Mathematics - Statistics Question with Solution | TestHub

Let the two unknown items be $x$ and $y$. Then, $\mathrm{Mean}=4\Rightarrow \frac{1+2+6+x+y}{5}=4$ $\Rightarrow x+y=11\dots (i)$ and variance $=5.2$ $\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}-(\text{ mean }{)}^2=5.2$ $\Rightarrow 41+x^2+y^2=5(5.2+16)$ $\begin{array}{ll}\Rightarrow & 41+x^2+y^2=106\\ \Rightarrow & x^2+y^2=65\end{array}$ On solving Eqs. (i) and (ii), we get $x=4,y=7\text{ or }x=7,y=4$