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TR Jain & Vk Ohri (2017) Solutions for Class 11 Science Economics Chapter 12 - Correlation

TR Jain & Vk Ohri (2017) Solutions for Class 11 Science Economics Chapter 12 Correlation are provided here with simple step-by-step explanations. These solutions for Correlation are extremely popular among class 11 Science students for Economics Correlation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the TR Jain & Vk Ohri (2017) Book of class 11 Science Economics Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s TR Jain & Vk Ohri (2017) Solutions. All TR Jain & Vk Ohri (2017) Solutions for class 11 Science Economics are prepared by experts and are 100% accurate.

Page No 312:

Question 1:

Explain the relation betwen price and quantity supplied through a scattered diagram.

Price (₹) 10 20 30 40 50 60
Quantity Supplied 25 50 75 100 125 150

Answer:

Price Quantity Supplied
10
20
30
40
50
60
25
50
75
100
125
150




Thus, there exists perfect positive correlation(+1) between price and quantity supplied.

Page No 312:

Question 2:

Show the relationship between X and Y through a scattered diagram.

X 8 16 24 31 42 50
Y 70 58 50 32 26 12

Answer:

X Y
8
16
24
31
42
50
70
58
50
32
26
12



Thus, there exists a negative relationship between X and Y.

Page No 312:

Question 3:

Visit your nearest mother dairy store. Get information on the daily price and quantity sold of bananas for the last 30 days. Draw a scattered diagram of the statistical information. Write your observation on the relationship (closeness) between price and quantity sold of bananas.

Answer:

Days Price
(Rs)
Quantity Sold (kg)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
20
20
20
22
22
22
22
25
25
25
25
25
28
28
28
28
30
30
30
30
30
30
35
35
35
35
35
40
40
40
70
71
70
69
68
68.5
69
65
64.5
65.5
65.4
64
60
59
59.5
60
56
57
57.5
57
57.2
56
50
51
49
49.5
50.5
46
47
47.5



Thus, there is a negative relationship between price and quantity sold of bananas.



Page No 318:

Question 1:

Find out coefficient of correlation between the age of Husband and Wife, using Karl Pearson's method based on actual mean value of the following series.

Age of Husband 20 23 27 31 35 38 40 42
Age of Wife 18 20 24 30 32 34 36 38

Answer:

Age of husband
(X)
Deviation
x=X-X
Square of deviation
x2
Age of wife
(Y)
Deviation
y=Y-Y
Square of deviation
y2
xy
20
23
27
31
35
38
40
42
−12
−9
−5
−1
3
6
8
10
144
81
25
1
9
36
64
100
18
20
24
30
32
34
36
38
−11
−9
−5
1
3
5
7
9
121
81
25
1
9
25
49
81
132
81
25
−1
9
30
56
90
ΣX = 256   x2=460 Y=232   y2=392 xy=422

X=ΣXN=2568=32Y=ΣYN=2328=29r=ΣxyΣx2×Σy2or, r=422460×392or, r=422180320or, r=422424.64=r=+0.994

Thus, the coefficient of correlation between husband's age and wife's age is +0.994.

Page No 318:

Question 2:

Calculate Karl Pearson's coefficient of correlation, between the age and weight of children.

Age (years) 1 2 3 4 5
Weight (kg) 3 4 6 7 10

Answer:

Age
X
Deviation
x=X-X
Square of deviation
x2
Weight
Y
Deviation
y=Y-Y
Square of deviation
y2
xy
1
2
3
4
5
−2
−1
0
1
2
4
1
0
1
4
3
4
6
7
10
−3
−2
0
1
4
9
4
0
1
16
6
2
0
1
8
Σx = 15   Σx2 = 10 ΣY = 30   Σy2 = 30 Σxy= 17

X=ΣXN=155=3Y=ΣYN=305=6r=ΣxyΣx2×Σy2or, r=1710×30or, r=17300or, r=1717.32r=+0.98

Thus, the coefficient of correlation between the age and weight of children is +0.98.

