Economy, asked by kunalsingh69191, 19 days ago

height62 72 68 58 65 70 weight 50 65 63 50 54 60 draw a scatter drigram

Answers

Answered by veerajagarwal
1

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)

Page No 312:

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.

Page No 312:

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.

Page No 313:

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.

Page No 313:

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.

Page No 313:

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:

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