WAEC 2018 · Paper 2 · Q9

Marks (%)	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	4	7	12	18	20	14	9	4	2

The table shows the distribution of marks scored by students in a test.

(a)

Construct a cumulative frequency table for the distribution.

Model answer

Marks (%)	Frequency	Upper class boundary	Cumulative frequency
10–19	4	19.5	4
20–29	7	29.5	11
30–39	12	39.5	23
40–49	18	49.5	41
50–59	20	59.5	61
60–69	14	69.5	75
70–79	9	79.5	84
80–89	4	89.5	88
90–99	2	99.5	90

Each cumulative frequency is the running total of the frequencies; the last one equals the total, 90.

(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 $(9.5, 0)$ where the cumulative frequency is 0 and ending at $(99.5, 90)$ . 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 45th mark. Read across from 45 to the curve and down: about 51.5. For distinction, read up from 74.5: about 80 students scored below 75, so about 10 of the 90 have a distinction.
(c)
Use the curve to estimate the: (i) median; (ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is $75\%$ .
Separate values with commas, e.g. 3, −2

The ogive with the median (purple) and the 74.5 reading (red).

Worked solution (try it first)

Running totals of the frequencies, against the upper class boundaries:
Marks 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Upper boundary 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5

Cumulative frequency 4 11 23 41 61 75 84 88 90

Marks	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Upper boundary	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
Cumulative frequency	4	11	23	41	61	75	84	88	90

Plot each cumulative frequency at its upper boundary, starting from $(9.5, 0)$ , and draw a smooth S-shaped curve through the points.

There are 90 students, so the median is at 45 on the cumulative frequency axis.
Go across to the curve and down: about 51.5.
(Check: 45 lies between 41 at 49.5 and 61 at 59.5, and $49.5 + \frac{45 - 41}{20} \times 10 = 51.5$ .)

A distinction is 75 or more.
Go up from 74.5 to the curve and across: about 80 students scored less than 75.
So about $90 - 80 = 10$ got a distinction, and the probability is about $\frac{10}{90} = \frac19$ .
(A careful reading gives about 79.5 below, so 10.5 above and $\frac{10.5}{90} \approx 0.12$ .
Either is a fair reading from a curve.)