Home » Commerce for 11th and 12th Standard Students » Sandeep Garg Solutions Class 11 Economics Chapter 5 – Measures of Central Tendency- Arithmetic Mean

# 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.

We at CoolGyan’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.

There are numerous concepts in economics, but here we provide you with the solution from the Tabular Presentation, which will be convenient for the students to score well in the board exams.

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) 12345678910 80100908090110190230210250 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 24589 363710 624155690 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 555860626465 10121812107 5506961080744640455 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 − 44 − 88 − 1212 − 1616 − 2020 − 24 2610141822 7916864 145416011210888 ΣfΣf=50 ΣfmΣfm=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

The above-provided solutions are considered to be the best solution for ‘Sandeep Garg Economics Class 11 Solutions Chapter 5 – Measures of Central Tendency- Arithmetic Mean. Stay tuned to CoolGyan’S to learn more.

 Sandeep Garg Solutions for other Chapters: