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 $8 .$