Question

Consider set of observations $x_1,x_2,x_3,\cdots ,x_{101}$. It is given that $x_1<x_2<x_3\cdots <x_{100}<x_{101}$, then the mean deviation of the set of observations about $k$ is minimum when $k$ equals

• A. $x_1$
• B. $x_{51}$
• C. $\frac{x_1+x_2+\cdots +x_{101}}{101}$
• D. $\frac{x_{50}+x_{51}}{2}$

Answer 1

The correct option is B $x_{51}$

Mean deviation is minimum when deviation is taken from "Median".
Number of observations $=101$
$\Rightarrow \text{median}=\frac{101+1}{2}\text{th observations}=x_{51}\therefore k=x_{51}$