Tr Jain & Vk Ohri 2017 Solutions for Class 11 Science Economics Chapter 7 Frequency Diagrams: Histogram, Polygon And Ogive are provided here with simple step-by-step explanations. These solutions for Frequency Diagrams: Histogram, Polygon And Ogive are extremely popular among Class 11 Science students for Economics Frequency Diagrams: Histogram, Polygon And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Tr Jain & Vk Ohri 2017 Book of Class 11 Science Economics Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Tr Jain & Vk Ohri 2017 Solutions. All Tr Jain & Vk Ohri 2017 Solutions for class Class 11 Science Economics are prepared by experts and are 100% accurate.

Page No 116:

Question 1:

Visit different schools in your locality. Collect information on the daily wage earners. Your information should include (i) daily earnings, and (ii) number of wage earners. Convert the raw data in to a continuous frequency distribution, and present your information in the form a histogram.

Answer:

Daily Earnings Number of wage earners
0−20
20−40
40−60
60−80
80−100
37
50
70
25
10



Page No 118:

Question 1:

Collect data on the CBSE result (2016) for class X in your school. Discuss with your teacher how can you present the information in terms of a frequency polygon. Write your observations on the performance of your school students.

Answer:

Marks
(in percentage)

 
Number of students
10−20 0
20−30 0
30−40 3
40−50 5
50−60 13
60−70 9
70−80 18
80−90 35
90−100 22

We can infer from the above table that most of the students have scored more than 80% marks and only eight students have scored less than 50% marks. Thus, it can be concluded that students have performed outstanding in the Board exams for class X.



Page No 121:

Question 1:

Get information from your class mates on the monthly pocket allowance they get from their parents. Present the raw data in terms of a frequency distribution. Show cumulative frequencies of your statistical series. Present your information in the form of (i) Less than ogive, and (ii) More than ogive.

Answer:

Continuous frequency distribution
 

Pocket allowance(Rs) Number of Students
 
0-10
10-20
20-30
30-40
40-50
50-60
60-70

5
7
10
25
35
10
8



For constructing a less than ogive , first the given frequency distribution must be converted into a less than cumulative frequency distribution as follows.
 
Pocket allowance Cumulative Frequency
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
5
5 + 7 = 12
12 + 10 = 22
22 + 25 = 47
47 + 35 = 82
82 + 10 = 92
92 + 8 = 100

For constructing a more than ogive , first the given frequency distribution must be converted into a more than cumulative frequency distribution as follows.
Pocket allowance Cumulative Frequency
More than 0
More than 10
More than 20
More than 30
More than 40
More than 50
More than 60
More than 70
100
100 − 5 = 95
95 − 7 = 88
88 − 10 = 78
78 − 25 = 53
53 − 35 = 18
18 − 10 = 8
8 − 8 = 0




Page No 128:

Question 1:

Represent the following frequency distribution graphically:

Number of Children Number of Families
0

1

2

3

4

5
5

10

20

25

30

10
  Total = 100

Answer:

The given distribution can be represented with the help of a bar diagram as follows.

Page No 128:

Question 2:

Construct histogram, frequency polygon and frequency curve from the following data:

Marks Obtained 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Number of Students 10 16 20 20 22 15 8 5

Answer:


For drawing a frequency polygon, we simply join the top mid-points of the rectangles of the histogram drawn above using a straight line.  



Similar to a frequency polygon, for drawing a frequency curve we again joint the top mid-points of the rectangles of the histogram drawn in the first part, but with a free hand (rather than a straight line).

Page No 128:

Question 3:

Make a frequency polygon and histogram using the given data:

Marks Obtained 10−20 20−30 30−40 40−50 50−60 60−70
Number of Students 5 12 15 22 14 4

Answer:



For drawing a frequency polygon, we simply join the top mid-points of the rectangle of the histogram as drawn above in the first part using a straight line.

Page No 128:

Question 4:

Draw 'less than' and 'more than' ogive curves from the following data:

Marks 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40
Number of Students 7 10 20 13 12 19 14 9

Answer:

(i) For constructing a less than ogive, first the given frequency distribution must be converted into a less than cumulative frequency distribution as follows. 

