Question
Consider any set of 201 observations $${x_1},{x_2},.....\,{x_{200}},\,{x_{201}}.$$ It is given that $${x_1}\, < \,{x_2}\, < \,.....\, < {x_{200}}\, < {x_{201}}.$$ Then the mean deviation of this set of observations about a point $$k$$ is minimum when $$k$$ equals
A.
$$\frac{{\left( {{x_1} + {x_2} + .....\, + {x_{200}} + \,{x_{201}}} \right)}}{{201}}$$
B.
$${x_{1}}$$
C.
$${x_{101}}$$
D.
$${x_{201}}$$
Answer :
$${x_{101}}$$
Solution :
Given that $${x_1}\, < \,{x_2} < {x_3}\, < \,.....\,\, < {x_{201}}$$
∴ Median of the given observation $$ = {\frac{{201 + 1}}{2}^{th}}$$ items
= $${101^{th}}$$ item
$$ = {x_{101}}$$
Now, deviations will be minimum if taken from the median
∴ Mean deviation will be min if $$k = {x_{101}}.$$