Tr Jain Vk Ohri 2019 Solutions for Class 11 Humanities Economics Chapter 12 Correlation are provided here with simple step-by-step explanations. These solutions for Correlation are extremely popular among Class 11 Humanities 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 2019 Book of Class 11 Humanities 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 2019 Solutions. All Tr Jain Vk Ohri 2019 Solutions for class Class 11 Humanities Economics are prepared by experts and are 100% accurate.

#### 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

 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.

#### 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

 X Y 8 16 24 31 42 50 70 58 50 32 26 12 Thus, there exists a negative relationship between X and Y.

#### 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.

 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.

#### 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

 Age of husband (X) Deviation $\mathbit{x}\mathbit{=}\mathbit{X}\mathbit{-}\overline{)\mathit{X}}$ Square of deviation x2 Age of wife (Y) Deviation $\mathbit{y}\mathbit{=}\mathbit{Y}\mathbit{-}\overline{)\mathit{Y}}$ 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 $\sum _{}^{}{x}^{2}=460$ $\sum _{}^{}Y=232$ $\sum _{}{y}^{2}=392$ ${\sum }_{}^{}xy=422$

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

#### 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

 Age X Deviation $\mathbit{x}\mathbit{=}\mathbit{X}\mathbit{-}\overline{)\mathit{X}}$ Square of deviation x2 Weight Y Deviation $\mathbit{y}\mathbit{=}\mathbit{Y}\mathbit{-}\overline{)\mathit{Y}}$ 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

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

#### 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

 Index of Industrial Production (X) Deviation $\mathbit{x}\mathbit{=}\mathbit{X}\mathbit{-}\overline{)\mathit{X}}$ Square of deviation (x2) No. of Unemployed People (Y) Deviation $\mathbit{y}\mathbf{=}\mathbit{Y}\mathbf{-}\overline{)\mathbit{Y}}$ 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

#### 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

 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

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

#### 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

 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

Hence, coefficient of rank correlation = −1

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

#### 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

 Husband (h) h2 Wife (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

Thus, there exists a high positive correlation between age of wife and age of husband.

#### 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

 Series A a2 Series 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

#### 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

 X dx2 Y dy2 dxdy 78 89 97 $\overline{)\mathrm{A}=69}$ 59 79 68 61 9 20 28 0 –10 10 –1 –8 81 400 784 0 100 100 1 64 $\overline{)\mathrm{B}=125}$ 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.

#### 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

 X dx2 Y dy2 dxdy 10 20 30 40 $\overline{)\mathrm{A}=50}$ 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 $\overline{)\mathrm{B}=15}$ 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

#### 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

 Economics (E) R1 History (H) R2 D = R1 – R2 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 $\sum {D}^{2}=11.50$

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.

Thus, there exists a positive correlation between marks scored in Economics and marks scored in History.

#### 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

 X R1 Y R2 D = R1 – R2 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 ${\sum }_{}{D}^{2}=141.5$

#### 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

Karl Pearson's Method

 Economics (X) dx2 History (Y) dy2 dxdy 77 54 27 52 14 $\overline{)\mathrm{A}=35}$ 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 $\overline{)\mathrm{B}=50}$ 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

Rank Difference Method
 Economics R1 History R2 D = R1 – R2 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 ${\sum }_{}{D}^{2}=280.5$

#### 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

 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 ${\sum }_{}{D}^{2}=28$
N = 7

#### 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

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 = R1 – R2 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

#### 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

 X dx2 Y dy2 dxdy 10 6 9 $\overline{)\mathrm{A}=10}$ 12 13 11 9 0 –4 –1 0 2 3 1 –1 0 16 1 0 9 4 1 1 9 4 6 9 $\overline{)\mathrm{B}=11}$ 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

#### 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

 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

#### 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

 Entry Ranks by Judge X (RX) Ranks by Judge Y (RY) D = RX – RY 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 ${\sum }_{}{D}^{2}=416$

#### 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.

 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 ${\sum }_{}{D}^{2}=32$

N = 8

#### 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.

 R1 R2 R3 D1 = R1 – R2 D2 = R1 – R3 D3 = R2 – R3 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 ${\sum }_{}{{D}_{1}}^{2}=200$ ${\sum }_{}{{D}_{2}}^{2}=60$ ${\sum }_{}{{D}_{3}}^{2}=214$

N = 10

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.

#### 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

 Age Group Mid Value (X) % of Regular Players (Y) ${{\mathbit{d}}^{\mathbf{\text{'}}}}_{\mathbf{X}}\mathbf{=}\frac{X\mathit{-}A}{h}\mathbf{=}\frac{\mathbit{X}\mathbf{-}\mathbf{37}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ ${\mathbit{d}}_{\mathbit{Y}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathit{Y}\mathit{-}\mathit{B}}{\mathit{i}}\mathbf{=}\frac{\mathbit{Y}\mathbf{-}\mathbf{28}}{\mathbf{5}}$ (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

#### 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

 Age Group Number of People Number of Players Percentage of Players (%) 15-16 250 200 $\frac{200}{250}×100=80%$ 16-17 200 150 $\frac{150}{200}×100=75%$ 17-18 150 90 $\frac{90}{150}×100=60%$ 18-19 120 48 $\frac{48}{120}×100=40%$ 19-20 100 30 $\frac{30}{100}×100=30%$ 20-21 80 12 $\frac{12}{80}×100=15%$

 Age Group Mid Value (X) Percentage of Players (%) (Y) ${\mathbit{d}}_{\mathbit{X}}\mathbf{=}\mathbit{X}\mathbit{-}\mathbit{A}\mathbf{=}\mathbit{X}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{5}$ ${\mathbit{d}}_{\mathbit{Y}}\mathbf{=}\mathbit{Y}\mathbit{-}\mathbit{B}\mathbf{=}\mathbit{Y}\mathbf{-}\mathbf{40}$ dXdY dX2 dY2 15-16 16-17 17-18 18-19 19-20 20-21 15.5 16.5 $\overline{)A=17.5}$ 18.5 19.5 20.5 80 75 60  $\overline{)B=40}$ 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

Hence, the coefficient of correlation between age group and playing habits is – 0.992

#### 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

 Cities Density Number of Deaths Population Death rate = 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 = X – A A = 500 dx2 Death Rate ​(Y) dy = Y – B 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

#### Question 19:

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

Hence, coefficient of correlation between X and Y series is + 0.89

#### Question 20:

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