WAEC 2012 · Paper 2 · Q12

The ages (in years) of men selected from a community are as follows:

44 28 58 50 93 35 34 52 57 61 40 63 90 67 64 56 51 82 73 43 73 73 44 71 95 52 71 25 35 79 28 40 72 88 82 63 53 48 98 65 63 44 73 70 68 46 54 62 41 70

(a)
Prepare a grouped frequency distribution table with the intervals $20$ – $29$ , $30$ – $39$ , …
Show the answer
Frequencies $3, 3, 9, 9, 9, 10, 3, 4$ (total 50)
(b)
Calculate the mean deviation of the distribution.

Worked solution (try it first)

(a)

Go through the list once, making a tally mark for each age in its class, then count:
Ages 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 3 3 9 9 9 10 3 4
The frequencies add up to 50, the number of ages in the list.

(b)

Class marks: $24.5, 34.5, \ldots, 94.5$ .
$\sum fx = 3(24.5) + 3(34.5) + 9(44.5) + 9(54.5) + 9(64.5) + 10(74.5) + 3(84.5) + 4(94.5)$
$= 3025$ , so the mean is $\frac{3025}{50} = 60.5$ .
The distances of the class marks from 60.5 are $36, 26, 16, 6, 4, 14, 24, 34$ .
Multiply by the frequencies: $108, 78, 144, 54, 36, 140, 72, 136$ , which add up to 768.
Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
$= \frac{768}{50}$
$= 15.36$ .

Report a problem with this question