WAEC 2009 · Paper 2 · Q14

The distribution of the lives (in days) of 40 transistor batteries is shown in the table.

Battery life (days)	26–30	31–35	36–40	41–45	46–50	51–55
Frequency	4	7	13	8	6	2

(a)
Draw a histogram for the distribution.
Model answer
Draw the bars on the class boundaries 25.5, 30.5, 35.5, 40.5, 45.5, 50.5, 55.5, with no gaps, and heights equal to the frequencies 4, 7, 13, 8, 6, 2. Label both axes.

For the mode, take the tallest bar (35.5–40.5). Join its top-left corner to the top-left corner of the bar on its right, and its top-right corner to the top-right corner of the bar on its left. Read down from the crossing: about 38.2 days.
(b)
Use your graph in (a) to determine the mode for the distribution.
(c)
Using an assumed mean of 43 days, calculate the mean of the distribution.
(d)
What percentage of the batteries will live for less than 31 days or more than 45 days?

Worked solution (try it first)

The classes all have width 5, so the bar heights are the frequencies $4, 7, 13, 8, 6, 2$ over the boundaries $25.5, 30.5, \ldots, 55.5$ .

On the tallest bar, join each top corner to the top corner of the opposite neighbour.
The lines cross above the mode.
By calculation: $35.5 + \dfrac{13 - 7}{(13 - 7) + (13 - 8)} \times 5 = 35.5 + \dfrac{6}{11} \times 5$
$= 38.2$ days.

Class mid-values are $28, 33, 38, 43, 48, 53$ , so the deviations $d = x - 43$ are $-15, -10, -5, 0, 5, 10$ .
Multiply by the frequencies: $fd = -60, -70, -65, 0, 30, 20$ , and $\sum fd = -145$ .
Mean $= 43 + \dfrac{\sum fd}{\sum f}$
$= 43 + \dfrac{-145}{40}$
$= 43 - 3.625$
$= 39.375$ days.