Draw a frequency polygon for the following data using histogram.
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Number of students 4 6 8 5 7 4 9 5 7
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
| Number of students | 4 | 6 | 8 | 5 | 7 | 4 | 9 | 5 | 7 |
Answer
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Hint:
We will draw a histogram for the given data. We will represent the frequency (number of students) for each class of marks using a bar. Then we will locate the midpoint of the upper horizontal side of each bar. Then we will join all the mid-points using line segments one after the other and this will give us the required frequency polygon.
Complete step by step solution:
We will make a histogram to represent the data in the table. On the X-axis, we will take all the class intervals (marks of students) and on the Y-axis, we will take the frequency (number of students).
There are no students who have got marks between 0 and 10, so we will make a class interval 0-10 (because the values on the x-axis start from 0.If we want to take values from 10, we will have to draw a kink)and take its frequency as 0.
Now we will find the midpoint of the upper horizontal side of each bar of the histogram. The midpoint of a class is given by \[M = \dfrac{{U + L}}{2}\] where M is the midpoint, U is the upper class limit and L lower class limit.
We will represent the data in the above table using a frequency polygon. We will represent the mid-points on the x-axis and the frequency on the y-axis.
Note:
Here, we have drawn a histogram because it becomes much easier to show different data of different ranges through bars. The taller bar represents more value in a particular range. We can also directly draw the polygon without making the histogram. We will directly find the mid-points of the given intervals and plot their corresponding frequencies on the y-axis. We will represent the mid-points on the x-axis.
We will draw a histogram for the given data. We will represent the frequency (number of students) for each class of marks using a bar. Then we will locate the midpoint of the upper horizontal side of each bar. Then we will join all the mid-points using line segments one after the other and this will give us the required frequency polygon.
Complete step by step solution:
We will make a histogram to represent the data in the table. On the X-axis, we will take all the class intervals (marks of students) and on the Y-axis, we will take the frequency (number of students).
There are no students who have got marks between 0 and 10, so we will make a class interval 0-10 (because the values on the x-axis start from 0.If we want to take values from 10, we will have to draw a kink)and take its frequency as 0.
Now we will find the midpoint of the upper horizontal side of each bar of the histogram. The midpoint of a class is given by \[M = \dfrac{{U + L}}{2}\] where M is the midpoint, U is the upper class limit and L lower class limit.
| Marks | Midpoint of the bar | Height of the bar (Frequency) |
| 0-10 | \[\dfrac{{10 + 0}}{2} = 5\] | 0 |
| 10-20 | \[\dfrac{{20 + 10}}{2} = 15\] | 4 |
| 20-30 | \[\dfrac{{30 + 20}}{2} = 25\] | 6 |
| 30-40 | \[\dfrac{{40 + 30}}{2} = 35\] | 8 |
| 40-50 | \[\dfrac{{50 + 40}}{2} = 45\] | 5 |
| 50-60 | \[\dfrac{{60 + 50}}{2} = 55\] | 7 |
| 60-70 | \[\dfrac{{70 + 60}}{2} = 65\] | 4 |
| 70-80 | \[\dfrac{{80 + 70}}{2} = 75\] | 9 |
| 80-90 | \[\dfrac{{90 + 80}}{2} = 85\] | 5 |
| 90-100 | \[\dfrac{{100 + 90}}{2} = 95\] | 7 |
We will represent the data in the above table using a frequency polygon. We will represent the mid-points on the x-axis and the frequency on the y-axis.
Note:
Here, we have drawn a histogram because it becomes much easier to show different data of different ranges through bars. The taller bar represents more value in a particular range. We can also directly draw the polygon without making the histogram. We will directly find the mid-points of the given intervals and plot their corresponding frequencies on the y-axis. We will represent the mid-points on the x-axis.
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