WAEC 2013 · Paper 2 · Q14✱✱

(a)
The mean of the numbers $2$ , $5$ , $6$ , $8$ , $x$ and $y$ is $6$ and their standard deviation is $\sqrt5$ . Find the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Mean: $\dfrac{2 + 5 + 6 + 8 + x + y}{6} = 6$ , so $21 + x + y = 36$ .
So $x + y = 15$ … (1).
The variance is $(\sqrt5)^2 = 5$ , so the squared deviations from 6 add up to $6 \times 5 = 30$ .
The known numbers give $(2 - 6)^2 + (5 - 6)^2 + (6 - 6)^2 + (8 - 6)^2 = 16 + 1 + 0 + 4$
$= 21$ .
So $(x - 6)^2 + (y - 6)^2 = 9$ … (2).
Put $y = 15 - x$ from (1) into (2): $(x - 6)^2 + (9 - x)^2 = 9$ .
Expand: $2x^2 - 30x + 117 = 9$ , so $x^2 - 15x + 54 = 0$ .
Factorise: $(x - 6)(x - 9) = 0$ , so $x = 6$ or $x = 9$ .
From (1), $y = 9$ or $y = 6$ : the two numbers are 6 and 9.