Question

# 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 BMarksFrequencyMarksFrequency0−1550−15315−301215−301630−452830−452545−603045−602760−753560−754075−901375−9010 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 marksFrequency0−157.550−157.5315−3022.51215−3022.51630−4537.52830−4537.52545−6052.53045−6052.52760−7567.53560−7567.54075−9082.51375−9082.510 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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