WAEC 2008 · Paper 2 · Q14

The following table shows the distribution of marks (%) obtained by some students in an examination.

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Number of students	50	50	40	60	100	100	50	25	15	10

(a)
Construct a cumulative frequency table for the distribution.
Separate values with commas, e.g. 3, −2
(b)
Draw an ogive for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary (9.5, 19.5, …, 99.5), starting from $(-0.5, 0)$ and ending at $(99.5, 500)$ . Join the points with a smooth rising curve and label both axes.

Readings from a hand-drawn curve vary a little: $Q_1 \approx 26$ (at 125) and $Q_3 \approx 57$ (at 375); about 185 students scored below 37; about 102 scored below 20 and about 402 below 60.
(c)(i)
Use your graph in (b) to determine the semi-interquartile range.
(c)(ii)
Use your graph in (b) to determine the number of students who failed, if the pass mark for the examination is 37.
(c)(iii)
Use your graph in (b) to determine the probability that a student selected at random scored between 20% and 60%.

Worked solution (try it first)

Add the frequencies as you go: $50, 100, 140, 200, 300, 400, 450, 475, 490, 500$ .
There are 500 students.
Pair each total with its upper class boundary: $9.5, 19.5, 29.5, \ldots, 99.5$ .

Plot the points $(9.5, 50), (19.5, 100), \ldots, (99.5, 500)$ , starting from $(-0.5, 0)$ , and join them with a smooth curve.

$Q_1$ is at the $\frac{500}{4} = 125$ th student.
Read across from 125 to the curve and down: $Q_1 \approx 25.8$ .
$Q_3$ is at the $375$ th student.
Read across from 375: $Q_3 \approx 57.0$ .
Semi-interquartile range $= \frac12(Q_3 - Q_1)$
$= \frac12(57.0 - 25.8)$
$\approx 15.6$ .