The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is

Let the remaining two observations be x and y. Then,

Mean =10

$\begin{array}{l}\Rightarrow \frac{1}{8}(5+7+10+12+14+15+x+y)=10\\ \Rightarrow 63+x+y=80\Rightarrow x+y=17\end{array}$

Variance =13.5

$\begin{array}{l}\Rightarrow \frac{1}{8}\left(25+49+100+144+196+225+x^2+y^2\right)-{10}^2=13.5\\ \Rightarrow x^2+y^2+739=908\\ \Rightarrow x^2+y^2=169\end{array}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x-y{)}^2+(x+y{)}^2=2\left(x^2+y^2\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x-y{)}^2=2\times 169-289=49\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}|x-y|=7\end{array}$