WAEC 2010 · Paper 2 · Q10

The table gives the distribution of marks for 360 candidates who sat for an examination.

Marks (%)	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89
Number of candidates	20	48	60	72	80	40	25	10	5

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary ( $9.5, 19.5, \ldots, 89.5$ ), starting from $(-0.5, 0)$ and ending at $(89.5, 360)$ , and join the points with a smooth S-shaped curve. Label both axes.

For (b) and (c): across from 90 and 270 the curve gives $Q_1 \approx 23.2$ and $Q_3 \approx 48.3$ ; up from 74.5 it reads about 350, so about 10 candidates scored $75\%$ or more.
(b)
Use your graph to estimate the semi-interquartile range.
(c)
If the minimum mark for distinction is $75\%$ , how many candidates passed with distinction?

The ogive: read the quartiles at 90 and 270, and the curve at 74.5.

Worked solution (try it first)

Add the frequencies as you go: $20, 68, 128, 200, 280, 320, 345, 355, 360$ .
Plot each total against its upper class boundary ( $9.5, 19.5, \ldots, 89.5$ ), start at $(-0.5, 0)$ , and join the points with a smooth curve.

Read the quartiles at $\frac{360}{4} = 90$ and $\frac{3 \times 360}{4} = 270$ : $Q_1 \approx 23.2$ and $Q_3 \approx 48.3$ .
Semi-interquartile range $= \frac12(Q_3 - Q_1)$
$= \frac12(48.3 - 23.2)$
$\approx 12.5$ marks.