WAEC 2023 · Paper 2 · Q5✱✱

The table shows the frequency distribution of a set of data.

$x$	1	2	3	4	5
$f$	$m + 2$	$m - 1$	$2m - 3$	$m + 1$	$3m - 4$

If the mean of the distribution is $\frac{31}{11}$ , find the median.

Worked solution (try it first)

$\sum f = 8m - 5$ and $\sum fx = (m + 2) + 2(m - 1) + 3(2m - 3) + 4(m + 1) + 5(3m - 4)$
$= 28m - 25$ .
The mean is $\dfrac{28m - 25}{8m - 5} = \dfrac{31}{11}$ , so $11(28m - 25) = 31(8m - 5)$ .
$308m - 275 = 248m - 155$ , so $60m = 120$ and $m = 2$ .
The frequencies are $4, 1, 1, 3, 2$ (total 11).
The median is the 6th value: the cumulative frequencies $4, 5, 6$ reach 6 at $x = 3$ , so the median is 3.