Page No 318:

Question 3:

Calculate coefficient of correlation, using Karl Pearson's formula based on actual mean value of the series given below.

Year Index of Industrial Production Number of Unemployed People in thousand
2010

2011

2012

2013

2014

2015

2016

2017
100

102

104

107

105

112

103

94
11.3

12.4

14.0

11.1

12.3

12.2

19.1

26.4

Answer:

Index of Industrial Production
(X)
Deviation
x=X-X
Square of deviation
(x2)
No. of Unemployed
People
(Y)

 
Deviation
y=Y-Y
Square deviation
(y2)
xy
100
102
104
107
105
112
103
94
−3.375
−1.375
0.625
3.625
1.625
8.625
−0.375
−9.375
11.39
1.89
0.39
13.14
2.64
74.39
0.14
87.89
11.3
12.4
14.0
11.1
12.3
12.2
19.1
26.4
−3.55
−2.45
−0.85
−3.75
−2.55
−2.65
4.25
11.55
12.60
6.00
0.72
14.06
6.50
7.02
18.06
133.40
+11.98
+3.37
−.53
−13.59
−4.14
−22.86
−1.59
−108.28
Σx = 827   Σx2 = 191.87 Σy = 118.8   Σy2 = 198.36 Σxy = −135.64

X=ΣXN=8278=103.375Y=ΣYN=118.88=14.85r=ΣxyΣx2×Σy2or, r=-135.64191.87×198.36or, r=-135.6438059.33or, r=-135.64195.09r=-0.69



Page No 322:

Question 1:

10 students obtained following ranks in their mathematics and statistics examinations. Find out the extent to which the knowledge of students is correlated in the two subjects.

Rank in Statistics 1 2 3 4 5 6 7 8 9 10
Rank in Mathematics 2 4 1 5 3 9 7 10 6 8

Answer:

Rank in statistics
(R1)
Rank in Mathematics
(R2
)
D = R1 − R2 D2
1
2
3
4
5
6
7
8
9
10
2
4
1
5
3
9
7
10
6
8
−1
−2
2
−1
2
−3
0
−2
3
2
1
4
4
1
4
9
0
4
9
4
N = 10     ΣD2= 40

rk=1-6ΣD2N3-N or, rk=1-6×40103-10 or, rk=1-2401000-10 or, rk=1-240990 or, rk =1-.24 rk=+0.76

Thus, there is a high degree of positive correlation between the marks of the students in statistics and mathematics.

Page No 322:

Question 2:

Calculate coefficient of rank correlation, given the following data set.

X 20 11 72 65 43 29 50
Y 60 63 26 35 43 51 37

Answer:

X Rank (R1) Y Rank (R2) D = R1 − R2 D2
20
11
72
65
43
29
50
2
1
7
6
4
3
5
60
63
26
35
43
51
37
6
7
1
2
4
5
3
−4
−6
6
4
0
−2
2
16
36
36
16
0
4
4
N = 7         ΣD2 = 112

rk=1-6ΣD2N3-Nor, rk =1-611273-7or, rk =1-672343-7or, rk =1-672336 or, rk =1-2 rk =-1
Hence, coefficient of rank correlation = −1



Page No 332:

Question 1:

Make a scattered diagram of the data given below. Does any relationship exist between the two?

X 4 5 6 7 8 9 10 11 12 13 14 15
Y 78 72 66 60 54 48 42 36 30 24 18 12

Answer:


Yes, there exists perfect negative correlation (–1) between X and Y.

Page No 332:

Question 2:

Calculate coefficient of correlation of the age of husband and wife using Karl Pearson's method.

