY=920 ΣX2=173348 ΣY2=85159 ΣXY=121255

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2   NΣY2-ΣY2or, r=10×121255-1316×92010×173348-1316210×85159-9202or, r=183040.29×72.04r =0.63

Thus, the correlation coefficient between the values  of X and Y is 0.63.

Page No 11.52:

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

Answer:

Height of Fathers
(X)
Height of Sons
(Y)
X2 Y2 XY
66
68
69
72
65
59
62
67
61
71
65
64
67
69
64
60
59
68
60
64
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=660 ΣY=640 ΣX2=43726 ΣY2=41068 ΣXY=42351

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r=10×42351-660×64010×43726-660210×41068-6402or, r=111040.74×32.86r=0.829

Therefore, coefficient of correlation= 0.829YNΣX2(ΣX)2NΣY2(ΣY)2r =10×1212551316×92010×173348(1316)210×85114(920)2r=183040.29×68.84r =0.

Page No 11.52:

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

Answer:

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 ΣX2=832 ΣY2=584 ΣXY=683

N = 8

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r =8×683-80×648×832-8028×584-642or, r =34416×24r=0.895

Therefore, coefficient of correlation= 0.895.



Page No 11.53:

Question 13:

The data on price and demand for a commodity is given below:

Price (₹) 14 16 17 18 19 20 21 22 23
Demand (kg.) 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.

Answer:

Price
(X)
demand
(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 ΣX2=3280 ΣY2=43318 ΣXY=11516

N = 9

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r =9×11516-170×6209×3280-17029×43318-6202or, r=-175624.89×73.90r =-0.954

Hence, there is a high degree of negative correlation between price and demand.

Page No 11.53:

Question 14:

Find coefficient of correlation from the following figures:

X 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5
Y 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000

Answer:

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
ΣX=22.5 ΣY=45000 ΣX2=71.25 ΣY2=285000000 ΣXY=142500

N = 9
r=NΣXY-ΣXΣYNΣX2-ΣX2NΣY2-ΣY2or, r =9×142500-22.5×450009×71.25-22.529×285000000-450002or, r =27000011.61×23237.90r =9.99=1

Thus, the coefficient of correlation is 1.

Page No 11.53:

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

Answer:

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 ΣX2=91 ΣY2=106 ΣXY=90

N = 6

r=NΣXY-ΣXΣYNΣX2-ΣX2NΣY2-ΣY2or, r =6×90-24×216×91-2126×106-242or, r =3610.24×7.74r =0.454

Hence, product moment correlation = 0.454

Page No 11.53:

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:

Answer:

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 ΣX2=43444 ΣY2=29809 ΣXY=35793

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r=10×35793-658×54310×43444-658210×29809-5432or, r=63638.41×56.92r=0.29

Hence, coefficient of correlation is 0.29.

Page No 11.53:

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

Answer:

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=424 ΣY=488 ΣX2=24688 ΣY2=30656 ΣXY=27232

N = 8

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r=8×27232-424×4888×24688-42428×30656-4882or, r=10944133.14×84.28r=0.9753

Hence, coefficient of correlation= 0.9753

Page No 11.53:

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

Answer:

Income
(X
)
Expenditure
(Y)
X2 Y2 XY
59
55
58
60
65
63
54
56
66
64
52
53
55
58
57
66
59
54
52
54
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=600 ΣY=560 ΣX2=36168 ΣY2=31524 ΣXY=33614

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r=10×33614-600×56010×36168-600210×31524-5602or, r=14040.98×40.49r=0.0843

Thus, coefficient of correlation = 0.0843

Page No 11.53:

Question 19:

Calculate Karl Pearson's coefficient between the marks (out of 30) 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

Answer:

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 ΣX2=3734 ΣY2=3440 ΣXY=3470

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2 NΣY2-ΣY2or, r=10×3470-184×18210×3734-184210×3440-1822or, r=121259.02×35.72r=0.574

Thus, coefficient of correlation = 0.574



Page No 11.54:

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

Answer:

Given:
dx=-6dy= 1 dx2 = 196 dy2 = 87dxdy=124N=6r=dxdy-(dx)×(dy)Ndx2-(dx)2N× dy2-(dy)2Nor, r= 124-(-6×1)6196-(-6)26× 87-(1)26or, r=124+113.78×9.31or, r=12513.78×9.31or, r=125128.29r=0.974

Hence, coefficient of correlation=0.974

Page No 11.54:

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.

Answer:

X Y R1 R2 D=R1−R2
 
D2
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
−11
−7
−3
−6
2
1
3
1
1
8
0
11
121
49
9
36
4
1
9
1
1
64
0
121
N = 12         ΣD2=416


rk=1-6ΣD2-112m3-m+112m3-m+ ...N3-N
Here, m = 0
rk=1-6ΣD2N3-N or, rk  =1-6×4161728-12  or, rk  =1-24961716rk  =-0.454

Hence, spearmen's rank coefficient of correlation = -0.454

Page No 11.54:

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 I J
Rank by Judge A 4 8 6 7 1 3 2 5 10 9
Rank by Judge B 3 9 6 5 1 2 4 7 8 10

Calculate the coefficient of rank correlation.

Answer:

Name of Worker Rank by Judge A
(RA)
Rank by
Judge B

(RB)
D= RA−RB
 
D2
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

1
-1
0
2
0
1
-2
-2
2
-1
 
1
1
0
4
0
1
4
4
4
1
        ΣD2=20

N = 10

rk=1-6ΣD2N3-Nor, rk =1-6×201000-10or, rk =1-0.121  rk=0.88

Hence, coefficient of rank correlation=0.88

Page No 11.54:

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
Ranks By 2nd 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

Using the ranking correlation method and state which pair of judges have the nearest approach.