Marks Cumulative Frequency
Less than 5
Less than 10
Less than 15
Less than 20
Less than 25
Less than 30
Less than 35
Less than 40
7
7 + 10 = 17
17 + 20 = 37
37 + 13 = 50
50 + 12 = 62
62 + 19 = 81
81 + 14 = 95
95 + 9 = 104
We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.



(ii) For constructing a less than ogive, first the given frequency distribution is converted into a more than cumulative frequency distribution as follows. ​
Marks Cumulative Frequency
More than 0
More than 5
More than 10
More than 15
More than 20
More than 25
More than 30
More than 35
More than 40
104
104 − 7 = 97
97 − 10 = 87
87 − 20 = 67
67 − 13 = 54
54 − 12 = 42
42 − 19 = 23
23 − 14 = 9
9 − 9 = 0

We now plot the cumulative frequencies against the lower limit of the class intervals. The curve obtained on joining the points so plotted is known as the more than ogive.

Page No 128:

Question 5:

What is meant by ogive or cumulative frequency curve? From the following distribution construct the 'less than' ogive:

Capital (in lakh) 0−10 10−20 20−30 30−40 40−50 50−60 60−70
Number of Companies 2 3 7 11 15 7 23

Answer:

Ogives or cumulative frequency curve is a smooth distribution curve that depicts cumulative frequency data on a graph paper. For constructing a less than ogive, first the less than cumulative frequencies must be calculated as follows.
 

Capital
(in lakh)
Cumulative Frequency
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
Less than 60
Less than 70
2
2 + 3 = 5
5 + 7 = 12
12 + 11 = 23
23 + 15 = 38
38 + 7 = 45
45 + 23 = 68

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive. 

Page No 128:

Question 6:

Present the data given in the table below in a histogram:

Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Frequency 4 10 16 22 26 18 8 2

Answer:

Page No 128:

Question 7:

Draw a histogram from the following data relating to the monthly pocket allowance of the students of Class XI of a school:

Size 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40
Number of Students 5 10 15 20 25 15 10 5

Answer:



Page No 154:

Question 8:

We are given the following marks secured by 25 students in an examination.
23, 28,30, 32, 35, 36, 36, 40, 41, 43, 44, 44, 45, 48, 49, 52, 53, 54, 56, 56, 58, 61, 62, 65, 68.
(i) Arrange this data in the form of a frequency distribution taking the following class intervals.
20−29, 30−39, 40−49, 50−59 and 60−69
(ii) Draw the frequency polygon and ogive for the above data.

Answer:

(i) The given data is arranged in the form of a frequency distribution as follows.

Class Interval
(Marks)
Frequency
(No. of Student)
20 − 29
30 − 39
40 − 49
50 − 59
60 − 69
2
5
8
6
4
  Σf = 25
 
In order to draw a frequency polygon without a histogram, we calculate the mid-points of each of the class intervals and plot them on a graph against their respective frequencies. The curve obtained on joining the points is the frequency polygon.
Marks Mid Value No. of Students
20 − 29 20 + 292=24.5 2
30 − 39 30 + 392=34.5 5
40 − 49 40 + 492=44.5 8
50 − 59 50 + 592=54.5 6
60 − 69 60 + 692=64.5 4
Total   25


In order to draw a less than ogive, we first convert the frequency distribution as prepared above in a less than cumulative frequency distribution as follows.
Less than Ogive Cumulative Frequency
Less than 29
Less than 39
Less than 49
Less than 59
Less than 69
2
2 + 5 = 7
7 + 8 = 15
15 + 6 = 21
21 + 4 = 25

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.



Page No 155:

Question 9:

Represent the following data in the form of a histogram:

Mid-point 115 125 135 145 155 165 175 185 195
Size 6 55 48 72 116 60 38 22 3

Answer:

In order to construct a histogram, we first require the class intervals corresponding to the various mid-points, which is calculated using the following 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 = 125 -115 2 = 5

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

For instance:

The lower limit of first class = 115 – 5 = 110

Upper limit of first class = 115 + 5 = 120.

Thus, the first class interval is (110-120). Similarly, we can calculate the remaining class intervals.

Mid Point Class Interval Size
115
125
135
145
155
165
175
185
195
110 − 120
120 − 130
130 − 140
140 − 150
150 − 160
160 − 170
170 − 180
180 − 190
190 − 200
6
55
48
72
116
60
38
22
3

Page No 155:

Question 10:

The frequency distribution of marks obtained by students in a class test is given below. Draw frequency polygon and ogive.

