Answer: Histogram with heights 8, 16, 22, 18, 14, 12; frequency polygon through (19, 0), (21, 8), (23, 16), (25, 22), (27, 18), (29, 14), (31, 12), (33, 0)
Histogram
| Time (minutes) | Number of students |
|---|
| 20–22 | 8 |
| 22–24 | 16 |
| 24–26 | 22 |
| 26–28 | 18 |
| 28–30 | 14 |
| 30–32 | 12 |
Steps: Take time (minutes) on the X-axis and number of students on the Y-axis, with a convenient scale on each. Since the first class does not start at 0, show a kink mark between the origin and the first class. On each class interval draw a rectangle whose height equals its frequency; adjacent rectangles touch each other (no gaps). The tallest rectangle is for the class 24–26 (frequency 22).
Frequency polygon: mark the mid-points of the upper sides of the rectangles, i.e. the points below, and join them in order.
| Class | Class mark | Frequency | Point |
|---|
| 20–22 | 21 | 8 | (21, 8) |
| 22–24 | 23 | 16 | (23, 16) |
| 24–26 | 25 | 22 | (25, 22) |
| 26–28 | 27 | 18 | (27, 18) |
| 28–30 | 29 | 14 | (29, 14) |
| 30–32 | 31 | 12 | (31, 12) |
Take class marks (time) on the X-axis and number of students on the Y-axis with a convenient scale. Plot the points and also the points (19, 0) and (33, 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:
(19, 0), (21, 8), (23, 16), (25, 22), (27, 18), (29, 14), (31, 12), (33, 0)
The closed figure obtained with the X-axis is the required frequency polygon.