Husband (Age) 23 27 28 29 30 31 33 35 36
Wife (Age) 18 20 22 27 29 27 29 28 29

Answer:

Husband
(
h)
h= H- H h2 Wife
(w)
w= W- W w2 hw
23
27
28
29
30
31
33
35
36
–7.22
–3.22
–2.22
–1.22
–.22
.78
2.78
4.78
5.78
52.12
10.36
4.92
1.48
0.04
0.60
7.72
22.84
33.40
18
20
22
27
29
27
29
28
29
–7.44
–5.44
–3.44
1.56
3.56
1.56
3.56
2.56
3.56
55.35
29.59
11.83
2.43
12.67
2.43
12.67
6.55
12.67
53.71
17.51
7.63
–1.90
–.78
1.21
9.89
12.23
20.57
h = 272   h2 = 133.48 w = 229   w2 = 146.19 wh = 120.07

Mean age of husbands H =hn=2729=30.22Mean age of wifes W =wn=2299=25.44r=whh2×w2=120.17133.48×146.19=+0.86
Thus, there exists a high positive correlation between age of wife and age of husband.

Page No 332:

Question 3:

Calculate correlation of the following data using Karl Pearson's method:

Series A 112 114 108 124 145 150 119 125 147 150
Series B 200 190 214 187 170 170 210 190 180 181

Answer:

Series A a= A- A a2 Series B b= B- B b2 ab
112
114
108
124
145
150
119
125
147
150
–17.4
–15.4
–21.4
–5.4
15.6
20.6
–10.4
–4.4
17.6
20.6
302.76
237.16
457.96
29.16
243.36
424.36
108.16
19.36
309.76
424.36
200
190
214
187
170
170
210
190
180
181
10.8
.8
24.8
–2.2
–19.2
–19.2
20.8
.8
–9.2
–8.2
116.64
.64
615.04
4.84
368.64
368.64
432.64
.64
84.64
67.24
–187.92
–12.32
–530.72
11.88
–299.52
–395.52
–216.32
–3.52
–161.92
–168.92
A = 1294   a2 = 2556.4 B = 1892   b2 = 2059.6 ab = –1964.8

Mean of Series A A =An=129410=129.4Mean  of Series B B =Bn=189210=189.2r=aba2×b2=-1964.82556.4×2059.6=-0.85

Page No 332:

Question 4:

Using assumed average in Karl Pearson's formula, calculate coefficient of correlation, given the following data:

X 78 89 97 69 59 79 68 61
Y 125 137 156 112 107 106 123 138

Answer:

X dx= X- A dx2 Y dy= Y- B dy2 dxdy
78
89
97
A=69
59
79
68
61
9
20
28
0
–10
10
–1
–8
81
400
784
0
100
100
1
64
B=125
137
156
112
107
106
123
138
0
12
31
–13
–18
–19
–2
13
0
144
961
169
324
361
4
169
0
240
868
0
180
–190
2
–104
N = 8 dx= 48 dx2 =1530 N = 8 dy = 4 dy2 = 2132 dxdy= 996

Suppose the assumed mean is 69 and 125 for series A and series B, respectively.
r=dxdy-dx×dyNdx2-dx2N×dy2-dy2Nor, r=996-48×481530-(48)282132-(4)28 =97235.24×46.15 r=+0.597

Page No 332:

Question 5:

Find out Karl Pearson's coefficient of correlation:

Capital Units (in '000) 10 20 30 40 50 60 70 80 90 100
Profit Receipt 2 4 8 5 10 15 14 20 22 30

Answer:

X dx= X- A dx2 Y dy= Y- B dy2 dxdy
10
20
30
40
A=50
60
70
80
90
100
–40
–30
–20
–10
0
10
20
30
40
50
1600
900
400
100
0
100
400
900
1600
2500
2
4
8
5
10
B=15
14
20
22
30
–13
–11
–7
–10
–5
0
–1
5
7
15
169
121
49
100
25
0
1
25
49
225
520
330
140
100
0
0
–20
150
280
750
N = 10 dx= 50 dx2 = 8500 N = 10 dy = –20 dy2 = 764 dxdy= 2250

r=dxdy-dx×dyNdx2-dx2N×dy2-dy2Nor, r=2250-50×-20108500-(50)28×764-(-20)210 =235090.82×26.90 r=+0.961

Page No 332:

Question 6:

Seven students of a class secured following marks in Economics and History. Calculate coefficient of correlation with the help of these data.

