Mathematics - Statistics Question with Solution | TestHub

Let the remaining two observations are $x$ and $y$.

Mean $\overline{x}=\frac{\sum x_i}{N}=\frac{2+4+10+12+14+x+y}{7}$

$\Rightarrow \frac{42+x+y}{7}=8$

$\Rightarrow x+y=14 ...\left(1\right)$
Variance, 

    ${\sigma }^2=\frac{\sum x_i^2}{N}-{\left(\overline{x}\right)}^2=\frac{4+16+100+144+196+x^2+y^2}{7}-64$

$\Rightarrow {\sigma }^2=\frac{460+x^2+y^2}{7}-64=16$

$\Rightarrow x^2+y^2=100 \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get,

$x=6,$ $y=8$ and $xy=48$