
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
587.1k+ views
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.
Recently Updated Pages
Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Trending doubts
A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

State and explain Ohms law class 10 physics CBSE

Distinguish between soap and detergent class 10 chemistry CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE

What is a "free hit" awarded for in limited-overs cricket?

Draw the diagram of the sectional view of the human class 10 biology CBSE