Economics 66 90 89 55 58 44 42
History 58 76 65 58 53 49 56

Answer:

Economics
(E)
R1 History
(H)
R2 D = R1R2 D2
66
90
89
55
58
44
42
3
1
2
5
4
6
7
58
76
65
58
53
49
56
3.5
1
2
3.5
6
7
5
–.5
0
0
1.5
–2
–1
2
.25
0
0
2.25
4
1
4
N = 7         D2=11.50

Here, note that for the marks scored in History, two ranks are tied. That is, two students scored 58 marks. Thus, we use the following formula for the calculation of correlation. 
rk=1-6D2+112M13-M1N3-NHere, M1= 2, as two students have scored 58 in History.or, rk=1-611.50+112(8-2)343-7, rk=336-72336=+0.79
Thus, there exists a positive correlation between marks scored in Economics and marks scored in History.

Page No 332:

Question 7:

Find out rank difference correlation of X and Y:

X 80 78 75 75 58 67 60 59
Y 12 13 14 14 14 16 15 17

Answer:

X R1 Y R2 D = R1R2 D2
80
78
75
75
58
67
60
59
1
2
3.5
3.5
8
5
6
7
12
13
14
14
14
16
15
17
8
7
5
5
5
2
3
1
–7
–5
–1.5
–1.5
3
3
3
6
49
25
2.25
2.25
9
9
9
36
N = 8         D2=141.5

rk=1-6D2+112M13-M1+112M23-M2N3-NHere,M1= 2, as the item 75 is appearing twice in X-series.M2= 3, as the item 14 is appearing thrice in Y-series.or, rk=1-6141.50+112(23-2)+112(33-3)512-8, rk=504-864504=-0.714

Page No 332:

Question 8:

Calculate coefficient of correlation of the following data with rank difference and Karl Pearson's methods:

Economics (Marks) 77 54 27 52 14 35 90 25 56 60
Hindi (Marks) 35 58 60 46 50 40 35 56 44 42

Answer:

Karl Pearson's Method

Economics
(X)
dx= X- A dx2 History
(Y)
dy= Y- B dy2 dxdy
77
54
27
52
14
A=35
90
25
56
60
42
19
–8
17
–21
0
55
–10
21
25
1764
361
64
289
441
0
3025
100
441
625
35
58
60
46
B=50
40
35
56
44
42
–15
8
10
–4
0
–10
–15
6
–6
–8
225
64
100
16
0
100
225
36
36
64
–630
152
–80
–68
0
0
–825
–60
–126
–200
N = 10 dx= 140 dx2 = 7110 N = 10 dy = –34 dy2 = 860 dxdy=–1837

r=dxdy-dx×dyNdx2-dx2N×dy2-dy2Nor, r=-1831-140×(-34)107110-(140)210860-(-34)210 =-135571.76×27.39 r=-0.689

Rank Difference Method
Economics R1 History R2 D = R1R2 D2
77
54
27
52
14
35
90
25
56
60
2
5
8
6
10
7
1
9
4
3
35
58
60
46
50
40
35
56
44
42
9.5
2
1
5
4
8
9.5
3
6
7
–7.5
3
7
1
6
–1
–8.5
6
–2
–4
56.25
9
49
1
36
1
72.25
36
4
16
N =10         D2=280.5

rk=1-6D2+112M13-M1N3-NHere,M1= 2, as two students have scored 35 marks in History.or, rk=1-6280.50+112(23-2)1000-10, rk=990-1686990=-0.703



Page No 333:

Question 9:

Seven methods of teaching Economics in two universities are shown below. Calculate rank difference correlation.

