Mathematics - Statistics Question with Solution | TestHub

$a,b,c\in Na<b<c$ $\overline{\mathrm{x}}=$ mean $=\frac{9+25+\mathrm{a}+\mathrm{b}+\mathrm{c}}{5}=18$ $a+b+c=56$ Mean deviation $=\frac{\Sigma |{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}|}{\mathrm{n}}=4$ $\begin{array}{rl}& =9+7+|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=20\\ & =|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=4\end{array}$ Variance $=\frac{\Sigma {|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}|}^2}{\mathrm{n}}=\frac{136}{5}$ $\begin{array}{rl}& =81+49+|18-a{|}^2+|18-b{|}^2+|18-c{|}^2=136\\ & =(18-a{)}^2+(18-b{)}^2+(18-c{)}^2=6\end{array}$ Possible values $(18-\mathrm{a}{)}^2=1,(18-\mathrm{b}{)}^2=1(18-\mathrm{c}{)}^2=4$ $\mathrm{a}<\mathrm{b}<\mathrm{c}$ so $\begin{array}{ccc}18-a=1& 18-b=-1& 18-c=-2\\ \mathrm{a}=17& \mathrm{b}=19& \mathrm{c}=20\end{array}$ $a+b+c=56$ $2\mathrm{a}+\mathrm{b}-\mathrm{c}34=19-20=33$