# 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.

 3 5 7 9 11 13 17 19 20 21

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. 1 2 3 4 5 6 7 8 9 10 Daily Income 100 120 80 85 95 130 200 250 225 275

 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.

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

 X f fX 2 4 5 8 9 3 6 3 7 10 6 24 15 56 90 N=Σf= 29 ΣfX= 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.) 55 58 60 62 64 65 Number of Flowers 10 12 18 12 10 7

 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
$\bar{X}\, =\, \frac{\sum fX}{\sum f}$

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.

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

 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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