Tr Jain Vk Ohri for Class 11 Commerce Economics Chapter 12

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.

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.

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.

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		$\sum_{}^{} x^{2} = 460$	$\sum_{}^{} Y = 232$		$\sum_{} y^{2} = 392$	$\sum_{}^{} x y = 422$

X = \frac{ΣX}{N} = \frac{256}{8} = 32 Y = \frac{ΣY}{N} = \frac{232}{8} = 29 r = \frac{Σ x y}{\sqrt{Σ x^{2} \times Σ y^{2}}} or, r = \frac{422}{\sqrt{460 \times 392}} or, r = \frac{422}{\sqrt{180320}} or, r = \frac{422}{424.64} = \Rightarrow r = + 0.994

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

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 = \frac{ΣX}{N} = \frac{15}{5} = 3 Y = \frac{ΣY}{N} = \frac{30}{5} = 6 r = \frac{Σ x y}{\sqrt{Σ x^{2} \times Σ y^{2}}} or, r = \frac{17}{\sqrt{10 \times 30}} or, r = \frac{17}{\sqrt{300}} or, r = \frac{17}{17.32} \Rightarrow r = + 0.98

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
2014 2015 2016 2017 2018 2019 2020 2021	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 = \frac{ΣX}{N} = \frac{827}{8} = 103.375 Y = \frac{ΣY}{N} = \frac{118.8}{8} = 14.85 r = \frac{Σ x y}{\sqrt{Σ x^{2} \times Σ y^{2}}} or, r = \frac{- 135.64}{\sqrt{191.87 \times 198.36}} or, r = \frac{- 135.64}{\sqrt{38059.33}} or, r = \frac{- 135.64}{195.09} \Rightarrow r = - 0.69

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

r_{k} = 1 - \frac{6 {ΣD}^{2}}{N^{3} - N} or, r_{k} = 1 - \frac{6 \times 40}{{(10)}^{3} - 10} or, r_{k} = 1 - \frac{240}{1000 - 10} or, r_{k} = 1 - \frac{240}{990} or, r_{k} = 1 - . 24 \Rightarrow r_{k} = + 0.76

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

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

r_{k} = 1 - \frac{6 {ΣD}^{2}}{N^{3} - N} or, r_{k} = 1 - \frac{6 (112)}{7^{3} - 7} or, r_{k} = 1 - \frac{672}{343 - 7} or, r_{k} = 1 - \frac{672}{336} or, r_{k} = 1 - 2 \Rightarrow r_{k} = - 1

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

Answer:

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

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) = \frac{\sum_{}^{} h}{n} = \frac{272}{9} = 30.22 Mean age of wifes (W) = \frac{\sum w}{n} = \frac{229}{9} = 25.44 r = \frac{\sum w h}{\sqrt{\sum {h^{2}}_{} \times \sum w^{2}}} = \frac{120.17}{\sqrt{133.48 \times 146.19}} = + 0.86

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

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) = \frac{\sum_{}^{} A}{n} = \frac{1294}{10} = 129.4 Mean of Series B (B) = \frac{\sum B}{n} = \frac{1892}{10} = 189.2 r = \frac{\sum a b}{\sqrt{\sum {a^{2}}_{} \times \sum b^{2}}} = \frac{- 1964.8}{\sqrt{2556.4 \times 2059.6}} = - 0.85

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	$d_{x} = X - A$	d_x²	Y	$d_{y} = Y - B$	d_y²	*d_xd_y*
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	∑d_x= 48	∑d_x² =1530	N = 8	∑d_y= 4	∑d_y² = 2132	∑d_xd_y= 996

Suppose the assumed mean is 69 and 125 for series A and series B, respectively.

r = \frac{\sum_{} d_{x} d_{y} - \frac{(\sum_{}^{} d_{x}) \times (\sum_{}^{} d_{y})}{N}}{\sqrt{\sum_{}^{} d_{x}^{2} - \frac{{(\sum_{}^{} d_{x})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{y}^{2} - \frac{{(\sum_{}^{} d_{y})}^{2}}{N}}} or, r = \frac{996 - \frac{48 \times 4}{8}}{\sqrt{1530 - \frac{(48)^{2}}{8}} \sqrt{2132 - \frac{(4)^{2}}{8}}} = \frac{972}{35.24 \times 46.15} ∴ r = + 0.597

