Question

The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations

Answer 1

Let two remaining observations be x and y. then

$\frac{6+7+10+12+12+13+x+y}{8}=9$

$\therefore 60+x+y=72\Rightarrow x+y=12.....\left(i\right)$

Aso $\frac{1}{8}(6^2+7^2+{10}^2+{12}^2+{12}^2+{13}^2+x^2+y^2)-(9{)}^2=9.25$

$\Rightarrow \frac{1}{8}(36+49+100+144+144+169+x^2+y^2)-81=9.25$

$\Rightarrow 642+x^2+y^2=722$

$\Rightarrow x^2+y^2=80....\left(ii\right)$

Now $(x+y{)}^2+(x-y{)}^2=2(x^2+y^2)$

$\Rightarrow (12{)}^2+(x-y{)}^2=2\times 80$

$\Rightarrow (x-y{)}^2=160-144$

$\Rightarrow (x-y{)}^2=16\Rightarrow x-y=\pm 4$

When x - y = 14

Solving x + y = 12 and x - y = 4 we get x = 8 and y = 4

When x -y = - 4

Solving x + y = 12 and x - y = - 4 we get x = 4 and y = 8