Teaching Methods I II III IV V VI VII
Rank of 'A's Students 2 1 5 3 4 7 6
Rank of 'B's Students 1 3 2 4 7 5 6

Answer:

Teaching Methods RA RB D = RA RB D2
I
II
III
IV
V
VI
VII
2
1
5
3
4
7
6
1
3
2
4
7
5
6
1
–2
3
–1
–3
2
0
1
4
9
1
9
4
0
        D2=28
N = 7
rk= 1-6D2N3 -Nor, rk=1-6×28343-7=336-168336=0.5Hence, rk=0.5

Page No 333:

Question 10:

Give three examples of perfect correlation. Find out rank difference coefficient of correlation with the help of the following data:

X 48 33 40 9 16 65 26 15 57
Y 13 13 22 6 14 20 9 6 15

Answer:

Three examples of perfect correlation are:
1. T.V. viewing and Study hours (-ve correlation). That is, as the hours spent in T.V. viewing increases, the numbers of hours that can be devoted to study decreases and vice-versa. 
2. Income used for consumption and amount of saving (-ve correlation). That is, greater the portion of income used for consumption purposes, smaller is the portion of income left for saving purposes and vice-versa.  
3. Amount deposited in bank and interest earned (+ve correlation). That is, as the amount deposited in the bank increases, the amount of interest that is earned increases and vice-versa. 
 

X R1 Y R2 D = R1R2 D2
48
33
40
9
16
65
26
15
57
3
5
4
9
7
1
6
8
2
13
13
22
6
14
20
9
6
15
5.5
5.5
1
8.5
4
2
7
8.5
3
–2.5
–.5
3
.5
3
–1
–1
–.5
.1
6.25
.25
9
.25
9
1
1
.25
1
N = 9         D2= 28

rk=1-6D2+112M13-M1+112M23-M2N3-NHere,M1= 2, as the item 13 is appearing twice in Y-series.M2= 2, as the item 6 is appearing twice in Y-series.or, rk=1-628+112(23-2)+112(23-2)729-9, rk=720-147720=0.758

Page No 333:

Question 11:

Calculate coefficient of correlation of the following data:

X 10 6 9 10 12 13 11 9
Y 9 4 6 9 11 13 8 4

Answer:

X   dx= X- A dx2 Y dy= Y- B dy2 dxdy
10
6
9
A=10
12
13
11
9
0
–4
–1
0
2
3
1
–1
0
16
1
0
9
4
1
1
9
4
6
9
B=11
13
8
4
–2
–7
–5
–2
0
2
–3
–7
4
49
25
4
0
4
9
49
0
28
5
0
0
6
–3
7
N = 8 dx= 0 dx2 = 32 N = 8 dy= –24 dy2 = 144 dxdy= 43

r=dxdy-dx×dyNdx2-dx2N×dy2-dy2Nor, r=43-0×-24832-(0)28×144-(-24)28 =4332×72=435.65×8.48 r=+0.896

Page No 333:

Question 12:

Deviation of two series of X and Y are shown. Calculate coefficient of correlation.

X +5 −4 −2 +20 −10 0 +3 0 −15 −5
Y +5 −12 −7 +25 −10 −3 0 +2 −9 −15

Answer:

dx  dx2 dy dy2 dxdy
5
–4
–2
20
–10
0
3
0
–15
–5
25
16
4
400
100
0
9
0
225
25
5
–12
–7
25
–10
3
0
2
–9
–15
25
144
49
625
100
9
0
4
81
225
25
48
14
500
100
0
0
0
135
75
dx= –8 dx2 = 804 dy= –24 dy2 = 1262 dxdy= 897

r=dxdy-dx×dyNdx2-dx2N×dy2-dy2Nor, r=897--8×-2410804-(-8)210×1262-(-24)210 =877.828.24×34.70 r=+0.895

Page No 333:

Question 13:

In a baby competition, two judges accorded following to 12 competitors. Find the coefficient of rank correlation.

