TR Jain & Vk Ohri (2017) 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 (2017) 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 (2017) Solutions. All TR Jain & Vk Ohri (2017) Solutions for class 11 Humanities 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 (â¹) | 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 |
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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 |
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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 |
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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 |
Square of deviation x2 |
Age of wife (Y) |
Deviation |
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 |
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 |
Square of deviation x2 |
Weight Y |
Deviation |
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.
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 |
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 |
Square of deviation (x2) |
No. of Unemployed People (Y) |
Deviation |
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 |
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 |
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 |
Hence, coefficient of rank correlation = −1
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) |
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.
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 | 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 |
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 | dx2 | Y | dy2 | dxdy | ||
78 89 97 59 79 68 61 |
9 20 28 0 –10 10 –1 –8 |
81 400 784 0 100 100 1 64 |
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.
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 | dx2 | Y | dy2 | dxdy | ||
10 20 30 40 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 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 |
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 = 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 |
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.
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) |
dx2 | History (Y) |
dy2 | dxdy | ||
77 54 27 52 14 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 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 |
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 = 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 |
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 |
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 = 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 |
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 | dx2 | Y | dy2 | dxdy | ||
10 6 9 12 13 11 9 |
0 –4 –1 0 2 3 1 –1 |
0 16 1 0 9 4 1 1 |
9 4 6 9 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 |
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 |
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 = 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 |
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 |
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 |
N = 8
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 |
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 |
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.
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) |
(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 |
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 | |
16-17 | 200 | 150 | |
17-18 | 150 | 90 | |
18-19 | 120 | 48 | |
19-20 | 100 | 30 | |
20-21 | 80 | 12 |
Age Group | Mid Value (X) |
Percentage of Players (%) (Y) |
dXdY | dX2 | dY2 | ||
15-16 16-17 17-18 18-19 19-20 20-21 |
15.5 16.5 18.5 19.5 20.5 |
80 75 60 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
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 = |
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 â(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 |
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:
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:
Hence, coefficient of correlation between X and Y series is + 0.89
Page No 392:
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:
![](/img/site_content/ask-answer/loader.gif)
Yes, there exists perfect negative correlation (–1) between X and Y.
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