Sandeep Garg 2018 Solutions for Class 11 Commerce Economics Chapter 1 Organisation Of Data are provided here with simple step-by-step explanations. These solutions for Organisation Of Data are extremely popular among class 11 Commerce students for Economics Organisation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Sandeep Garg 2018 Book of class 11 Commerce Economics Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Sandeep Garg 2018 Solutions. All Sandeep Garg 2018 Solutions for class 11 Commerce Economics are prepared by experts and are 100% accurate.

Page No 4.49:

Question 1:

Following are the figures of marks obtained by 40 students. You ae required to arrange them in ascending and in descending order.

15 18 16 14 10 6 5 3 8 7
22 18 14 19 17 8 6 4 10 3
12 16 15 13 11 10 18 22 14 19
11 18 22 14 25 21 17 8 9 10

Answer:

Marks in Ascending Order:
3, 3, 4, 5, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 10, 11, 11, 12, 13, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 18, 18 ,19, 19, 21, 22, 22, 22, 25

Marks in Descending Order:
25, 22, 22, 22, 21, 19, 19, 18, 18, 18, 18, 17, 17, 16, 16, 15, 15, 14, 14, 14, 14, 13, 12, 11, 11, 10, 10, 10, 10, 9, 8, 8, 8, 7, 6, 6, 5, 4, 3, 3

Page No 4.49:

Question 2:

Heights (in inches) of 35 students of a class in given below. Classify the following data in a discrete frequency series

58 60 72 68 58 55 58 72 55 68 66 60
60 55 76 66 58 72 68 60 55 68 55 58
62 66 72 58 60 68 55 58 65 72 68  

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Height
(in inches)
Tally Marks Frequency
55
58
60
62
65
66
68
72
ΙΙΙΙ Ι
ΙΙΙΙ  ΙΙ
ΙΙΙΙ 
Ι
Ι
ΙΙΙ
ΙΙΙΙ Ι
ΙΙΙΙ Ι
6
7
5
1
1
3
6
6
    f=35

Page No 4.49:

Question 3:

The following are the marks of the 30 students in statistics. prepare a frequency distribution taking the class−intervals as 10−20, 20−30 and so on.

12 33 23 25 18 35 37 49 54 51 37 15 33 42 45
47 55 69 65 63 46 29 18 37 46 59 29 35 45 27

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Marks Tally Marks Number of students
(Frequency)
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
ΙΙΙΙ
ΙΙΙΙ  
ΙΙΙΙ ΙΙ
ΙΙΙΙ ΙΙ
ΙΙΙΙ
ΙΙΙ
4
5
7
7
4
3
    f=30



Page No 4.50:

Question 4:

Following are the marks (out of 100) obtained by 50 students in statistics:

70 55 51 42 57 45 60 47 63 53
33 65 39 82 55 64 58 61 65 42
50 52 53 45 45 25 36 59 63 39
65 54 49 54 64 75 42 41 52 35
30 35 15 48 26 20 40 55 46 18
make a frequency distribution taking the first class interval as 0-10.

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Marks Tally Marks Number of students
(Frequency)
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
70 - 80
80  - 90
-
ΙΙ
ΙΙΙ
ΙΙΙΙ ΙΙ
ΙΙΙΙ ΙΙΙΙ ΙΙ
ΙΙΙΙ ΙΙΙΙ ΙΙΙΙ
ΙΙΙΙ ΙΙΙΙ
ΙΙ
Ι
0
2
3
7
12
14
9
2
1
    f=50

Page No 4.50:

Question 5:

Prepare a frequency table taking class intervals 20−24, 25−29, 30−34 and so on, from the following data:

21 20 55 39 48 46 36 54 42 30
29 42 32 40 34 31 35 37 52 44
39 45 37 33 51 53 52 46 43 47
41 26 52 48 25 34 37 33 36 27
54 36 41 33 23 39 28 44 45 38

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Class Intervals Tally Marks Frequency
20 - 24
25 - 29
30 - 34
35 - 39
40 - 44
45 - 49
50 - 54
55 - 59
ΙΙΙ
ΙΙΙΙ 
ΙΙΙΙ ΙΙΙ
ΙΙΙΙ ΙΙΙΙ Ι
ΙΙΙΙ ΙΙΙ
ΙΙΙΙ ΙΙ
ΙΙΙΙ ΙΙ
Ι
3
5
8
11
8
7
7
1
    f=50

Page No 4.50:

Question 6:

From the following data, calculate the lower limit of the first class and upper limit of the last class.

Daily Wages Less than 120 120−140 140−160 160−180 Above 180
No. of Workers 35 12 10 40 13

Answer:

In the given question, the size of each class is 20. Thus, maintaining the uniformity, we take the first class as 100-120 and the last class as 180-200. Therefore, the lower limit of the first class-interval is 120 − 20 = 100 and the upper limit of the last class is 180 + 20 = 200.

