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प्रश्न
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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उत्तर
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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Represent the above data by drawing a suitable bar graph.
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?