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	$d_{x} = X - A$	d_x²	Y	$d_{y} = Y - B$	d_y²	d_xd_y
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	∑d_x= 50	∑d_x² = 8500	N = 10	∑d_y= –20	∑d_y² = 764	∑d_xd_y= 2250

r = \frac{\sum_{} d_{x} d_{y} - \frac{(\sum_{}^{} d_{x}) \times (\sum_{}^{} d_{y})}{N}}{\sqrt{\sum_{}^{} d_{x}^{2} - \frac{{(\sum_{}^{} d_{x})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{y}^{2} - \frac{{(\sum_{}^{} d_{y})}^{2}}{N}}} or, r = \frac{2250 - \frac{50 \times (- 20)}{10}}{\sqrt{8500 - \frac{(50)^{2}}{8}} \times \sqrt{764 - \frac{(- 20)^{2}}{10}}} = \frac{2350}{90.82 \times 26.90} ∴ r = + 0.961

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 = 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.

r_{k} = 1 - \frac{6 [\sum D^{2} + \frac{1}{12} (M_{1}^{3} - M_{1})]}{N^{3} - N} Here, M_{1} = 2, as two students have scored 58 in History . or, r_{k} = 1 - \frac{6 [11.50 + \frac{1}{12} (8 - 2)]}{343 - 7} ∴, r_{k} = \frac{336 - 72}{336} = + 0.79

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

Answer:

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					$\sum_{} D^{2} = 141.5$

r_{k} = 1 - \frac{6 [\sum D^{2} + \frac{1}{12} (M_{1}^{3} - M_{1}) + \frac{1}{12} (M_{2}^{3} - M_{2})]}{N^{3} - N} Here, M_{1} = 2, as the item 75 is appearing twice in X - series . M_{2} = 3, as the item 14 is appearing thrice in Y - series . or, r_{k} = 1 - \frac{6 [141.50 + \frac{1}{12} (2^{3} - 2) + \frac{1}{12} (3^{3} - 3)]}{512 - 8} ∴, r_{k} = \frac{504 - 864}{504} = - 0.714

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)	$d_{x} = X - A$	d_x²	History (Y)	$d_{y} = Y - B$	d_y²	d_xd_y
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	∑d_x= 140	∑d_x² = 7110	N = 10	∑d_y= –34	∑d_y² = 860	∑d_xd_y=–1837

r = \frac{\sum_{} d_{x} d_{y} - \frac{(\sum_{}^{} d_{x}) \times (\sum_{}^{} d_{y})}{N}}{\sqrt{\sum_{}^{} d_{x}^{2} - \frac{{(\sum_{}^{} d_{x})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{y}^{2} - \frac{{(\sum_{}^{} d_{y})}^{2}}{N}}} or, r = \frac{- 1831 - \frac{140 \times (- 34)}{10}}{\sqrt{7110 - \frac{(140)^{2}}{10}} \sqrt{860 - \frac{(- 34)^{2}}{10}}} = \frac{- 1355}{71.76 \times 27.39} ∴ r = - 0.689

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					$\sum_{} D^{2} = 280.5$

r_{k} = 1 - \frac{6 [\sum D^{2} + \frac{1}{12} (M_{1}^{3} - M_{1})]}{N^{3} - N} Here, M_{1} = 2, as two students have scored 35 marks in History . or, r_{k} = 1 - \frac{6 [280.50 + \frac{1}{12} (2^{3} - 2)]}{1000 - 10} ∴, r_{k} = \frac{990 - 1686}{990} = - 0.703

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	*R_A*	*R_B*	D = R_A – *R_B*	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

r_{k} = 1 - \frac{6 \sum_{} D^{2}}{N^{3} - N} or, r_{k} = 1 - \frac{6 \times 28}{343 - 7} = \frac{336 - 168}{336} = 0.5 Hence, r_{k} = 0.5

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 = 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					$\sum_{} D^{2} = 28$

r_{k} = 1 - \frac{6 [\sum D^{2} + \frac{1}{12} (M_{1}^{3} - M_{1}) + \frac{1}{12} (M_{2}^{3} - M_{2})]}{N^{3} - N} Here, M_{1} = 2, as the item 13 is appearing twice in Y - series . M_{2} = 2, as the item 6 is appearing twice in Y - series . or, r_{k} = 1 - \frac{6 [28 + \frac{1}{12} (2^{3} - 2) + \frac{1}{12} (2^{3} - 2)]}{729 - 9} ∴, r_{k} = \frac{720 - 147}{720} = 0.758

