WAEC 2013 · Paper 2 · Q11

Marks (%)	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Frequency	2	3	5	13	19	31	13	9	4	1

The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary ( $10.5, 20.5, \ldots, 100.5$ ), starting from $(0.5, 0)$ and ending at $(100.5, 100)$ , and join the points with a smooth S-shaped curve. Label both axes.

For (b): across from 60 the curve gives the 60th percentile, about 56; up from 34.5 it reads about 15, so about 85 of the 100 passed and the probability is about $0.85$ .
(b)
Use the graph to find the: (i) 60th percentile; (ii) probability that a student passed the test if the pass mark was fixed at $35\%$ .
Separate values with commas, e.g. 3, −2

The ogive with the 60th-percentile reading.

Worked solution (try it first)

Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Upper boundary	10.5	20.5	30.5	40.5	50.5	60.5	70.5	80.5	90.5	100.5
Cumulative frequency	2	5	10	23	42	73	86	95	99	100

Plot each cumulative frequency at its upper boundary, starting from $(0.5, 0)$ , and join the points with a smooth S-shaped curve.

The 60th percentile is at $\frac{60}{100} \times 100 = 60$ on the cumulative frequency axis.
Go across to the curve and down: about 56.
(As a check, 60 lies between 42 at 50.5 and 73 at 60.5: $50.5 + \frac{60 - 42}{31} \times 10 \approx 56.3$ .)

A pass is 35 or more.
Go up from 34.5 (the boundary below 35) to the curve and across: about 15 students scored less than 35.
So about $100 - 15 = 85$ passed, and $P(\text{passed}) \approx \frac{85}{100}$
$= 0.85$ .
A reading close to this from your own curve is fine.