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Page No 72:

Question 1:

Prepare a statistical table from the following data taking the class width as 7 by inclusive method:

28 17 15 22 29 21 23 27 18 12
7 2 9 4 6 1 8 3 10 5
20 16 12 8 4 33 27 21 15 9
3 36 27 18 9 2 4 6 32 31
29 18 14 13 15 11 9 7 1 5
37 32 28 26 24           

Answer:

Class Interval Tally Bars Frequency
(f)
0 − 7 15
8 − 15 15
16 − 23 10
24 − 31 10
32 − 39 5
    Ʃf = 55

Page No 72:

Question 2:

Prepare a discrete series from the following data:

62 50 57 58 51 53 62 64 60 61
60 51 64 55 55 52 60 65 58 60
59 52 63 56 56 58 64 63 62 60

Answer:

Value Tally Bar Frequency
(f)
50 1
51 2
52 2
53 1
55 2
56 2
57 1
58 3
59 1
60 5
61 1
62 3
63 2
64 3
65 1
    Ʃf = 30

Page No 72:

Question 3:

Arrange the following marks in frequency table taking the lowest class interval as (10-20).

85 70 36 80 78 51 69
96 73 73 92 64 49 78
85 75 80 42 85 29 72
72 77 65 73 95 53 43
91 84 72 75 57 59 61
70 61 75 85 23    

Answer:

Marks Tally Bar Frequency
(f)
10 − 20    
20 − 30 2
30 − 40 1
40 − 50 3
50 − 60 4
60 − 70 5
70 − 80 14
80 − 90 7
90 − 100 4
    Ʃf = 40

Page No 72:

Question 4:

Change the following into continuous series and convert the series into 'less than' and 'more than' cumulative series :

Marks (mid-values)  5  15 25 35 45 55
No.of students 8 12 15 9 4 2

 

Answer:

Mid Value Class Interval Number of Students
(f)
5
15
25
35
45
55
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
8
12
15
9
4
2
    Ʃf = 50

Cumulative Series
 
Less Than More Than
Marks Number of Students Marks Number of Students
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
8
8 + 12 = 20
20 + 15 = 35
35 + 9 = 44
44 + 4 = 48
48 + 2 = 50
More than 0
More than 10
More than 20
More than 30
More than 40
More than 50
50
50 − 8 = 42
42 − 12 = 30
30 − 15 = 15
15 − 9 = 6
6 − 4 = 2

Page No 72:

Question 5:

Marks obtained by 24 students in English and Statistics in a class  are given below . Prepare two-way frequency distribution table.

S.No 1 2 3 4 5 6 7 8 9 10 11 12
Marks in English 22 23 23 23 23 24 23 25 22 23 24 24
Marks in Statistics 16 16 18 16 16 17 16 19 16 18 18 17
 
S.No 13 14 15 16 17 18 19 20 21 22 23 24
Marks in English 23 25 23 22 27 27 26 28 25 24 23 25
Marks in Statistics 16 17 17 17 15 16 18 19 19 16 17 19
 

Answer:

Marks in English
Marks in Statistics 22 23 24 25 26 27 28 Total
15 1
16 9
17 6
18 4
19 - 4
Total 3 9 4 4 1 2 1 24



Page No 73:

Question 6:

In a survey ,it was found that 64 families bought milk ( in litres) in the following quantities in a particular month.

19 16 22 9 22 12 39 19 14 23 6 24 16 18 7
17 20 25 28 18 10 24 20 21 10 7 18 28 24 20
14 23 25 34 22 5 33 23 26 29 13 36 11 26 11
37 30 13 8 15 22 21 32 21 31 17 16 23 12 9
15 27 17 21                      

Convert the above data in a frequency distribution making classes of 5-9, 10-14 and so on.
 

