In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.
| No. of days | 8–10 | 10–12 | 12–14 | 14–16 | 16–18 | 18–20 |
|---|---|---|---|---|---|---|
| No. of workers | 5 | 16 | 30 | 40 | 35 | 14 |
Answer: Plot (9, 5), (11, 16), (13, 30), (15, 40), (17, 35), (19, 14) and join them, closing at (7, 0) and (21, 0)
| Class | Class mark | Frequency | Point |
|---|---|---|---|
| 8–10 | 9 | 5 | (9, 5) |
| 10–12 | 11 | 16 | (11, 16) |
| 12–14 | 13 | 30 | (13, 30) |
| 14–16 | 15 | 40 | (15, 40) |
| 16–18 | 17 | 35 | (17, 35) |
| 18–20 | 19 | 14 | (19, 14) |
Take class marks (number of days) on the X-axis and number of workers on the Y-axis with a convenient scale. Plot the points and also the points (7, 0) and (21, 0), which are the class marks of the imaginary classes just before the first and just after the last class (frequency 0). Join all the points in order by line segments:
(7, 0), (9, 5), (11, 16), (13, 30), (15, 40), (17, 35), (19, 14), (21, 0)
The closed figure obtained with the X-axis is the required frequency polygon.
Log in or create a free account to answer this question.
