WAEC 2019 · Paper 2 · Q7

The data show the marks obtained by students in a Biology test.

50 56 25 56 68 73 66 64 56 48 20 39 9 50 46 54 54 40 50 96 36 44 18 97 65 21 60 44 54 32 92 49 37 94 72 88 89 35 59 34 15 88 53 16 84 52 72 46 60 42

(a)
Construct a frequency distribution table using the class intervals $0$ – $9$ , $10$ – $19$ , $20$ – $29$ , …
Model answer

Class interval Frequency

0–9 1

10–19 3

20–29 3

30–39 6

40–49 8

50–59 12

60–69 6

70–79 3

80–89 4

90–99 4

Total 50

Tally each score into its class, then count; the frequencies must add up to 50.
(b)
Draw a cumulative frequency curve 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 $(99.5, 50)$ . 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): the median is the 25th mark. Read across from 25: about 52.9. Read up from 65.5: about 37 students scored below 66, so about 13 of the 50 (26%) scored at least 66.
(c)
Use the graph to estimate the: (i) median; (ii) percentage of students who scored at least 66 marks, correct to the nearest whole number.
Separate values with commas, e.g. 3, −2

The ogive: cumulative frequency at each upper class boundary.

Worked solution (try it first)

Tally each mark into its class:
Marks 0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 1 3 3 6 8 12 6 3 4 4
The frequencies add up to 50.

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	1	3	3	6	8	12	6	3	4	4

The cumulative frequencies are $1, 4, 7, 13, 21, 33, 39, 42, 46, 50$ .
Plot them at the upper class boundaries $9.5, 19.5, \ldots, 99.5$ , start at $(-0.5, 0)$ , and draw a smooth curve.

The median is at $\frac{50}{2} = 25$ .
Go across to the curve and down: about 53.
(Check: 25 lies between 21 at 49.5 and 33 at 59.5, and $49.5 + \frac{4}{12} \times 10 \approx 52.8$ .)

"At least 66" means 66 or more.
Go up from 65.5 and across: about 36.6, so about 37 students scored less than 66 and about $50 - 37 = 13$ scored at least 66.
As a percentage: $\frac{13}{50} \times 100 = 26\%$ .