The mean and variance of 7

Let two remaining observations are $x_{1}, x_{2}$.

So, $\bar{x}=\frac{2+4+10+12+14+x_{1}+x_{2}}{7}=8$ (given)

$\Rightarrow x_{1}+y_{1}=14$ ......(i)

Now, $\sigma^{2}=\frac{\sum x_{i}^{2}}{N}-\left(\frac{\sum x_{i}}{N}\right)^{2}=16$ (given)

$=\frac{4+16+100+144+196+x_{1}^{2}+x_{2}^{2}}{7}-64=16$

$\Rightarrow 460+x_{1}^{2}+x_{2}^{2}=(16+64) \times 7$

$\Rightarrow x_{1}^{2}+x_{2}^{2}=100$ .....(ii)

$\because(x+y)^{2}=x^{2}+y^{2}+2 x y \Rightarrow x y=48$........(iii)

$\because(x-y)^{2}=(x+y)^{2}-4 x y=196-192=4$

$\Rightarrow x-y=2 \Rightarrow|x-y|=2$