WAEC 2010 · Paper 2 · Q12

The frequency distribution of the weight of 100 participants in a high jump competition is as shown below.

Weight (kg)	20–29	30–39	40–49	50–59	60–69	70–79
Number of participants	10	18	22	25	16	9

(a)

Construct the cumulative frequency table.

Model answer

Weight (kg)	Frequency	Upper class boundary	Cumulative frequency
$20\text{–}29$	$10$	$29.5$	$10$
$30\text{–}39$	$18$	$39.5$	$28$
$40\text{–}49$	$22$	$49.5$	$50$
$50\text{–}59$	$25$	$59.5$	$75$
$60\text{–}69$	$16$	$69.5$	$91$
$70\text{–}79$	$9$	$79.5$	$100$

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

(b)
Draw the cumulative frequency curve.
Model answer
Plot each cumulative frequency against its upper class boundary, starting from $(19.5, 0)$ , and join the points with a smooth S-shaped curve. For (c): reading across from 50 gives the median, about $49.5\text{ kg}$ , and from 25 and 75 the quartiles, about $37.8\text{ kg}$ and $59.5\text{ kg}$ .
(c)
From the curve, estimate the: (i) median; (ii) semi-interquartile range; (iii) probability that a participant chosen at random weighs at least $60\text{ kg}$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive: read the median at 50 and the quartiles at 25 and 75.

Worked solution (try it first)

(a)

Add the frequencies as you go: $10, 28, 50, 75, 91, 100$ .
Pair each total with the upper class boundary: $29.5, 39.5, 49.5, 59.5, 69.5, 79.5$ .

(b)

Plot cumulative frequency against upper class boundary, starting at $(19.5, 0)$ , and join the points with a smooth curve.

(c)(i)

(ii)

Read the quartiles at $\frac{100}{4} = 25$ and $\frac{3 \times 100}{4} = 75$ : $Q_1 \approx 37.8\text{ kg}$ and $Q_3 \approx 59.5\text{ kg}$ .
The semi-interquartile range is $\frac12(Q_3 - Q_1) = \frac12(59.5 - 37.8)$
$\approx 10.8\text{ kg}$ .

(iii)

Weights of at least $60\text{ kg}$ start at the boundary $59.5$ , where the curve reads 75.
So $100 - 75 = 25$ participants weigh at least $60\text{ kg}$ .
The probability is $\frac{25}{100} = \frac14$ .