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WAEC 2008 · Paper 2 · Q6

The table shows the distribution of marks obtained by some candidates in a test.

Marks 10–14 15–24 25–29 30–39 40–44 45–49
Number of candidates 14 30 22 18 12 4
  1. (a)

    Draw a histogram for the distribution.

    Model answer
    9.514.524.529.539.544.549.5510152025MarksFrequency per 5 marks

    The classes have unequal widths (5 or 10 marks), so the bar heights are frequency densities, not frequencies. Use the class boundaries 9.5, 14.5, 24.5, 29.5, 39.5, 44.5 and 49.5, with no gaps between bars.

    Taking 5 marks as the standard width, the heights are 14, 15, 22, 9, 12 and 4: the two classes of width 10 have their frequencies halved. The area of each bar is then in proportion to its frequency.

Worked solution (try it first)
  1. Find the class boundaries: 9.5,14.5,24.5,29.5,39.5,44.5,49.59.5, 14.5, 24.5, 29.5, 39.5, 44.5, 49.5.
  2. Find the class widths: 5,10,5,10,5,55, 10, 5, 10, 5, 5.
  3. They are not all equal.
  4. With unequal widths, the height of a bar is the frequency density.
  5. Take 5 marks as the standard width: height =frequencywidth×5= \dfrac{\text{frequency}}{\text{width}} \times 5.
  6. For 15–24: 3010×5=15\frac{30}{10} \times 5 = 15.
  7. For 30–39: 1810×5=9\frac{18}{10} \times 5 = 9.
  8. The other classes already have width 5, so their heights are their frequencies: 14, 22, 12 and 4.
  9. Draw bars on the boundaries with heights 14, 15, 22, 9, 12, 4, no gaps, and label the axes "Marks" and "Frequency per 5 marks".

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