WAEC 2011 · Paper 2 · Q14

The table shows the frequency distribution of marks scored by some candidates in an examination.

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	2	5	8	18	20	15	5	4	2	1

(a)
Draw the cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary of its class, starting from $(-0.5, 0)$ where the cumulative frequency is 0 and ending at $(99.5, 80)$ . Join the points with a smooth rising S-shaped curve (an ogive), not straight lines. Label both axes. Readings from a hand-drawn curve differ a little from person to person; examiners accept a small range, usually about ±1.

For (b): read across from 20 and 60 (a quarter and three-quarters of 80) to the curve and down: $Q_1 \approx 32.7$ and $Q_3 \approx 53.6$ , so the semi-interquartile range is about $10.4$ . Read up from 72 to the curve: about 74 candidates scored less, so about 6 (roughly 7.4%) had a distinction.
(b)
Use your graph to estimate the: (i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.
Separate values with commas, e.g. 3, −2

Cumulative frequency against upper class boundary. Readings: Q₁ at 20, Q₃ at 60, and the mark 72.

Worked solution (try it first)

Upper class boundaries $9.5, 19.5, \ldots, 99.5$ with cumulative frequencies $2, 7, 15, 33, 53, 68, 73, 77, 79, 80$ .
Plot each against its upper boundary and join with a smooth curve.

Read up from 72: about 74 candidates scored less than 72, so about $80 - 74 = 6$ passed with distinction: $\frac{6}{80} \times 100 = 7.5\%$ .