 # Sandeep Garg Solutions Class 11 Economics Chapter 6 - Measures of Central Tendency- Median and Mode

Sandeep Garg Class 11 Economics Solutions Chapter 6 – Measures of Central Tendency- Median and Mode are 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 6

Question 1

Find out the median.

 S. No. 1 2 3 4 5 6 7 8 9 Marks Obtained 10 12 14 17 18 20 21 30 32

10, 12, 14, 17, 18, 20, 21 30, 32

N = 9

Median = $\left ( \frac{N+1}{2} \right )$th

Median = $\left ( \frac{9+1}{2} \right )$th = 5th item

Therefore, the median is given by the size of the 5th items. Therefore, the Median of the data so given is 18.

Question 2

Calculate the value of median from the following data:

 Income 12,00 1,800 5,000 2,500 3,000 1,600 3,500 No. of Persons 12 16 2 10 3 15 7

 Income Number of People (f) Cumulative Frequency (c.f.) 1200 1600 12 15 12 27 1800 16 43 2500 3000 3500 5000 10 3 7 2 53 56 63 65 N=65

Median = size of $\left ( \frac{N+1}{2} \right )$th item

Median = $\frac{65+1}{2}$ = 33 rd

While locating the item in the Cumulative Frequency column, the item that exceeded 33rd is 43 that is corresponding to 1800. Thus, the median is 1800.

Question 3

Find out the median size from the following:

 X 160 150 152 161 156 f 5 8 6 3 7

 X f Cumulative Frequency (cf) 150 152 8 6 8 14 156 7 21 160 161 5 3 26 29 N=29

Median = size of $\left ( \frac{N+1}{2} \right )$th item

Median = $\frac{29+1}{2}$ = 15 th

When locating the item inCumulative Frequency coloumn. The item that exceeded 15th is 21 corresponding to 156. Therefore, median is 156.

Question 4

Evaluate lower quartile and upper quartile from the following series.

 Variable 5 10 15 20 25 30 35 40 Frequency 16 18 22 21 24 14 11 9

 Variable Frequency (f) Cumulative Frequency (c.f.) 5 16 16 10 18 34 15 22 56 20 21 77 25 24 101 30 14 115 35 11 126 40 9 135 N=135

Q1 = size of $\left ( \frac{N+1}{4} \right )$th item

Or, Q1 = size of $\left ( \frac{135+1}{4} \right )$th item

= Q1 = 34th item

Upper Quartile

Q1 = size of 3 $\left ( \frac{N+1}{4} \right )$th item

Or, Q1 = size of 3 $\left ( \frac{135+1}{4} \right )$th item

= Q1 = 102nd item

This corresponds to 115 in the cumulative frequency.

Hence, upper quartile is 30.

Question 5

Age of 11 students of class XI is given below. Find the modal age by: (i) Observation Method; (ii) Frequency distribution Method.

 Age (in years) 15 16 16 17 15 16 18 17 15 17 17