WAEC 2019 · Paper 2 · Q12

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Number of students	5	5	10	18	23	23	9	4	2	1

The table shows the marks obtained by students in an examination.

(a)

Construct a cumulative frequency table for the distribution.

Model answer

Marks	Frequency	Upper class boundary	Cumulative frequency
$0\text{–}9$	$5$	$9.5$	$5$
$10\text{–}19$	$5$	$19.5$	$10$
$20\text{–}29$	$10$	$29.5$	$20$
$30\text{–}39$	$18$	$39.5$	$38$
$40\text{–}49$	$23$	$49.5$	$61$
$50\text{–}59$	$23$	$59.5$	$84$
$60\text{–}69$	$9$	$69.5$	$93$
$70\text{–}79$	$4$	$79.5$	$97$
$80\text{–}89$	$2$	$89.5$	$99$
$90\text{–}99$	$1$	$99.5$	$100$

The last cumulative frequency, 100, is the total number of students.

(b)
Draw an ogive 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 $(99.5, 100)$ . 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 (c): read across from 50 for the median (about 44.8), and from 25 and 75 for the quartiles (about 32.7 and 55.2), giving a semi-interquartile range of about 11.2.
(c)
Use the ogive to determine the: (i) median mark; (ii) semi-interquartile range.
Separate values with commas, e.g. 3, −2
(d)
If a student is selected at random, what is the probability that he obtained at least 60 marks?

The ogive with the quartile and median readings.

Worked solution (try it first)

Upper boundaries $9.5, 19.5, \ldots, 99.5$ with cumulative frequencies $5, 10, 20, 38, 61, 84, 93, 97, 99, 100$ .

Read across from 50: median $\approx 39.5 + \dfrac{12}{23} \times 10$
$\approx 44.7$ .

$Q_1 \approx 29.5 + \dfrac{5}{18} \times 10$
$\approx 32.3$ and $Q_3 \approx 49.5 + \dfrac{14}{23} \times 10$
$\approx 55.6$ , so the semi-interquartile range is $\frac12(55.6 - 32.3) \approx 11.7$ .

At least 60 marks: $9 + 4 + 2 + 1 = 16$ students, so $P = \dfrac{16}{100} = 0.16$ .