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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.

\(\left ( \bar{X} \right ) =\, \frac{EX}{N}\)

 

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

 

\(\bar{X}\, =\, \frac{125}{10}\)

 

\(\bar{X}\, =\, 12.5\)

 

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
\(\left ( \bar{X} \right )\, =\, \frac{\sum X}{N}\)

 

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

 

\(\bar{X}\, =\, \frac{1430}{10}\)

 

\(\bar{X}\, =\, 143\)

 

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
\(\bar{X}\, =\, \frac{\sum fX}{\sum f}\)

 

\(\bar{X}\, =\, \frac{191}{29}\)

 

\(\bar{X}\, =\, 6.59\)

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
\(\bar{X}\, =\, \frac{\sum fX}{\sum f}\)

 

Or, \(\bar{X}\, =\, \frac{3265}{69}\)

 

\(\bar{X}\, =\, 60.37\)

 

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

\(\bar{X}\, \frac{\sum fm}{\sum f}\)

 

\(\bar{X}\, =\, \frac{536}{50}\)

 

\(\bar{X}\, =\, 10.72\)

Therefore, the mean marks acquired by the students are 10.72

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