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 economics educators from the contemporary edition of Sandeep Garg Economics Class 11 textbook solutions.

We at BYJU’S provide Sandeep Garg Economics Class 11 Solutions to give a comprehensive insight into the subject to the students. These insights will be a valuable advantage to students while completing their homework or while studying for their exams.

In this article we are providing the solutions from Tabular Presentation, which will help the students to score well in the board exams.

Sandeep Garg Solutions Class 11 – Chapter 5

Question 1

Find the average from the following data:

3 5 7 9 11 13 17 19 20 21

Answer:

Here, the average is calculated using a direct method.

\(\begin{array}{l}\left ( \bar{X} \right ) =\, \frac{EX}{N}\end{array} \)

 

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

 

\(\begin{array}{l}\bar{X}\, =\, \frac{125}{10}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, 12.5\end{array} \)

 

Question 2

Following are the daily income of ten families. Evaluate the average daily income.

S. No. 1 2 3 4 5 6 7 8 9 10
Daily Income 100 120 80 85 95 130 200 250 225 275

Answer:

S.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
\(\begin{array}{l}\left ( \bar{X} \right )\, =\, \frac{\sum X}{N}\end{array} \)

 

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

 

\(\begin{array}{l}\bar{X}\, =\, \frac{1430}{10}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, 143\end{array} \)

 

Question 3

Find the average from the series.

X 2 4 5 8 9
f 3 6 3 7 10

Answer:

X f fX
2

4

5

8

9

 3

6

3

7

10

 6

24

15

56

90
N=Σf= 29 ΣfX= 191
\(\begin{array}{l}\bar{X}\, =\, \frac{\sum fX}{\sum f}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, \frac{191}{29}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, 6.59\end{array} \)

The average of the given series is 6.59.

 

Question 4

Find the arithmetic mean from the frequency table.

Height (in cm) 55 58 60 62 64 65
Number of Flowers 10 12 18 12 10 7

Answer:

Height

(X)

 Flowers

(f)

 fX
55

58

60

62

64

65

 10

12

18

12

10

7

 550

696

1080

744

640

455
N=Σf= 69 ΣfX=4165
\(\begin{array}{l}\bar{X}\, =\, \frac{\sum fX}{\sum f}\end{array} \)

 

Or,

\(\begin{array}{l}\bar{X}\, =\, \frac{3265}{69}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, 60.37\end{array} \)

 

Question 5

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

Marks 0−4 4−8 8−12 12−16 16−20 20−24
No. of Students 7 9 16 8 6 4

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
\(\begin{array}{l}\bar{X}\, \frac{\sum fm}{\sum f}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, \frac{536}{50}\end{array} \)

 

\(\begin{array}{l}\bar{X}\, =\, 10.72\end{array} \)

The mean marks acquired by the students are 10.72

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