Page No 4.50:

Question 7:

Calculate the missing class-intervals from the following distribution:

Class-Interval Below 20 20−50 50−90 90−140 More than 140
Frequency 7 12 16 2 4

Answer:

In the given question, the class-intervals are of different width. Thus, the missing class-intervals cannot be determined.

Page No 4.50:

Question 8:

Convert the following 'more than' cumulative frequency distribution into a 'less than' cumulative frequency distribution

Class-Interval (More than) 10 20 30 40 50 60 70 80
Frequency 124 119 107 84 55 31 12 2

Answer:

The 'more than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.
 

Class-Interval Frequency
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
70 - 80
80 - 90
124 − 119 = 5
119 − 107 = 12
107 − 84 = 23
84 − 55 = 29
55 − 31 = 24
31 − 12 = 19
12 − 2 = 10
                2

From the above distribution, we can make the 'less than' cumulative frequency distribution as follows.
 
Class-Interval Frequency
Less then 20
Less then 30
Less then 40
Less then 50
Less then 60
Less then 70
Less then 80
Less then 90
5
5 + 12 = 17
17 + 23 = 40
40 +29 = 69
69 + 24 = 93
93 + 19 = 112
112 + 10 = 122
122 + 2 = 124

Page No 4.50:

Question 9:

Convert the following cumulative frequency series into simple frequency series

Marks No. of Students
Less than 20 10
Less than 40 18
Less than 60 25
Less than 80 45
Less than 100 55

Answer:


The 'less than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.
 

Class-Interval Frequency
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100
10
18 − 10 = 8
25 − 18 = 7
45 − 25 = 20
55 − 45 = 10



Page No 4.51:

Question 10:

Prepare a frequency distribution from the following data:

Mid-Points 25 35 45 55 65 75
Frequency 6 10 9 12 8 5

Answer:

The class interval can be calculated from the mid-points using the following adjustment formula. 

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
Value of adjustment = 35 - 252= 5

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 25 – 5 = 20

Upper limit of first class = 25 + 5 = 30

Thus, the first class interval is (20 30). Similarly, we can calculate the remaining class intervals.


Frequency Distribution
 
Mid Value Class Interval Frequency
25
35
45
55
65
75
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
6
10
9
12
8
5
    f=50
 

Page No 4.51:

Question 11:

The ages of 20 husbands and wives are given below. Form a two-way frequency table showing the relationship between the ages of husbands and wives with the class-intervals 20−25, 25−30, etc.

S.No Age of Husband Age of Wife S. No. Age of Husband Age of Wife
1 28 23 11 27 24
2 37 30 12 39 34
3 42 40 13 23 20
4 25 26 14 33 31
5 29 25 15 36 29
6 47 31 16 32 35
7 37 35 17 22 23
8 35 25 18 29 27
9 23 21 19 38 34
10 41 38 20 48 47

Answer:


The given data can be presented in the form of a bivariate frequency distribution as follows.
 


Age of wife (Y)
Age of husband (X) Total
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50
             
20 − 25         5
25 − 30         5
30 − 35       5
35 − 40       3
40 − 45           1
45 − 50           1
Total 3 5 2 6 3 1 20

Page No 4.51:

Question 12:

The following data give the points scored in a tennis match by two player X and Y at the end of twenty games:

(10,12) (2,11) (7,9) (15,19) (17,21) (12,8) (16,10) (14,14) (22,18) (16,4)
(15,16) (22,20) (19,15) (7,18) (11,11) (12,18) (10,10) (5,13) (11,7) (10,10)
Taking class intervals as: 5−9, 10−14, 15−19..., for both X and Y, construct a Bivariate Frequency Distribution.

Answer:

The given data can be presented in the form of a bivariate frequency distribution as follows.
 


Player Y
Player X Total
0 − 4 5 − 9 10 − 14 15 − 19 20 − 24
           
0 − 4         1
5 − 9       3
10 − 14   8
15 − 19   6
20 − 24       2
Total 1 3 8 6 2 20

Page No 4.51:

Question 13:

In a survey, it was found that 64 families bought milk in the following quantities in a particular month. Prepare a frequency distribution with classes as 5−9, 10−14 etc.

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

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Quantity Tally Marks Frequency
5 − 9 6
10 − 14 11
15 − 19 13
20 − 24   18
25 − 29 8
30 − 34 5
35 − 39 3
40 − 44    
    f=64

Page No 4.51:

Question 14:

You are given below a mid value series, convert it into a continuous series.

Mid-Value 15 25 35 45 55
Frequency 8 12 15 10 5

Answer:

The class interval can be calculated from the mid-points using the following adjustment formula. 

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
Value of adjustment = 25 - 152= 5

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 15 – 5 = 10

Upper limit of first class = 15 + 5 = 20

Thus, the first class interval is (10 20). Similarly, we can calculate the remaining class intervals.


 
Mid value Class Interval Frequency
15
25
35
45
55
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
8
12
15
10
5
    f=50



Page No 4.52:

Question 15:

For the data given below, prepare a frequency distribution table with classes 100−110, 110-120, etc.

