WAEC 2008 · Paper 2 · Q8✱✱

The data below are the ages, in years, of 45 people.

(a)
Form a frequency distribution of the data using the intervals $21 - 25$ , $26 - 30$ , $31 - 35$ , etc.
Show the answer
$21 - 25$ : 6; $26 - 30$ : 9; $31 - 35$ : 9; $36 - 40$ : 9; $41 - 45$ : 6; $46 - 50$ : 6 (total 45)
(b)
Draw the histogram of the distribution.
Model answer
Draw the bars on the class boundaries $20.5, 25.5, \dots, 50.5$ with no gaps, heights 6, 9, 9, 9, 6, 6. For (c), the three middle bars are equally tall, so treat $25.5 - 40.5$ as one modal block: join its top corners to the tops of the neighbouring bars, crossing over. The lines meet above $33$ .
(c)
Use your histogram to estimate the mode.
(d)
Calculate the mean age.

Worked solution (try it first)

Tally each age into its class.
The frequencies are: $21 - 25$ : 6, $26 - 30$ : 9, $31 - 35$ : 9, $36 - 40$ : 9, $41 - 45$ : 6, $46 - 50$ : 6.
They add up to 45.

A histogram uses the class boundaries: $20.5, 25.5, 30.5, 35.5, 40.5, 45.5, 50.5$ .
Draw touching bars of heights 6, 9, 9, 9, 6, 6.

The three bars from $25.5$ to $40.5$ are equally tall, so take them together as the modal block.
Join the top-left corner of the block to the top of the next bar on the right, $(40.5, 6)$ , and the top-right corner to the top of the bar on the left, $(25.5, 6)$ .
The two lines cross above $33$ , so the mode is about $33$ years.

Use the class midpoints $x = 23, 28, 33, 38, 43, 48$ and find $fx$ : $138, 252, 297, 342, 258, 288$ .
Add them: $\sum fx = 1575$ , and $\sum f = 45$ .
Mean $= \frac{\sum fx}{\sum f}$
$= \frac{1575}{45}$
$= 35$ .
The mean age is $35$ years.