Question

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

Answer 1

Let the remaining two observation be $x$ and $y$
The observation are $2,4,10,12,14,x,y$
Mean $=\overline{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$\Rightarrow 56=42+x+y$
$\Rightarrow x+y=14$ ......(1)
Variance $=16=\frac{1}{n}\sum _{i=l}^7(X_i-\overline{X}{)}^2$
$16=\frac{1}{7}\left[\right(-6{)}^2+(-4{)}^2+(2{)}^2+(4{)}^2+(6{)}^2+x^2+y^2-2\times 8(x+y)+2\times \left(8{)}^2\right]$
$16=\frac{1}{7}[36+16+4+6+36+x^2+y^2-16(14)+2(64\left)\right]$ ...........[using (1)]
$16=\frac{1}{7}[108+x^2+y^2-224+128]$
$16=\frac{1}{7}[12+x^2+y^2]$
$\Rightarrow x^2+y^2=112-12=100$
$x^2+y^2=100$ ..........(2)
From (1), we obtain
$x^2+y^2+2xy=196$ ....(3)
From (2) and (3), we obtain
$2xy=196-100$
$\Rightarrow 2xy=96$ ...........(4)
subtracting (4) from (2), we obtain
$x^2+y^2-2xy=100-96$
$\Rightarrow (x-y{)}^2=4$
$\Rightarrow x-y=\pm 2$ ............(5)
Therefore, from (1) and (5) we obtain
$x=8$ and $y=6$ when $x-y=2$
$x=6$ and $y=8$ when $x-y=-2$
Thus the remaining observation are $6$ and $8$