Answer:

 

Competitors R1 R2 R3 D1= R1 - R2
 
D12 D2= R1 - R3 D22 D3= R2 - R3 D32
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
3
3
1
−5
2
5
4
−3
−5
−5
9
9
1
25
4
25
16
9
25
25
−3
2
2
0
2
2
−3
1
−3
0
9
4
4
0
4
4
9
1
9
0
−6
−1
1
5
0
−3
−7
4
2
5
36
1
1
25
0
9
49
16
4
25
          ΣD12=148   ΣD22=44   ΣD32=166

N = 10

Rank Correlation between Judge 1 and Judge 2
rk1,2=1- 6D12N3-N=1-6×1481000-10or, rk1,2=1-0.896rk1,2=0.103
Rank Correlation between Judge 1 and Judge 3

rk1,3=1- 6D22N3-N=1-6×441000-10or, rk1,3=1-0.266rk1,3=0.73
Rank Correlation between Judge 2 and Judge 3

rk2,3=1- 6D32N3-N=1-6×1661000-10or, rk2,3=1-1.006rk2,3=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.

Page No 11.54:

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.

Answer:

Marks in Maths Marks in Accountancy R1 R2 D= R1−R2
 
D2
50
60
65
70
75
40
80
85
80
71
60
75
90
82
70
50
2
3
4
5
6
1
7
8
6
4
2
5
8
7
3
1
−4
−1
2
0
−2
−6
4
7
16
1
4
0
4
36
16
49
          ΣD2=126

N = 8
rk=1-6ΣD2N3-Nor, rk   =1-6×126512-8rk=-0.5

Hence, coefficient of rank correlation = -0.5



Page No 11.55:

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 (in lakh ₹) 84 51 91 60 68 62 86 58 53 47

Answer:

Advertising Cost
(X)
Sales
(Y)
R1 R2 D= R1−R2
 
D2
78
36
98
25
75
82
90
62
65
39
84
51
91
60
68
62
86
58
53
47
7
2
10
1
6
8
9
4
5
3
8
2
10
5
7
6
9
4
3
1
−1
0
0
−4
−1
2
0
0
2
2
1
0
0
16
1
4
0
0
4
4
          ΣD2=30

N= 1

rk=1-6ΣD2N3-Nor, rk=1-6×301000-10or, rk= 1- 0.1818 =-0.818rk=-0.82

Hence, coefficient of rank correlation = -0.82

Page No 11.55:

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

Answer:

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

rk=1-6ΣD2+112m13-m1+112m23-m2+112m33-m3N3-Nor, rk=1-641+11233-3+11223-2+11223-21000-10or, rk=1-641+2+0.5+0.5990or, rk=1-0.266rk=0.733
Hence, coefficient of rank correlation = 0.733

Page No 11.55:

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.

Answer:

Given, rk = 0.5
N = 10

Now,
rk=1-6ΣD2N3-NSubstituting the given values in the formula we get,0.5=1-6ΣD21000-10or, 495=990-6ΣD2or, ΣD2=4956or, ΣD2=82.5Wrong ΣD2=82.5Correct ΣD2=82.5-32+72                    =122.5Substituting the value of correct ΣD2 in the formula for correlation coefficient:Correct Rank Correlation Coefficient ( rk) = 1-6×122.51000-10 rk=0.257

Hence, correct coefficient of rank correlation is 0.257.

Page No 11.55:

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.

Answer:

Given,  r = 0.35
Cov (x,y) = 10.5
σy2=56.25σy=7.5
Substituting the given values in the formula we get,

r=Cov x, yσx σyor, 0.35=10.57.5 σxσx=10.57.5×0.35σx=4

Hence, standard deviation of X series is 4.

Page No 11.55:

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?

Answer:

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
rk=1-6ΣD2N3-Nor, rk=1-6×2141000-10 rk =-0.2969

Hence, coefficient of rank correlation is -0.297.

Page No 11.55:

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

Answer:

Given, σy=10σy2=100r=0.6Σxy=150Σx2=125 σy2=Σy2N100=Σy2NΣy2=100 NNow,r=ΣxyΣx2×Σy20.6=150125100 N0.6=15012500 N

Squaring both sides, we get
0.36=2250012500 NN=5

Hence, number of items = 5

Page No 11.55:

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

Answer:

Marks in Maths Marks in Statistics R1 R2 D = R1−R2 D2
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
m1= m2 = m3 = 2

rk=1- 6ΣD2+112m13-m1+112m23-m2+112m33-m3N3-Nor, rk=1- 65+1128-2+1128-2+1128-2990rk=0.960

Hence, the value of rank correlation coefficient is 0.960.



Page No 11.56:

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

Answer:

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=636 ΣY=1270 ΣX2=40668 ΣY2=162568 ΣXY=81291

N = 10

r=NΣXY-ΣXΣYX2-ΣX2 NΣY2-ΣY2or, r =10×81291-636×127010×40668-636210×162568-12702or, r =519046.733×113.04r=0.982

Hence, coefficient of correlation is 0.982.

Page No 11.56:

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

Answer:

Given, Cov(x,y) = 9.2

σx2=11.5 σx=3.39σy2=14.2σy=3.76Now,r=Cov x, yσxσySubstituting the given values in the formula we get,or, r=9.23.39×3.76r=0.721

Hence, coefficient of correlation is 0.721.

Page No 11.56:

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

Answer:

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 ΣY=640 ΣX2=40700 ΣY2=45350 ΣXY=42575

N = 10

r=NΣXY-ΣXΣYNΣX2-ΣX2  NΣY2-ΣY2or, r=10×42575-610×64010×40700-610210×45350-6402or, r=35350186.81 × 209.52r =0.9031

Hence, Karl Pearson's coefficient is 0.9031.



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