WAEC 2015 · Paper 2 · Q13

Marks (%)	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	7	11	17	20	29	34	30	25	21	6

The table shows the marks scored by some candidates in an examination.

(a)
Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
Model answer

Upper boundary 9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5

Cumulative frequency 7 18 35 55 84 118 148 173 194 200

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

For (b): across from 190 ( $95\%$ of 200) the curve gives about 87.6; up from 45.5 it reads about 72, so the probability is about $\frac{72}{200} = 0.36$ .
(b)
Use the curve to estimate, correct to one decimal place, the: (i) lowest mark for distinction if $5\%$ of the candidates passed with distinction; (ii) probability of selecting a candidate who scored at most $45\%$ .
Separate values with commas, e.g. 3, −2

Upper boundary	9.5	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
Cumulative frequency	7	18	35	55	84	118	148	173	194	200

The ogive.

Worked solution (try it first)

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Upper boundary	9.5	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
Cumulative frequency	7	18	35	55	84	118	148	173	194	200

Plot $(9.5, 7), (19.5, 18), \ldots, (99.5, 200)$ , starting from $(-0.5, 0)$ , and draw a smooth S-shaped curve through them.

If the top $5\%$ got a distinction, then $95\%$ scored below the lowest distinction mark.
$95\%$ of 200 is 190.
Go across from 190 to the curve and down: about 87.6.
(Check: 190 lies between 173 at 79.5 and 194 at 89.5, and $79.5 + \frac{190 - 173}{21} \times 10 \approx 87.6$ .)

"At most $45\%$ " means 45 or less, so read the curve at 45.5: go up from 45.5 and across, about 72 candidates.
The probability is about $\frac{72}{200} = 0.36$ .
A reading close to this from your own curve is fine.