Mathematics - Statistics Question with Solution | TestHub

$\begin{array}{ccccccc}\mathrm{x}& \mathrm{C}& 2\mathrm{C}& 3\mathrm{C}& 4\mathrm{C}& 5\mathrm{C}& 6\mathrm{C}\\ \mathrm{f}& 2& 1& 1& 1& 1& 1\end{array}$ $\bar{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}$ $\begin{array}{rl}& \mathrm{Var}(x)=\frac{{\mathrm{c}}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7}\\ & -{\left(\frac{22\mathrm{c}}{7}\right)}^2\\ & =\frac{92{\mathrm{c}}^2}{7}-{\mathrm{c}}^2\times \frac{484}{49}\\ & =\frac{(644-484){\mathrm{c}}^2}{49}=\frac{160{\mathrm{c}}^2}{49}\\ & 160=\frac{160\times {\mathrm{c}}^2}{49}\Rightarrow \mathrm{c}=7\end{array}$