Question

The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14, find the remaining two observations.

Answer 1

Let two remaining observations be x and y. Then

$\frac{2+4+10+12+14+x+y}{7}=8$

$\therefore 42+x+y=56\Rightarrow x+y=14.....\left(i\right)$

Also $\frac{1}{7}(2^2+4^2+{10}^2+{12}^2+{14}^2+x^2+y^2)-(8{)}^2=16$

$\Rightarrow \frac{1}{7}(4+16+100+144+196+x^2+y^2)-64=16$

$\Rightarrow 460+x^2+y^2=560\Rightarrow x^2+y^2=100...\left(ii\right)$

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

$\Rightarrow (14{)}^2+(x-y{)}^2=2\times 100$

$\Rightarrow (x-y{)}^2=200-196$

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

When x - y = 2

Solving x + y = 14 and x - y = 2 we get x = 8 and y = 6

When x - y = - 2

Solving x + y = 14 and x - y = - 2 we get x = 6 and y = 8