WAEC 2017 · Paper 2 · Q8

Marks	1	2	3	4	5
Number of students	$m + 2$	$m - 1$	$2m - 3$	$m + 5$	$3m - 4$

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

(a)
If the mean mark is $3\frac{6}{23}$ , find the value of $m$ .
(b)
Find the: (i) interquartile range; (ii) probability of selecting a student who scored at least 4 marks in the test.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$\sum f = (m + 2) + (m - 1) + (2m - 3) + (m + 5) + (3m - 4)$
$= 8m - 1$ and $\sum fx = (m + 2) + 2(m - 1) + 3(2m - 3) + 4(m + 5) + 5(3m - 4)$
$= 28m - 9$ .
The mean is $3\frac{6}{23} = \frac{75}{23}$ , so $\frac{28m - 9}{8m - 1} = \frac{75}{23}$ .
Cross-multiply: $23(28m - 9) = 75(8m - 1)$ , so $644m - 207 = 600m - 75$ , $44m = 132$ and $m = 3$ .

(b)(i)

With $m = 3$ the frequencies are $5, 2, 3, 8, 5$ , a total of 23.
Running totals: $5, 7, 10, 18, 23$ .
$Q_1$ is at position $\frac{23}{4} = 5.75$ , which rounds up to the 6th mark: 2.
$Q_3$ is at position $\frac{3 \times 23}{4} = 17.25$ , which rounds up to the 18th mark: 4.
Interquartile range $= 4 - 2 = 2$ .

(ii)

"At least 4 marks" means 4 or 5: $8 + 5 = 13$ students.
The probability is $\frac{13}{23}$ .

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