WAEC 2014 · Paper 2 · Q11

Score	1	2	3	4	5	6
Frequency	2	5	13	11	9	10

The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the:

(a)
mean deviation of the distribution;
(b)
probability that a score selected at random is at least 4.

Worked solution (try it first)

(a)

First the mean: $\sum fx = 1(2) + 2(5) + 3(13) + 4(11) + 5(9) + 6(10)$
$= 2 + 10 + 39 + 44 + 45 + 60$
$= 200$ , and $\sum f = 50$ , so $\bar x = \frac{200}{50} = 4$ .
The distances from 4 are $|x - 4| = 3, 2, 1, 0, 1, 2$ .
Multiply by the frequencies: $2(3) + 5(2) + 13(1) + 11(0) + 9(1) + 10(2) = 6 + 10 + 13 + 0 + 9 + 20$
$= 58$ .
Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
$= \frac{58}{50}$
$= 1.16$ .

(b)

"At least 4" means a score of 4, 5 or 6.
That happened $11 + 9 + 10 = 30$ times out of 50, so the probability is $\frac{30}{50} = \frac35$ .

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