The following table gives the distribution of students of two sections according to the marks obtained by them:
Section A | Section B | ||
Marks | Frequency | Marks | Frequency |
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 | 3 9 17 9 | 0 – 10 10 20 20 - 30 30 - 40 40 - 50 | 5 19 15 10 1 |
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.
The class marks can be found by
= 
Now,
For section A,
MARKS | CLSS MARKS | FREQUENCY |
0-10 | 5 | 3 |
10-20 | 15 | 9 |
20-30 | 25 | 17 |
30-40 | 35 | 12 |
40-50 | 45 | 9 |
For section B,
MARKS | CLSS MARKS | FREQUENCY |
0-10 | 5 | 5 |
10-20 | 15 | 19 |
20-30 | 25 | 15 |
30-40 | 35 | 10 |
40-50 | 45 | 1 |
Now, the frequency polygon for the given data:
28