The marks scored by 35 students in a test are given in the table below. Marks 1-5 6-10 11-15 16-20 21-25 26-30 Frequency 2 7 12 8 5 1 Draw a histogram for t...
Question 1 Report
The marks scored by 35 students in a test are given in the table below.
| Marks | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 |
| Frequency | 2 | 7 | 12 | 8 | 5 | 1 |
Draw a histogram for the distribution.
Answer Details
All six classes have the same width. Each class spans five marks (\(1\text{-}5,\ 6\text{-}10,\ \dots,\ 26\text{-}30\)), so the class widths are equal and an ordinary frequency histogram may be drawn: the bar heights are simply the frequencies, and there is no need to adjust for frequency density. First convert the class limits to continuous class boundaries by subtracting \(0.5\) from each lower limit and adding \(0.5\) to each upper limit.
| Marks | Class boundaries | Width | Frequency |
|---|---|---|---|
| 1 - 5 | 0.5 - 5.5 | 5 | 2 |
| 6 - 10 | 5.5 - 10.5 | 5 | 7 |
| 11 - 15 | 10.5 - 15.5 | 5 | 12 |
| 16 - 20 | 15.5 - 20.5 | 5 | 8 |
| 21 - 25 | 20.5 - 25.5 | 5 | 5 |
| 26 - 30 | 25.5 - 30.5 | 5 | 1 |
The bars are drawn touching (no gaps), because the class boundaries are continuous. The frequencies total \(2+7+12+8+5+1 = 35\), matching the 35 students.
The tallest bar, over the class \(11\text{-}15\), is the modal class. Reading the peak of the histogram gives an estimate of the mode within that interval.
Examination note: do not merge or re-band the classes. The intervals are already equal in width, so each bar height equals its frequency; changing the boundaries to unequal widths would distort the distribution and lose marks.
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