Mathematics - Statistics Question with Solution | TestHub

$$\begin{array}{rl}& \bar{x}=\frac{7+8+9+7+8+7+\lambda +8}{8}=8\\ & \Rightarrow \frac{54+\lambda }{8}=8\Rightarrow \lambda =10\end{array}$$ Now variance $={\sigma }^2$ $$\begin{array}{rl}& (7-8{)}^2+(8-8{)}^2+(9-8{)}^2+(7-8{)}^2+(8-8{)}^2\\ & =\frac{+(7-8{)}^2+(10-8{)}^2+(8-8{)}^2}{8}\\ & \Rightarrow {\sigma }^2=\frac{1+0+1+1+0+1+4+0}{8}=\frac{8}{8}=1\\ & \end{array}$$ Hence, the variance is 1 .