Marks 0−10 10−20 20−30 30−40 40−50
Number of Students 5 10 14 10 3

Answer:

In order to construct a frequency polygon without a histogram, we first calculate the mid-points corresponding to the class intervals. The mid-points are then plotted against their respective frequencies. The curve obtained on joining the points is the frequency polygon.

Marks Mid Point Students
0 − 10 0+102=5 3
10 − 20 10+202=15 10
20 − 30 20+302=25 14
30 − 40 30+402=35 10
40 − 50 40+502=45 3


In order to construct the ogives, we first need to calculate the less than and the more than cumulative frequencies as follows.
 
Marks Cumulative Frequency
Less than 10
Less than 20
Less than 30
Less than 40
Less than 50
3
3 + 10 = 13
13 + 14 = 27
27 + 10 = 37
37 + 3 = 40
              
Marks Cumulative Frequency
More than 0
More than 10
More than 20
More than 30
More than 40
40
40 − 3 = 37
37 − 10 = 27
27 − 14 = 13
13 − 10 = 3

Page No 155:

Question 11:

(i) Construct a histogram and frequency polygon of the following distribution:

Marks 0−10 10−20 20−30 30−40 40−50
Number of Students 8 18 35 25 14
(ii) Show that the area under frequency polygon is equal to the area under histogram.

Answer:

(a)

(b)
The area under a histogram and under a frequency polygon is the same because of the fact that we extend the first class interval to the left by half the size of class interval as the starting point of the frequency polygon. Similarly, the last class interval is extended to the right by the same amount as the end point of the frequency polygon. This ensures that the area that was excluded while joining the mid-points is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same. 

Page No 155:

Question 12:

Draw a frequency polygon from the following data by using (i) histogram, and (ii) without using histogram:

Daily Wages (in ₹) 10−15 15−20 20−25 25−30 30−35
Number of Workers 40 70 60 80 60

Answer:

Using a histogram, a frequency polygon is obtained by joining the top mid-points of the rectangles of the histogram.


When frequency polygon is to be drawn without a histogram then, we simply plot the mid-points of the class intervals with their respective frequencies. The curve obtained on joining the points is the frequency polygon.
 

Daily Wages Mid Point No. of Workers
10 − 15 10+152=12.5 40
15 − 20 10+152=17.5 70
20 − 25 10+152=22.5 60
25 − 30 10+152=27.5 80
30 − 35 10+152=32.5 60
 

Page No 155:

Question 13:

Draw 'less than' as well as 'more than' ogives for the following data:

Weight (in Kg) 30−34 35−39 40−44 45−49 50−54 55−59 60−64
Frequency 3 5 12 18 14 6 2

Answer:

Before proceeding to construct the ogives, we first need to convert the given inclusive series into an exclusive series using the following formula. 

The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class.
Here, the value of adjustment = 35 - 342 = 0.5

Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.

Weight Frequency
29.5 − 34.5
34.5 − 39.5
39.5 − 44.5
44.5 − 49.5
49.5 − 54.5
54.5 − 59.5
59.5 − 64.5
3
5
12
18
14
6
2

Now for constructing a less than ogive, we convert the frequency distribution into a less than cumulative frequency distribution as follows. 
Weight Cumulative Frequency
Less than 34.5
Less than 39.5
Less than 44.5
Less than 49.5
Less than 54.5
Less than 59.5
Less than 64.5
3
3 + 5 = 8
8 + 12 = 20
20 + 18 = 38
38 + 14 = 52
52 + 6 = 58
58 + 2 = 60

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.


For constructing a more than ogive, we convert the frequency distribution into a more than cumulative frequency distribution as follows.
Weight Cumulative Frequency
More than 0
More than 34.5
More than 39.5
More than 44.5
More than 49.5
More than 54.5
More than 59.5
More than 64.5
60
60 − 3 = 57
57 − 5 = 52
52 − 12 = 40
40 − 18 = 22
22 − 14 = 8
8 − 6 = 2
2 − 2 = 0
We now plot the cumulative frequencies against the lower limit of the class intervals. The curve obtained on joining the points so plotted is known as the more than ogive.



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