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.
Let the remaining two observations be $x$ and $y$.
Therefore, the observations are $6,7,10,12,12,13, x, y$
Mean, $\bar{x}=\frac{6+7+10+12+12+13+x+y}{8}=9$
$\Rightarrow 60+x+y=72$
$\Rightarrow x+y=12$ ...........$(1)$
Variance $ = 9.25 = \frac{1}{n}\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \bar x} \right)}^2}} $
$9.25=\frac{1}{8}[(-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(4)^{2}$
$+x^{2}+y^{2}-2 \times 9(x+y)+2 \times(9)^{2}]$
$9.25=\frac{1}{8}\left[9+4+1+9+9+16+x^{2}+y^{2}-18(12)+162\right]$ ........[ using $(1)$ ]
$9.25=\frac{1}{8}\left[48+x^{2}+y^{2}-216+162\right]$
$9.25=\frac{1}{8}\left[x^{2}+y^{2}-6\right]$
$\Rightarrow x^{2}+y^{2}=80$ .........$(2)$
From $(1),$ we obtain
$x^{2}+y^{2}+2 x y=144$ ........$(3)$
From $(2)$ and $(3),$ we obtain
$2 x y=64$ ..........$(4)$
Subtracting $(4)$ from $(2),$ we obtain
$x^{2}+y^{2}-2 x y=80-64=16$
$\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 observations are $4$ and $8$
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