Mathematics - Statistics Question with Solution | TestHub

Assume, the two remaining observations are$x$and$y$

$\overline{x}=\frac{2+4+10+12+14+x+y}{7}=8$

$\Rightarrow 42+x+y=56$

$\Rightarrow x+y=14 \dots \dots \left(1\right)$

${\sigma }^2=\frac{\Sigma x_1^2}{n}-{\left(\frac{\Sigma x_1}{n}\right)}^2$

$16=\frac{4+16+100+144+196+x^2+y^2}{7}-{\left(8\right)}^2$

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

$\Rightarrow 560=460+x^2+y^2$

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

$\Rightarrow xy=48$, solving equations $\left(1\right) \& \left(2\right)$

${\left(x-y\right)}^2={\left(x+y\right)}^2-4xy=4$

$\therefore \left|x-y\right|=2$