Entry A B C D E F G H I J K L
jJudge X 1 2 3 4 5 6 7 8 9 10 11 12
Judge Y 12 9 6 10 3 5 4 7 8 2 11 1

Answer:

Entry Ranks by Judge X
(
RX)
Ranks by Judge Y
(
RY)
D = RXRY D2
A
B
C
D
E
F
G
H
I
J
K
L
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       D2=416

rk= 1-6D2N3 -Nor, rk=1-6×4161728-12=1-24961716=-0.455Hence, rk=-0.455

Page No 333:

Question 14:

In a Fancy-dress competition, two judges accorded the following ranks to eight participants:

Judge X 8 7 6 3 2 1 5 4
Judge Y 7 5 4 1 3 2 6 8
Calculate coefficient of rank correlation.

Answer:

RX RY D = RX RY D2
8
7
6
3
2
1
5
4
7
5
4
1
3
2
6
8
1
2
2
2
–1
–1
–1
–4
1
4
4
4
1
1
1
16
      D2=32

N = 8
rk= 1-6D2N3 -Nor, rk=1-6×32512-8=504-192504=0.619Hence, rk=0.619

Page No 333:

Question 15:

In a beauty contest, three judges accorded following ranks to 10 participants:

Judge I 1 6 5 10 3 2 4 9 7 8
Judge II 3 5 8 4 7 10 2 1 6 9
Judge III 6 4 9 8 1 2 3 10 5 7
Find out by Spearman's Rank Difference Method which pair of judges has a common taste in respect of beauty.

Answer:

R1 R2 R3 D1 = R1R2 D2 = R1R3 D3 = R2R3 D12 D22 D32
1
6
5
10
3
2
4
9
7
8
3
5
8
4
7
10
2
1
6
9
6
4
9
8
1
2
3
10
5
7
–2
1
–3
6
–4
–8
2
8
1
–1
– 5
2
–4
2
2
0
1
–1
2
1
–3
1
–1
–4
6
8
–1
–9
1
8
4
1
9
36
16
64
4
64
1
1
25
4
16
4
4
0
1
1
4
1
9
1
1
16
36
64
1
81
1
64
            D12=200 D22=60 D32=214

N = 10

Rank Correlation between Judge 1 and Judge 2rk1,2=1-6D12N3-N=1-6×2001000-10=990-1200990=-0.212Rank Correlation between Judge 1 and Judge 3rk1,3=1-6D22N3-N=1-6×601000-10=990-360990=+0.636Rank Correlation between Judge 2 and Judge 3rk2,3=1-6D32N3-N=1-6×2141000-10=990-1284990=-0.296


Observation and Conclusion:
As the rank correlation coefficient between Judge 1 and Judge 3 is highest and positive, so it can be regarded that they have a common taste in respect of beauty.



Page No 334:

Question 16:

Following data relates to age group and percentage of regular players. Calculate Karl Pearson's coefficient of correlation.

Age Group 20−25 25−30 30−35 35−40 40−45 45−50
% of Regular Players 40 35 28 20 15 5

Answer:

Age Group Mid Value
(X)
% of Regular Players
(Y)
d'X=X-Ah=X-37.55 dY'=Y-Bi=Y-285 (dX') (dY') (dX')2 (dY')2
20-25
25-30
30-35
35-40
40-45
45-50
22.5
27.5
32.5
37.5
42.5
47.5
40
35
28
20
15
5
–3
–2
–1
0
1
2
2.4
1.4
0
–1.6
–2.6
–4.6
–7.2
–2.8
0
0
–2.6
–9.2
9
4
1
0
1
4
5.76
1.96
0
2.56
6.76
21.16
      dX' = –3 dY' = –5 ∑(dX') (dY') = – 21.8 ∑(dX')2 = 19 ∑(dX')2 = 38.2

r=dX'dY'-dX'×dY'NdX'2 -dX'2N×dY'2 -dY'2Nor, r=-21.8--3×-5619--326×38.2--526=-24.319-96×38.2-256or, r=--24.317.5 × 34.03=-24.3595.525=-24.324.403Hence, r=-0.996

Page No 334:

Question 17:

From the following data, relating to playing habits in various age group of 900 students. Calculate coefficient of correlation between age group and playing habits.

