Sandeep Garg Solutions Class 11 Economics Chapter 5 – Measures of Central Tendency- Arithmetic Mean


Sandeep Garg Class 11 Economics Solutions Chapter 5 – Measures of Central Tendency- Arithmetic Mean is illustrated by the professional economic educator from the contemporary edition of Sandeep Garg Economics Class 11 textbook solutions.

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Sandeep Garg Solutions Class 11 – Chapter 5

Question 1

From the following data find an average.

3579111317192021

Answer:

Here, the average is calculated using a direct method.

(left ( ar{X} ight ) =, frac{EX}{N})

 

(ar{X}, =, frac{3+5+7+9+11+13+17+19+20+21}{10})

 

(ar{X}, =, frac{125}{10})

 

(ar{X}, =, 12.5)

 

Question 2

Given below are the daily income of ten families. Evaluate the average daily income.

S. No.12345678910
Daily Income100120808595130200250225275

Answer:

Serial No. Daily Income (in Rs)

(X)

1

2

3

4

5

6

7

8

9

10

80

100

90

80

90

110

190

230

210

250

N =10∑X= 1430
(left ( ar{X} ight ), =, frac{sum X}{N})

 

(ar{X}, =, frac{80+100+90+80+90+110+190+230+210+255}{10})

 

(ar{X}, =, frac{1430}{10})

 

(ar{X}, =, 143)

 

Question 3

Find an average from the below series.

X24589
f363710

Answer:

XffX
2

4

5

8

9

3

6

3

7

10

6

24

15

56

90

N=Σf= 29ΣfX= 191
(ar{X}, =, frac{sum fX}{sum f})

 

(ar{X}, =, frac{191}{29})

 

(ar{X}, =, 6.59)

Therefore, the average of the above series is 6.59

 

Question 4

Prepare arithmetic mean from the below frequency table.

Height (in cms.)555860626465
Number of Flowers10121812107

Answer:

Height

(X)

Flowers

(f)

fX
55

58

60

62

64

65

10

12

18

12

10

7

550

696

1080

744

640

455

Nf= 69ΣfX=4165
(ar{X}, =, frac{sum fX}{sum f})

 

Or, (ar{X}, =, frac{3265}{69})

 

(ar{X}, =, 60.37)

 

Question 5

From the following data arrange the mean marks acquired by the students using direct method.

Marks0−44−88−1212−1616−2020−24
No. of Students7916864

Answer:

Class Interval

(Marks)

Mid-Values

(m)

Students

(f)

fm
0 − 4

4 − 8

8 − 12

12 − 16

16 − 20

20 − 24

2

6

10

14

18

22

7

9

16

8

6

4

14

54

160

112

108

88

Σf

Σf=50

Σfm

Σfm=586

(ar{X}, frac{sum fm}{sum f})

 

(ar{X}, =, frac{536}{50})

 

(ar{X}, =, 10.72)

Therefore, the mean marks acquired by the students are 10.72

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