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Question
The following table gives the distribution of students of two sections according to the mark obtained by them:-
| Section A | Section B | ||
| Marks | Frequency | Marks | Frequency |
| 0 - 10 | 3 | 0 - 10 | 5 |
| 10 - 20 | 9 | 10 - 20 | 19 |
| 20 - 30 | 17 | 20 - 30 | 15 |
| 30 - 40 | 12 | 30 - 40 | 10 |
| 40 - 50 | 9 | 40 - 50 | 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.
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Solution
We can find the class marks of the given class intervals by using the following formula.
Class mark = `"Upper class limit + Lower class limit"/2`
| Section A | Section B | ||||
| Marks | Class-marks | Frequency | Marks | Class-marks | Frequency |
| 0 - 10 | 5 | 3 | 0 - 10 | 5 | 5 |
| 10 - 20 | 15 | 9 | 10 - 20 | 15 | 19 |
| 20 - 30 | 25 | 17 | 20 - 30 | 25 | 15 |
| 30 - 40 | 35 | 12 | 30 - 40 | 35 | 10 |
| 40 - 50 | 45 | 9 | 40 - 50 | 45 | 1 |
Taking class marks on the x-axis and frequency on the y-axis and choosing an appropriate scale (1 unit = 3 for the y-axis), the frequency polygon can be drawn as follows:
It can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.
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