WAEC 2019 · Paper 2 · Q7

The following are the scores of 48 students in a promotion test.

52 56 25 56 68 73 66 64 56 48 15 88 20 39 9 50 98 54 54 40 50 96 53 16 36 44 18 97 65 21 60 44 54 32 84 52 92 49 37 94 72 88 89 35 59 34 72 60

(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 5

50–59 12

60–69 6

70–79 3

80–89 4

90–99 5

Total 48

Tally each score into its class, then count; the frequencies must add up to 48.
(b)
Draw a histogram for the distribution.
Model answer
Draw bars on the class boundaries, not the class limits (−0.5, 9.5, …, 99.5), with no gaps between them. The height of each bar is the frequency, and each axis is labelled.

To estimate the mode, take the tallest bar. Join its top-left corner to the top-left corner of the bar on its right, and its top-right corner to the top-right corner of the bar on its left. Read down from where the two lines cross: the mode is about 54.9.
(c)
(i) Use the histogram to estimate the modal score. (ii) If the pass mark for promotion is 30, find, correct to one decimal place, the percentage of students who will be promoted.
Separate values with commas, e.g. 3, −2

Histogram bars (class boundaries 9.5, 19.5, …) with the crossed lines that locate the mode.

Worked solution (try it first)

Tally each score into its class:
Scores 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 5 12 6 3 4 5
The frequencies add up to 48.

Scores	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	5	12	6	3	4	5

Draw the histogram on the class boundaries $-0.5, 9.5, 19.5, \ldots, 99.5$ , with touching bars of heights $1, 3, 3, 6, 5, 12, 6, 3, 4, 5$ .

The tallest bar is 49.5–59.5.
Draw the two crossed lines from its top corners to the neighbouring bars' top corners, and read down: about 55.
(By formula: $49.5 + \frac{12 - 5}{(12 - 5) + (12 - 6)} \times 10 \approx 54.9$ .)

Promoted means a score of 30 or more.
Below 30 there are $1 + 3 + 3 = 7$ students, so $48 - 7 = 41$ are promoted: $\frac{41}{48} \times 100 \approx 85.4\%$ .