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	$d_{x} = X - A$	d_x²	Y	$d_{y} = Y - B$	d_y²	d_xd_y
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	∑d_x= 0	∑d_x² = 32	N = 8	∑d_y= –24	∑d_y² = 144	∑d_xd_y= 43

r = \frac{\sum_{} d_{x} d_{y} - \frac{(\sum_{}^{} d_{x}) \times (\sum_{}^{} d_{y})}{N}}{\sqrt{\sum_{}^{} d_{x}^{2} - \frac{{(\sum_{}^{} d_{x})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{y}^{2} - \frac{{(\sum_{}^{} d_{y})}^{2}}{N}}} or, r = \frac{43 - \frac{0 \times (- 24)}{8}}{\sqrt{32 - \frac{(0)^{2}}{8}} \times \sqrt{144 - \frac{(- 24)^{2}}{8}}} = \frac{43}{\sqrt{32} \times \sqrt{72}} = \frac{43}{5.65 \times 8.48} ∴ r = + 0.896

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:

d_x	d_x²	d_y	d_y²	d_xd_y
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
∑d_x= –8	∑d_x² = 804	∑d_y= –24	∑d_y² = 1262	∑d_xd_y= 897

r = \frac{\sum_{} d_{x} d_{y} - \frac{(\sum_{}^{} d_{x}) \times (\sum_{}^{} d_{y})}{N}}{\sqrt{\sum_{}^{} d_{x}^{2} - \frac{{(\sum_{}^{} d_{x})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{y}^{2} - \frac{{(\sum_{}^{} d_{y})}^{2}}{N}}} or, r = \frac{897 - \frac{(- 8) \times (- 24)}{10}}{\sqrt{804 - \frac{(- 8)^{2}}{10}} \times \sqrt{1262 - \frac{(- 24)^{2}}{10}}} = \frac{877.8}{28.24 \times 34.70} ∴ r = + 0.895

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 (R_X)	Ranks by Judge Y (R_Y)	D = R_X – R_Y	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				$\sum_{} D^{2} = 416$

r_{k} = 1 - \frac{6 \sum_{} D^{2}}{N^{3} - N} or, r_{k} = 1 - \frac{6 \times 416}{1728 - 12} = 1 - \frac{2496}{1716} = - 0.455 Hence, r_{k} = - 0.455

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:

*R_X*	*R_Y*	D = R_X – *R_Y*	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

r_{k} = 1 - \frac{6 \sum_{} D^{2}}{N^{3} - N} or, r_{k} = 1 - \frac{6 \times 32}{512 - 8} = \frac{504 - 192}{504} = 0.619 Hence, r_{k} = 0.619

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 = 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

Rank Correlation between Judge 1 and Judge 2 r_{k_{1, 2}} = 1 - \frac{6 \sum_{} {D_{1}}^{2}}{N^{3} - N} = 1 - \frac{6 \times 200}{1000 - 10} = \frac{990 - 1200}{990} = - 0.212 Rank Correlation between Judge 1 and Judge 3 r_{k_{1, 3}} = 1 - \frac{6 \sum_{} {D_{2}}^{2}}{N^{3} - N} = 1 - \frac{6 \times 60}{1000 - 10} = \frac{990 - 360}{990} = + 0.636 Rank Correlation between Judge 2 and Judge 3 r_{k_{2, 3}} = 1 - \frac{6 \sum_{} {D_{3}}^{2}}{N^{3} - N} = 1 - \frac{6 \times 214}{1000 - 10} = \frac{990 - 1284}{990} = - 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.

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)	${d^{'}}_{X} = \frac{X - A}{h} = \frac{X - 37.5}{5}$	$d_{Y}^{'} = \frac{Y - B}{i} = \frac{Y - 28}{5}$	*(d_X') (d_Y')*	*(d_X')²*	*(d_Y')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
			∑d_X' = –3	∑d_Y' = –5	∑(d_X') (d_Y') = – 21.8	∑(d_X')² = 19	∑(d_X')² = 38.2

r = \frac{\sum d_{X}^{'} d_{Y}^{'} - \frac{(\sum d_{X}^{'}) \times (\sum d_{Y}^{'})}{N}}{\sqrt{[\sum {d_{X}^{'}}^{2} - \frac{{(\sum d_{X}^{'})}^{2}}{N}] \times [\sum {d_{Y}^{'}}^{2} - \frac{{(\sum d_{Y}^{'})}^{2}}{N}]}} or, r = \frac{(- 21.8) - \frac{(- 3) \times (- 5)}{6}}{\sqrt{[19 - \frac{{(- 3)}^{2}}{6}] \times [38.2 - \frac{{(- 5)}^{2}}{6}]}} = \frac{- 24.3}{\sqrt{[19 - \frac{9}{6}] \times [38.2 - \frac{25}{6}]}} or, r = - \frac{- 24.3}{\sqrt{17.5 \times 34.03}} = - \frac{24.3}{\sqrt{595.525}} = - \frac{24.3}{24.403} Hence, r = - 0.996

