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

Let the remaining two observations be $x$ and $y$
Therefore, the observations are $6,7,10,12,12,13,x,y$
Thus mean  $\bar{X}=\frac{6+7+10+12+12+13+x+y}{8}=9$
$\Rightarrow 60+x+y=12$
$\Rightarrow x+y=12$  ..........(i)
Variance $=9.25=\frac{1}{N}\sum _{i=l}^8(x_i-\bar{x}{)}^2$
$9.25=\frac{1}{8}\left[(-3{)}^2+(-2{)}^2+(1{)}^2+(3{)}^2+(3{)}^2+(4{)}^2+x^2+y^2-2\times 9(x+y)+2\times (9{)}^2\right]$
$9.25=\frac{1}{8}\left[9+4+1+9+9+16+x^2+y^2-18(12)+162\right]$   .........[Using (i) ]
  $9.25=\frac{1}{8}[48+x^2+y^2-216+162]$
  $9.25=\frac{1}{8}[x^2+y^2-6]$ 
  $\Rightarrow x^2+y^2=80$     ......(2)
From (1) we obtain
  $x^2+y^2+2xy=144$   ......(3)
From (2) and (3),  we obtain 
$2xy=64$ .......(4)
Subtracting (4) from (2), we obtain 
  $x^2+y^2-2xy=80-64=16$
  $\Rightarrow x-y=\pm 4$   ..........(5)
Therefore from (1) and (5), we obtain 
$x=8$ and $y=4$ when $x-y=4$
$x=4$ and $y=8$ when $x-y=-4$
Thus the remaining observation are $4$ and $8$