Answer:

Quantity of Milk
(kg)
Tally Bar Frequency
(f)

 
5 − 9 7
10 − 14 10
15 − 19 13
20 − 24 18
25 − 29 8
30 − 34 5
35 − 39 3
    Ʃf = 64

Page No 73:

Question 7:

The marks obtained by 20 students in Statistics and Economics are given below . Prepare a bivariate frequency distribution.

Marks in Statistics 10 11 10 11 11 14 12 12 13 10 13 12 11 12 10 14 14 12 13 10
Marks in Economics 20 21 22 21 23 23 22 21 24 25 24 23 22 23 22 22 24 20 24 23
 

Answer:

Marks in Economics
Marks in Statistics 20 21 22 23 24 25 Total
10 5
11 4
12 5
13 3
14   3
Total 2 3 5 5 4 1 20

Page No 73:

Question 8:

Prepare 'less than' and 'more than' cumulative frequency distributions of the following date:

Wages (₹) : 140-150 150-160 160-170 170-180 180-190 190-200
No. of workers : 5 10 20 9 6 2
 

Answer:

Less than Cumulative Frequency Distribution
 

Marks Number  of Students
Less than 150
Less than 160
Less than 170
Less than 180
Less than 190
Less than 200
5
5 + 10 = 15
15 + 20=35
35 + 9 = 44
44 + 6 = 50
50 + 2 = 52

More than Cumulative Frequency Distribution
 
Marks Number of Students
More than 140
More than 150
More than 160
More than 170
More than 180
More than 190
52
52 − 5 = 47
47 − 10 = 37
37 − 20 = 17
17 − 90 = 8
8 − 6 = 2

Page No 73:

Question 9:

Find out the frequency distribution and 'more than' cumulative frequency table:

Price (₹) below : 10 20 30 40 50 60
Quantity(kg) : 17 22 29 37 50 60
 

Answer:

The frequency distribution for the given data is as follows.
 

Price Class Interval Quantity
(f)

 
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
17
22 − 17 = 5
29 − 22 = 7
37 − 29 = 8
50 − 37 = 13
60 − 50 = 10

More than Cumulative Frequency Distribution
 
More than Quantity
(f)
More than 0
More than 10
More than 20
More than 30
More than 40
More than 50
60
60 − 17 = 43
43 − 5 = 38
38 − 7 = 31
31 − 8 = 23
23 − 13 = 10

Page No 73:

Question 10:

If class mid-points in a frequency distribution  of a group of persons are: 125, 132, 139, 146, 153, 160, 167, 174, 181 pounds, find:
(a) size of the class intervals, and
(b) the class boundaries.

Answer:

From the mid-values the class size is calculated using the following formula.
Class Size = Mid value of one class - Mid value of the preceding class
Class Size = 132 - 125 = 7
Now, the upper and lower limit for each of the class intervals is calculated as follows.
Lower limit = Mid value - Class Size2or, Lower limit = Mid value - 72Upper limit = Mid value + Class Size2or, Upper limit = Mid value + 72

 

Mid Value Class Interval
125
132
139
146
153
160
167
174
181
121.5 − 128.5
128.5 − 135.5
135.5 − 142.5
142.5 − 149.5
149.5 − 156.5
156.5 − 163.5
163.5 − 170.5
170.5 − 177.5
177.5 − 184.5

Page No 73:

Question 11:

Prepare a continuous series from the following data:

Mid values : 125 135 145 155 165
Frequency : 10 9 12 8 6
 

Answer:

From the mid-values the class size is calculated using the following formula.
Class Size = Mid value of one class - Mid value of the preceding class
Class Size = 135 - 125 = 10
Now, the upper and lower limit for each of the class intervals is calculated as follows.
Lower limit = Mid value - Class Size2or, Lower limit = Mid value - 102Upper limit = Mid value + Class Size2or, Upper limit = Mid value + 102

 

Mid Value Class Interval f
125
135
145
155
165
120 − 130
130 − 140
140 − 150
150 − 160
160 − 170
10
9
12
8
​6



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