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	$\frac{200}{250} \times 100 = 80 %$
16-17	200	150	$\frac{150}{200} \times 100 = 75 %$
17-18	150	90	$\frac{90}{150} \times 100 = 60 %$
18-19	120	48	$\frac{48}{120} \times 100 = 40 %$
19-20	100	30	$\frac{30}{100} \times 100 = 30 %$
20-21	80	12	$\frac{12}{80} \times 100 = 15 %$

Age Group	Mid Value (X)	Percentage of Players (%) (Y)	$d_{X} = X - A = X - 17.5$	$d_{Y} = Y - B = Y - 40$	d_Xd_Y	d_X²	d_Y²
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	∑d_X= 3	∑d_Y = 60	∑d_Xd_Y== –210	∑d_X² = 19	∑d_Y²=∑3950

r = \frac{\sum_{} d_{X} d_{Y} - \frac{(\sum_{}^{} d_{X}) \times (\sum_{}^{} d_{Y})}{N}}{\sqrt{\sum_{}^{} d_{X}^{2} - \frac{{(\sum_{}^{} d_{X})}^{2}}{N}} \times \sqrt{\sum_{}^{} d_{Y}^{2} - \frac{{(\sum_{}^{} d_{Y})}^{2}}{N}}} or, r = \frac{- 210 - \frac{3 \times (60)}{6}}{\sqrt{19 - \frac{(3)^{2}}{6}} \times \sqrt{3950 - \frac{(60)^{2}}{6}}} = \frac{- 240}{\sqrt{17.5} \times \sqrt{3350}} or, r = \frac{- 240}{4.18 \times 57.88} = \frac{- 240}{241.94} = - 0.992 ∴ r = - 0.992

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

Answer:

Cities	Density	Number of Deaths	Population	Death rate = $\frac{Number of Deaths}{Population} \times 100$
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 = 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

r = \frac{\sum d x d y - \frac{(\sum d x) (\sum d y)}{n}}{\sqrt{\sum d x^{2} - \frac{{(\sum d x)}^{2}}{n}} \sqrt{\sum d y^{2} - \frac{{(\sum d y)}^{2}}{n}}} = \frac{100 - \frac{400 \times 3}{6}}{\sqrt{300000 - \frac{{(400)}^{2}}{6}} \sqrt{2.1 - \frac{{(3)}^{2}}{6}}} or, r = \frac{100 - 200}{\sqrt{273333.34} \times \sqrt{0.6}} = \frac{- 100}{522.81 \times 0.77} = \frac{- 100}{402.56} = - 0.248 Hence, Karl Pearson' s Coefficient of Correlation between density of population and death rate is - 0.248

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

Answer:

$G i v e n : N = 15 X = 25 Y = 18 σ_{x} = 3.01 σ_{y} = 3.03 Σ x^{2} = 136 Σ y^{2} = 138 Σ x y = 122 r = \frac{Σ x y}{N σ_{x} σ_{y}} = \frac{122}{15 \times 3.01 \times 3.03} = 0.89$
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

Answer:

$G i v e n : N = 15 X = 80 Y = 120 Σ x^{2} = 56 Σ y^{2} = 156 Σ x y = 92 σ_{x} = \sqrt{\frac{Σ x^{2}}{N}}, σ_{y} = \sqrt{\frac{Σ y^{2}}{N}} σ_{x} = \sqrt{\frac{56}{15}} = 1.93, σ_{y} = \sqrt{\frac{156}{15}} = 3.22 r = \frac{Σ x y}{N σ_{x} σ_{y}} = \frac{92}{15 \times 1.93 \times 3.22} = 0.98$
Hence, Karl Pearson's coefficient of correlation is +0.98

Tr Jain Vk Ohri for Class 11 Commerce Economics Chapter 12 - Correlation

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