# The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

**Solution:**

Frequency polygons can be drawn independently without drawing histograms. This requires the midpoints of the class intervals used in the data. The mid-points are called class-marks.

Class Mark = (Upper Limit + Lower Limit)/2

Now, the data table with the inclusion of class marks is as follows:

The frequency polygon for the above data can be constructed by:

- Taking class marks on the
*x*-axis. - Taking ‘frequency’ on the
*y*-axis with an appropriate scale of 1 unit = 2 students as the lowest frequency value observed is 1 and the highest frequency value is 19.

It can be observed that the performance of students of Section ‘A’ is better than the students of Section B as section ‘A’ shows more students securing marks between class intervals ‘40 – 50’ and ‘30 – 40’.

**Video Solution:**

## The following table gives the distribution of students of two sections according to the marks obtained by them: Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

### NCERT Solutions for Class 9 Maths - Chapter 14 Exercise 14.3 Question 6:

It is given that distribution of students of two sections according to the marks obtained. The marks of the students of both the sections has been represented on the same graph by two frequency polygons. It can be observed that the performance of students of Section ‘A’ is better than the students of Section B as section ‘A’ shows more students securing marks between class intervals ‘40 – 50’ and ‘30 – 40’ class.