WAEC 2008 · Paper 2 · Q5

Mark ( $x$ )	2	3	4	5	6	7	8	9
Frequency ( $f$ )	1	3	1	$m$	4	$n$	2	1

Worked solution (try it first)

The frequencies add up to 20: $1 + 3 + 1 + m + 4 + n + 2 + 1 = 20$ , so $m + n = 8$ .
The mean is $\frac{\sum fx}{\sum f} = 5.4$ , so $\sum fx = 5.4 \times 20 = 108$ .
Work out $\sum fx$ : $2 + 9 + 4 + 5m + 24 + 7n + 16 + 9 = 64 + 5m + 7n$ .
So $64 + 5m + 7n = 108$ , which gives $5m + 7n = 44$ .
From the first equation $m = 8 - n$ .
Substitute: $5(8 - n) + 7n = 44$ , so $40 + 2n = 44$ and $n = 2$ .
Then $m = 8 - 2 = 6$ .
So $m = 6$ and $n = 2$ .