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Question 9
Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.

Section ASection BMarksFrequencyMarksFrequency01550153153012153016304528304525456030456027607535607540759013759010

Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?


Solution

Firstly, we find the mid marks of the given sections A and B by using the formula,
Class mark =Lower limit+Upper limit2

The new table for section A and B is shown below.

Section ASection BMarksMid marksFrequencyMarksMid marksFrequency0157.550157.53153022.512153022.516304537.528304537.525456052.530456052.527607567.535607567.540759082.513759082.510

We can draw a frequency polygon by plotting the class marks along the horizontal axis and the frequency along the vertical axis.
Now, plotting all the points A(7.5, 5), B(22.5, 12), C(37.5, 28), D(52.5, 30), E(67.5, 35), F(82.5, 13) for section A.
Also, plotting all the points H(7.5, 3), l(22.5, 16), J(37.5, 25), K(52.5, 27), L(67.5, 40) and M(82.5, 10) for section B.


It is clear from the graph that maximum marks 67.5 scored by 40 students in section B.

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