Question

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

Section A		Section B
Marks	Frequency	Marks	Frequency
0 – 15	5	0 – 15	3
15 – 30	12	15 – 30	16
30 – 45	28	30 – 45	25
45 – 60	30	45 – 60	27
60 –75	35	60 – 75	40
75 – 90	13	75 – 90	10

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

Answer 1

Firstly, we find the mid marks of the given sections A and B by using the formula

Class mark = `("Lower limit" + "Upper limit")/2`

So, the new table for section A and section B is shown below:

Section A			Section B
Marks	Mid marks	Frequency	Marks	Mid marks	Frequency
0 – 15	7.5	5	0 – 15	7.5	3
15 – 30	22.5	12	15 – 30	22.5	16
30 – 45	37.5	28	30 – 45	37.5	25
45 – 60	52.5	30	45 – 60	52.5	27
60 –75	67.5	35	60 – 75	67.5	40
75 – 90	82.5	13	75 – 90	82.5	10

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), I(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 score by 40 students in section B.