Sandeep Garg Solutions for 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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

ANSWER:

Age

(X)

Tally

Marks

Frequency

(f)

15 III 3
16 III 3
17 IIII 4 Modal Class
18 I 1
N=11

Since 17 occurs the highest number of times in the series i.e. 4 times.

Hence, Mode = 17

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