WAEC 2012 · Paper 2 · Q14

The marks scored by forty candidates in an examination are shown in the table.

Marks	1	2	3	4	5	6	7	8	9
Number of candidates	2	3	$m$	8	10	5	3	3	$n$

(a)
If the mean of the distribution is $4.725$ , find the values of $m$ and $n$ .
Separate values with commas, e.g. 3, −2
(b)
What is the probability that a candidate chosen at random scored more than 5?

Worked solution (try it first)

(a)

The frequencies add to 40: $34 + m + n = 40$ , so $m + n = 6$ .
Add up $fx$ : $2 + 6 + 3m + 32 + 50 + 30 + 21 + 24 + 9n = 165 + 3m + 9n$ .
Mean: $\dfrac{165 + 3m + 9n}{40} = 4.725$ , so $165 + 3m + 9n = 189$ and $m + 3n = 8$ .
Subtract $m + n = 6$ : $2n = 2$ , so $n = 1$ and $m = 5$ .

(b)

Scored more than 5 (marks 6 to 9): $5 + 3 + 3 + 1 = 12$ of the 40.
Probability $= \dfrac{12}{40} = \dfrac{3}{10}$ .

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