125 108 112 126 110 132 136 130 140 155
120 130 136 138 125 111 112 125 140 148
147 137 145 150 142 135 137 132 165 154

Answer:

In the form of frequency distribution, the given data can be arranged as follows.
 

Class Interval Telly Marks Frequency
100 − 110 1
110 − 120 4
120 − 130 5
130 − 140 10
140 − 150   6
150 − 160 3
160 − 170 1
    f=30

Page No 4.52:

Question 16:

Prepare a bivariate frequency distribution for the following data for 20 students:

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

Answer:

The data can be presented in the form of a bivariate distribution as follows.
 

Marks in Statistics Marks in Maths  
10 11 12 13 14 Total
           
20       2
21       3
22   5
23   6
24       4
Total 5 4 5 3 3 20

Page No 4.52:

Question 17:

In a school, no student has scored less than 25 marks or more than 60  marks in an examination. Their cumulative frequencies are as follows:

Less than 60 55 50 45 40 35 30
Total frequency 120 115 90 75 65 45 32
Find the frequencies in the class intervals 25−30, 30−35,....55−60.

Answer:

Marks Cumulative Frequency
Less than 60
Less than 55
Less than 50
Less than 45
Less than 40
Less than 35
Less than 30
120
115
90
75
65
45
32

This can be presented in the form of a simple frequency distribution as follows.
 
Marks Number of Students
25 − 30
30 − 35
35 − 40
40 − 45
45 − 50
50 − 55
55 − 60
32
45 − 32 = 13
65 − 45 = 20
75 − 65 = 10
90 − 75 = 15
115 − 90 = 25
120 − 115 = 5

Page No 4.52:

Question 18:

Marks scored by 50 students are given below:

40 45 38 24 46 42 45 18 53 64
45 32 52 54 78 65 52 64 66 43
48 55 50 43 48 20 27 65 37 55
51 55 62 66 38 16 60 58 46 35
72 62 54 58 30 36 43 82 46 53
(a) Arrange the marks in ascending order.
(b) Represent the marks in the form of discrete frequency distribution.
(c) Construct an inclusive frequency distribution with first class as 10−19. Also construct class boundaries.
(d) Construct a frequency distribution with exclusive class-intervals, taking the lowest class as 10−20.
(e) Convert the exclusive series constructed in (d) into 'less than' and 'more than' cumulative frequency distribution.
 

Answer:

(a)

Marks arranged in ascending order
16
18
20
24
27
30
32
35
36
37
38
38
40
42
43
43
43
45
45
45
46
46
46
48
48
50
51
52
52
53
53
54
54
55
55
55
58
58
60
62
62
64
64
65
65
66
66
72
78
82

(b)

Discrete Frequency Distribution
Marks
Frequency
(f)

 
16
18
20
24
27
30
32
35
36
37
38
40
42
43
45
46
48
50
51
52
53
54
55
58
60
62
64
65
66
72
78
82
1
1
1
1
1
1
1
1
1
1
2
1
1
3
3
3
2
1
1
2
2
2
3
2
1
2
2
2
2
1
1
1
  f=50

(c)
Inclusive Frequency Distribution
Class Interval
(Marks)
No. of Students
10 − 19
20 − 29
30 − 39
40 − 49
50 − 59
60 − 69
70 − 79
80 − 89
2
3
7
13
13
9
2
1
  f=50

(d) 
Exclusive Frequency Distribution
Class Interval No. of Students
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
80 −90
2
3
7
13
13
9
2
1
  f=50

(e)
Less than frequency distribution
Marks Cumulative Frequency
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
Less than 80
Less than 90
2
2 + 3 = 5
5 + 7 = 12
12 + 13 = 25
25 + 13 = 38
38 + 9 = 47
47 + 2 = 49
49 + 1 = 50

More than frequency distribution
Marks Cumulative Frequency
More than 10
More than 20
More than 30
More than 40
More than 50
More than 60
More than 70
More than 80
50
50 − 2 = 48
48 − 3 = 45
45 − 7 = 38
38 − 13 = 25
25 − 13 = 12
12 − 9 = 3
3 − 2 =1

Page No 4.52:

Question 19:

From the following data of the ages of different persons, prepare less than and more than cumulative frequency distribution.

Age (in years) 10−20 20−30 30−40 40−50 50−60 60−70 70−80
No. of Persons 5 12 10 6 4 11 2

Answer:

Age
(in years)

 
No. of Persons
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
5
12
10
6
4
11
2
  f=50

Less than frequency distribution
Age Cumulative frequency
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
Less than 80
5
5 + 12 = 17
17 + 10 = 27
27 + 6 = 33
33 + 4 = 37
37 + 11 = 48
48 + 2 = 50

More than frequency distribution
Age Cumulative frequency
More than 10
More than 20
More than 30
More than 40
More than 50
More than 60
More than 70
50
50 − 5 = 45
45 − 12 = 33
33 − 10 = 23
23 − 6 = 17
17 − 4 = 13
13 − 11 = 2



View NCERT Solutions for all chapters of Class 13