height62 72 68 58 65 70 weight 50 65 63 50 54 60 draw a scatter drigram
Answers
Answer:ANSWER:
a) Scatter Diagram
b) The points obtained on the scatter diagram lie close to each other and reflect an upward trend. Thus, there exists a high degree of positive correlation between capital employed and profits earned.
c)
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Question 2:
Plot the following data as a scatter diagram and comment on the result obtained:
X : 11 10 15 13 10 16 13 8 17 14
Y : 6 7 9 9 7 11 9 6 12 11
ANSWER:
Scatter Diagram
Thus, there exists a positive correlation of moderate degree between X and Y.
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Question 3:
Following are the heights and weights of 10 students in a class. Draw a scatter diagram and indicate whether the correlation is positive or negative.
Height (in inches) : 72 60 63 66 70 75 58 78 72 62
Weight (in kg) : 65 54 55 61 60 54 50 63 65 50
ANSWER:
Thus, there exists a very low degree of positive correlation between height and weight of students.
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Question 4:
Draw a scatter diagram for the data given below and interpret it:
X : 10 20 30 40 50 60 70 80
Y : 32 20 24 36 40 28 48 44
ANSWER:
Thus, there exists a moderate degree of correlation between X and Y.
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Question 5:
Draw a scatter diagram of the following data:
X : 15 18 30 27 25 23 30
Y : 7 10 17 16 12 13 9
ANSWER:
Thus, there exists a moderate degree of positive correlation between X and Y.
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Question 6:
From the following data compute the product moment correlation between X and Y.
X series Y series
Arithmetic Mean 25 18
Sum of Square of deviations from Arithmetic Mean 136 138
Summation of products of deviations of X and Y series from their respective means = 122
Number of points of values = 15
ANSWER:
Given:X=25Y=18 Σx2=136 Σy2=138 Σxy=122 n=15Now, r=ΣxyΣx2√ ×Σy2√r=122136√ ×138√=122136×138√=12211.66×11.74r=122136.996=0.89
Thus, the product moment correlation between X and Y is 0.89.
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Question 7:
Calculate Karl Pearson's Coefficient of Correlation on the following data:
X : 15 18 21 24 27 30 36 39 42 48
Y : 25 25 27 27 31 33 35 41 41 45
ANSWER:
X Y (X−X)
(x) x2 (Y−Y)
(y) y2 xy
15 25 −15 225 −8 64 120
18 25 −12 144 −8 64 96
21 27 −9 81 −6 36 54
24 27 −6 36 −6 36 36
27 31 −3 9 −2 4 6
30 33 0 0 0 0 0
36 35 6 36 2 4 12
39 41 9 81 8 64 72
42 41 12 144 8 64 96
48 45 18 324 12 144 216
ƩX = 300 ƩY =330 Ʃx2 =1080 Ʃy2 =480 Ʃxy =708
N = 10
X=ΣXN=30010=30Y=ΣYN=33010=33r=ΣxyΣx2√ ×Σy2√ =7081080√× 480√=70832.86×21.91 =708720 =0.983
Thus, the value of correlation coefficient is 0.983.
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Question 9:
Calculate Karl Pearson's Coefficient of Correlation between the sales and expenses of the following 10 firms:
Firms : 1 2 3 4 5 6 7 8 9 10
Sales(in ₹ '000) : 50 50 55 60 65 65 65 60 60 50
Expenses (in ₹ '000) : 11 13 14 16 16 15 15 14 13 13
ANSWER:
Sales
(X) Expenses
(Y) (X−X)
x x2 (Y−Y)
y y2 xy
50 11 −8 64 −3 9 24
50 13 −8 64 −1 1 8
55 14 −3 9 0 0 0
60 16 2 4 2 4 4
65 16 7 49 2 4 14
65 15 7 49 1 1 7
65 15 7 49 1 1 7
60 14 2 4 0 0 0
60 13 2 4 −1 1 −2
50 13 −8 64 −1 1 8
580 140 360 22 70
N = 10
X=ΣXN=58010=58Y=ΣYN=14010=14r=ΣxyΣx2√ Σy2√=70360√ 22√=7018.973×4.690=7088.9833=0.786
Note: As per the textbook, coefficient of correlation is 0.67. However, as per the above solution coefficient of correlation should be 0.786.
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Question 10:
Calculate correlation coefficient between X, the number of rainy days per month and Y, the number of rain coats sold in that month in a certain shop for 12 months. Interpret the results.
X : 14 8 18 10 22 9 3 5 6 11 13 13
Y : 15 11 20 12 15 7 3 4 7 10 11 29
ANSWER:
X Y (X−X)
(x) x2 (Y−Y)
(y) y2 xy
14 15 3 9 3 9 9
8 11 −3 9 −1 1 3
18 20 7 49 8 64 56
10 12 −1 1 0 0 0
22 15 11 121 3 9 33
9 7 −2 4 −5 25 10
3 3 −8 64 −9 81 72
5 4 −6 36 −8 64 48
6 7 −5 25 −5 25 25
11 10 0 0 −2 4 0
13 11 2 4 −1 1 −2
13 29 2 4 17 289 34
ƩX =132 ƩY =144 Ʃx2 =326 Ʃy2 =572 Ʃxy =288
N = 12
X=ΣXN=13212=11Y=ΣYN=14412=12r=ΣxyΣx2× Σy2√ =288326×572√=28818.05×23.91 =288431.82 =0.6669 =0.67(approx.)
There is a moderate degree of (+) correlation between the number of rainy days and the number of rain coats sold. In other words, as the number of rainy days increases in a month, the number of rain coats sold in that month increases moderately.
Note: As per the textbook, coefficient of correlation is −0.67. However, as per the above solution coefficient of correlation should be +0.67.
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Question 11:
The deviations from their means of two series (X and Y) are given below:
X : −4 −3 −2 −1 0 +1 +2 +3 +4
Y : +3 −3 −4 0 +4 +1 +2 −2 −1
Explanation: