WAEC 2016 · Paper 2 · Q12

Statistics & correlation

Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Number of students	3	17	41	85	97	115	101	64	21	6

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

(a)

(i) Construct a cumulative frequency table for the distribution. (ii) Draw a cumulative frequency curve for the distribution.

Model answer

(i)

Marks	Upper class boundary	Cumulative frequency
$1\text{–}10$	$10.5$	$3$
$11\text{–}20$	$20.5$	$20$
$21\text{–}30$	$30.5$	$61$
$31\text{–}40$	$40.5$	$146$
$41\text{–}50$	$50.5$	$243$
$51\text{–}60$	$60.5$	$358$
$61\text{–}70$	$70.5$	$459$
$71\text{–}80$	$80.5$	$523$
$81\text{–}90$	$90.5$	$544$
$91\text{–}100$	$100.5$	$550$

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

(b)
Use the curve to estimate the: (i) number of students who scored marks between 32 and 74; (ii) pass mark, if $18\%$ of the students failed; (iii) lowest mark for distinction, if $8\%$ of the students passed with distinction.
Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive, with the readings at 99 (pass mark) and 506 (distinction).

Worked solution (try it first)

(a)(i)

Cumulative frequencies $3, 20, 61, 146, 243, 358, 459, 523, 544, 550$ at the upper boundaries $10.5, 20.5, \ldots, 100.5$ .

(ii)

Plot these points, starting from $(0.5, 0)$ , and join them with a smooth curve.

(b)(i)

Read up from 32: about 74 students scored less.
Read up from 74: about 481.
So about $481 - 74 \approx 408$ students scored between 32 and 74.

(ii)

18% of 550 is 99.
Read across from 99: the pass mark is about 35.

(iii)

8% of 550 is 44, so read across from $550 - 44 = 506$ : the lowest distinction mark is about 78.

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