WAEC 2014 · Paper 2 · Q7

The table shows the distribution of the lengths of 20 iron rods measured in metres.

Length (m)	1.0 – 1.1	1.2 – 1.3	1.4 – 1.5	1.6 – 1.7	1.8 – 1.9
Frequency	2	3	8	5	2

Worked solution (try it first)

The class mid-points are 1.05, 1.25, 1.45, 1.65 and 1.85.
With $A = 1.45$ , the deviations $d = x - A$ are $-0.4$ , $-0.2$ , $0$ , $0.2$ and $0.4$ .
Multiply by the frequencies: $fd = -0.8$ , $-0.6$ , $0$ , $1.0$ and $0.8$ .
Add: $\sum fd = 0.4$ and $\sum f = 20$ .
Mean $= A + \dfrac{\sum fd}{\sum f}$ , so the mean is $1.45 + \dfrac{0.4}{20}$ .
That is $1.45 + 0.02 = 1.47$ .
The mean length is $1.47\text{ m}$ .