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# T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 12 - Correlation

Exercise/Page

## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 12 - Correlation Page/Excercise 332

Solution SAQ 1

X and Y series show a perfect negative relationship between each other.

Solution SAQ 2

Solution SAQ 3

Solution SAQ 4

Solution SAQ 5

Solution SAQ 6

 Economics (E) Rank R1 History (H) Rank R2 D=  R1 - R2 D2 66 3 58 3.5 -5 0.25 90 1 76 1 0 0 89 2 65 2 0 0 55 5 58 3.5 1.5 2.25 58 4 53 6 -2 4 44 6 49 7 -1 1 42 7 56 5 2 4 N = 7 𝛴 D2 = 11.50

58 is repeated two times in series 2. Thus, m1= 2 and following formula is used to calculate correlation.

Solution SAQ 7

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

75 is repeated two times in series 1 and 14 is repeated three times in series 2. Thus, m1= 2 and m2= 3 and following formula is used to calculate correlation.

Solution SAQ 8

Karl Pearson's Method:

 Economics (X) dx = X - 35 dx2 History (Y) dy = Y - 50 dy2 dxdy 77 42 1764 35 -15 225 -630 54 19 361 58 8 64 152 27 -8 64 60 10 100 -80 52 17 289 46 -4 16 -68 14 -21 441 50 0 0 0 35 0 0 40 -10 100 0 90 55 3025 35 -15 225 -825 25 -10 100 56 6 36 -60 56 21 441 44 -6 36 -126 60 25 625 42 -8 64 -200 N = 10 𝛴 dx = 140 𝛴dx2 = 7110 N = 10 𝛴dy = -34 𝛴dy2 = 860 𝛴dxdy=-1837

Rank Difference Method:

 Economics R1 History R2 D = R1 - R2 D2 77 2 35 9.5 -7.5 56.25 54 5 58 2 3 9 27 8 60 1 7 49 52 6 46 5 1 1 14 10 50 4 6 36 35 7 40 8 -1 1 90 1 35 9.5 -8.5 72.25 25 9 56 3 6 36 56 4 44 6 -2 4 60 3 42 7 -4 16 N = 10 𝛴D2 = 280.5

35 is repeated two times in series 2. Thus, m1= 2 and following formula is used to calculate correlation.

## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 12 - Correlation Page/Excercise 333

Solution SAQ 9

 Teaching Method Rank of A RA Rank of B RB D = RA - RB D2 I 2 1 1 1 II 1 3 -2 4 III 5 2 3 9 IV 3 4 -1 1 V 4 7 -3 9 VI 7 5 2 4 VII 6 6 0 0 N=7 𝛴D2 = 28

Solution SAQ 10

Examples of perfect correlation:

1. Relationship between study hours and marks
2. Relationship between consumption and saving from fixed income
3. Relationship between amount of loan taken from bank and interest paid
 X R1 Y R2 D=R1- R2 D2 48 3 13 5.5 -2.5 6.25 33 5 13 5.5 -0.5 0.25 40 4 22 1 3 9 9 9 6 8.5 0.5 0.25 16 7 14 4 3 9 65 1 20 2 -1 1 26 6 9 7 -1 1 15 8 6 8.5 -0.5 0.25 57 2 15 3 0.1 1 N = 9 𝛴 D2 = 28

13 and 6 are repeated two times in series 2. Thus, m1= 2 and m2= 2 and following formula is used to calculate correlation.

Solution SAQ 11

 X dx = X - 10 dx2 Y dy = Y - 11 dy2 dxdy 10 0 0 9 -2 4 0 6 -4 16 4 -7 49 28 9 -1 1 6 -5 25 5 10 0 0 9 -2 4 0 12 2 9 11 0 0 0 13 3 4 13 2 4 6 11 1 1 8 -3 9 -3 9 -1 1 4 -7 49 7 N = 8 𝛴dx = 0 𝛴dx2 = 32 N = 8 𝛴dy = -24 𝛴dy2 = 144 𝛴dxdy = 43

Solution SAQ 12

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

Solution SAQ 13

 Entry Judge X (RX) Judge Y (RY) D = RX - RY D2 A 1 12 -11 121 B 2 9 -7 49 C 3 6 -3 9 D 4 10 -6 36 E 5 3 2 4 F 6 5 1 1 G 7 4 3 9 H 8 7 1 1 I 9 8 1 1 J 10 2 8 64 K 11 11 0 0 L 12 1 11 121 N = 12 𝛴 D2 = 416

Solution SAQ 14

 RX RY D = RX - RY D2 8 7 1 1 7 5 2 4 6 4 2 4 3 1 2 4 2 3 -1 1 1 2 -1 1 5 6 -1 1 4 8 -4 16 𝛴D2 = 32

Solution SAQ 15

 R1 R2 R3 D1 = R1- R2 D2 = R1 - R3 D3 = R2 - R3 D12 D22 D32 1 3 6 -2 -5 -3 4 25 9 6 5 4 1 2 1 1 4 1 5 8 9 -3 -4 -1 9 16 1 10 4 8 6 2 -4 36 4 16 3 7 1 -4 2 6 16 4 36 2 10 2 -8 0 8 64 0 64 4 2 3 2 1 -1 4 1 1 9 1 10 8 -1 -9 64 1 81 7 6 5 1 2 1 1 4 1 8 9 7 -1 1 8 1 1 64 𝛴D12= 200 𝛴D22 = 60 𝛴D32 = 214

Judges I and III have a common taste in respect of beauty as they have the highest positive rank correlation coefficient.

## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 12 - Correlation Page/Excercise 334

Solution SAQ 16

Solution SAQ 17

 Age Group Mid Value (X) Percentage of Players (%) (Y) dX = X-17.5 dY = Y-40 dXdY dX2 dY2 15-16 15.5 200/250×100 = 80 -2 40 -80 4 1600 16-17 16.5 150/2-00×100 = 75 -1 35 -35 1 1225 17-18 17.5 90/150×100 = 60 0 20 0 0 400 18-19 18.5 48/120×100 = 40 1 0 0 1 0 19-20 19.5 30/100×100 = 30 2 -10 -20 4 100 20-21 20.5 12/80×100 = 15 3 -25 -75 9 625 N = 6 𝛴dX = 3 𝛴dY = 60 𝛴 dXdY = -210 𝛴dX2 = 19 𝛴 dY2 = 𝛴3950

Solution SAQ 18

 Density (X) dx = X - 500 dx2 dy = Y-1 dy2 dxdy 200 -300 90000 2 1 1 -300 500 0 0 1 0 0 0 700 200 40000 1.3 0.3 0.09 60 500 0 0 1.4 0.4 0.16 0 600 100 10000 1.6 0.6 0.36 60 900 400 160000 1.7 0.7 0.49 280 𝛴dx=400 𝛴dx2=300000 𝛴dy=3 𝛴dy2 = 2.1 𝛴dxdy = 100

Solution SAQ 19

Solution SAQ 20

# Text Book Solutions

CBSE XI Commerce - Statistics for Economics

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