Tr Jain Vk Ohri 2019 Solutions for Class 11 Humanities Economics Chapter 7 Frequency Diagrams – Histogram, Polygon And Ogive are provided here with simple stepbystep explanations. These solutions for Frequency Diagrams – Histogram, Polygon And Ogive are extremely popular among Class 11 Humanities 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 2019 Book of Class 11 Humanities Economics Chapter 7 are provided here for you for free. You will also love the adfree experience on Meritnation’s Tr Jain Vk Ohri 2019 Solutions. All Tr Jain Vk Ohri 2019 Solutions for class Class 11 Humanities Economics are prepared by experts and are 100% accurate.
Page No 149:
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 151:
Question 1:
Collect data on the CBSE result (2018) 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 155:
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 
010
1020 2030 3040 4050 5060 6070 
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 166:
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 166:
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 midpoints 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 midpoints of the rectangles of the histogram drawn in the first part, but with a free hand (rather than a straight line).
Page No 166:
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 midpoints of the rectangle of the histogram as drawn above in the first part using a straight line.
Page No 166:
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 
(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 167:
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 167:
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 167:
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 167:
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 
$\Sigma f=25$ 
In order to draw a frequency polygon without a histogram, we calculate the midpoints 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  $\frac{20+29}{2}=24.5$  2 
30 − 39  $\frac{30+39}{2}=34.5$  5 
40 − 49  $\frac{40+49}{2}=44.5$  8 
50 − 59  $\frac{50+59}{2}=54.5$  6 
60 − 69  $\frac{60+69}{2}=64.5$  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 167:
Question 9:
Represent the following data in the form of a histogram:
Midpoint  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 midpoints, 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 midpoint to obtain the lower limit.
For the given data, the class interval is calculated by the following value of adjustment.
$\mathrm{Value}\mathrm{of}\mathrm{Adjustment}=\frac{125115}{2}=5$
Thus, we add and subtract 5 to each midpoint 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 (110120). 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 167:
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 midpoints corresponding to the class intervals. The midpoints are then plotted against their respective frequencies. The curve obtained on joining the points is the frequency polygon.
Marks  Mid Point  Students 
0 − 10  $\frac{0+10}{2}=5$  3 
10 − 20  $\frac{10+20}{2}=15$  10 
20 − 30  $\frac{20+30}{2}=25$  14 
30 − 40  $\frac{30+40}{2}=35$  10 
40 − 50  $\frac{40+50}{2}=45$  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 167:
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 
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 midpoints is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same.
Page No 168:
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 midpoints of the rectangles of the histogram.
When frequency polygon is to be drawn without a histogram then, we simply plot the midpoints 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  $\frac{10+15}{2}=12.5$  40 
15 − 20  $\frac{10+15}{2}=17.5$  70 
20 − 25  $\frac{10+15}{2}=22.5$  60 
25 − 30  $\frac{10+15}{2}=27.5$  80 
30 − 35  $\frac{10+15}{2}=32.5$  60 
Page No 168:
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.
$\mathrm{Here},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{adjustment}=\frac{3534}{2}=0.5$
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 
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