Age Group 15−16 16−17 17−18 18−19 19−20 20−21
Number of Students 250 200 150 120 100 80
Regular Players 200 150 90 48 30 12

Answer:

Age Group Number of People Number of Players Percentage of Players (%)
15-16 250 200 200250×100=80%
16-17 200 150 150200×100=75%
17-18 150 90 90150×100=60%
18-19 120 48 48120×100=40%
19-20 100 30 30100×100=30%
20-21 80 12 1280×100=15%
 
Age Group Mid Value
(X)
Percentage of Players (%)
(Y)
dX=X-A=X-17.5 dY=Y-B=Y-40 dXdY dX2 dY2
15-16
16-17
17-18
18-19
19-20
20-21
15.5
16.5
A=17.5
18.5
19.5
20.5
80
75
60
 B=40
30
15
–2
–1
0
1
2
3
40
35
20
0
–10
–25
–80
–35
0
0
–20
–75
4
1
0
1
4
9
1600
1225
400
0
100
625
  N = 6 N = 6 dX= 3 dY = 60 dXdY== –210 dX2 = 19 dY2=∑3950

r=dXdY-dX×dYNdX2-dX2N×dY2-dY2Nor, r=-210-3×60619-(3)26×3950-(60)26 =-24017.5×3350or, r=-2404.18×57.88=-240241.94=-0.992 r=-0.992
Hence, the coefficient of correlation between age group and playing habits is – 0.992

Page No 334:

Question 18:

Following data relates to density of population, number of deaths and population of various cities. Calculate death rate and Karl Pearson coefficient between density of population and death rate.

Cities P Q R S T U
Density of Population 200 500 700 500 600 900
Number of Deaths 840 300 312 560 1,140 1,224
Population 42,000 30,000 24,000 40,000 90,000 72,000

Answer:

Cities Density Number of Deaths Population Death rate = Number of DeathsPopulation×100
P
Q
R
S
T
U
200
500
700
500
600
900
840
300
312
560
1440
1224
42000
30000
24000
40000
90000
72000
2
1
1.3
1.4
1.6
1.7
 
Density
(X)
dx = XA
A = 500
dx2 Death Rate
​(Y)
dy = YB
B = 1
dy2 dxdy
200
500
700
500
600
900
–300
0
200
0
100
400
90000
0
40000
0
10000
160000
2
1
1.3
1.4
1.6
1.7
1
0
0.3
0.4
0.6
0.7
1
0
0.09
0.16
0.36
0.49
–300
0
60
0
60
280
  dx = 400 dx2 = 300000   dy = 3 dy2 = 2.1 dxdy = 100

r=dxdy-dxdyndx2-dx2ndy2-dy2n=100-400×36300000-400262.1-326or, r=100-200273333.34 ×0.6=-100522.81×0.77=-100402.56=-0.248Hence, Karl Pearson's Coefficient of Correlation between density of population and death rate is -0.248 

Page No 334:

Question 19:

From the following data, determine Karl Pearson's coefficient of correlation between X and Y series for 15 paris:

  X-Series Y-Series
Mean 80 120
Sum of Squares of deviation from Arithmetic Mean 56 156
Sum of product of deviation of X and Y from their respective Means 92

Answer:

Given:N =15X =80Y = 120Σx2=56Σy2=156Σxy=92σx=Σx2N,σy=Σy2Nσx=5615=1.93, σy=15615=3.22r=ΣxyNσxσy=9215×1.93×3.22 =0.98
Hence, Karl Pearson's coefficient of correlation is +0.98

Page No 334:

Question 20:

From the following information, determine coefficient of correlation between X and Y series:

  X-Series Y-Series
Number of Items 15 15
Mean 25 18
SD 3.01 3.03
Sum of Squares of deviation from Mean 136 138
Sum of product of deviation of X and Y from their respective Means 122

Answer:

Given:N =15X =25Y = 18σx=3.01σy=3.03Σx2 =136Σy2 =138Σxy=122r=ΣxyNσxσy=12215×3.01×3.03 =0.89
Hence, coefficient of correlation between X and Y series is + 0.89



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