WAEC 2017 · Paper 2 · Q6

Height (cm) 36–40 41–45 46–50 51–55 56–60
Frequency 3 9 21 12 5

The table shows the heights, in cm, of some seedlings in a certain garden.

  1. (a)

    Draw a cumulative frequency curve for the distribution.

    Model answer
    35.540.545.550.555.560.51020304050Height (cm)Cumulative frequency

    Plot the cumulative frequencies 3,12,33,45,503, 12, 33, 45, 50 against the upper class boundaries 40.5,…,60.540.5, \ldots, 60.5, starting from (35.5,0)(35.5, 0), and join them with a smooth S-shaped curve. For (b): reading across from 12.5 and 37.5 gives Q1≈45.6Q_1 \approx 45.6 and Q3≈52.4Q_3 \approx 52.4. Readings from a hand-drawn curve vary a little; examiners accept a small range.

  2. (b)

    Using the curve, find the semi-interquartile range.

Try it on a graph

The ogive with the quartile readings.

Worked solution (try it first)

(a)

  1. Plot the cumulative frequencies 3,12,33,45,503, 12, 33, 45, 50 at the upper boundaries 40.5,45.5,50.5,55.5,60.540.5, 45.5, 50.5, 55.5, 60.5, starting from (35.5,0)(35.5, 0), and join with a smooth curve.

(b)

  1. N=50N = 50.
  2. Read across from 12.5 and 37.5: Q1≈45.6Q_1 \approx 45.6 and Q3≈52.4Q_3 \approx 52.4.
  3. Semi-interquartile range =12(52.4−45.6)≈3.4= \frac12(52.4 - 45.6) \approx 3.4.

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