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.
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.
Let the remaining two observations be $x$ and $y$.
The observations are $2,4,10,12,14, x , y$
Mean, $\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$\Rightarrow 56=42+x+y$
$\Rightarrow x+y=14$
Varaiance $ = 16 = \frac{1}{n}\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \bar x} \right)}^2}} $
$16=\frac{1}{7}[(-6)^{2}+(-4)^{2}+(2)^{2}$
$+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2 \times(8)^{2}]$
$16=\frac{1}{7}\left[36+16+4+16+36+x^{2}+y^{2}-16(14)+2(64)\right]$ .......[ using $(1)$ ]
$16=\frac{1}{7}\left[108+x^{2}+y^{2}-224+128\right]$
$16=\frac{1}{7}\left[12+x^{2}+y^{2}\right]$
$\Rightarrow x^{2}+y^{2}=112-12=100$
$\Rightarrow x^{2}+y^{2}=100$ ........$(2)$
From $(1),$ we obtain
$x^{2}+y^{2}+2 x y=196$ .........$(3)$
From $(2)$ and $(3),$ we obtain
$2 x y=196-100$
$\Rightarrow 2 x y=96$ .........$(4)$
Subtracting $(4)$ from $(2),$ we obtain
$x^{2}+y^{2}-2 x y=100-96$
$\Rightarrow(x-y)^{2}=4$
$\Rightarrow x-y=\pm 2$ .........$(5)$
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 observations are $6$ and $8 .$
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