Question

For the values ${x_1}$, ${x_2}$,…………${x_{101}}$ of a distribution ${x_1} < {x_2} < {x_3} < ....... < {x_{100}} < {x_{101}}$. The mean deviation of this distribution with respect to a number $k$ will be minimum when $k$ is equal to:
(A) ${x_1}$
(B) ${x_{51}}$
(C) ${x_{50}}$
(D) $\dfrac{{{x_1} - {x_2} - ..... - {x_{101}}}}{{101}}$

Answer 1

Hint: Given a distribution and we have to find the mean deviation of this distribution with respect to a number will be minimum. Mean deviation is the arithmetic average of the deviations observed values from an average of the observed values. Mean deviation from the median calculates the arithmetic average of the deviations of the observed values from median of the observed values. mean deviation is minimum when deviations are taken from median as median is the middle most value of the series and disperse the whole series into minimal value.
To find the value of median we will first group the data in ascending order. If the middle observation at odd place than median =$\dfrac{{n + 1}}{2}$
Here n= total observations
If the middle observations at even place than median $ = \dfrac{n}{2} + \dfrac{n}{2} + 1$

Complete step-by-step answer:
Step1: The given observations are ${x_1}$,${x_2}$,…………${x_{101}}$, firstly we will arrange them in ascending order. On arranging in the order we will get ${x_1} < {x_2} < {x_3} < ....... < {x_{100}} < {x_{101}}$
Step2: Hence total no of observations or terms are $101$. As we know that the minimum value of mean deviation depends when it is taken by median so if the $y$ is the median of given observation. Applying the formulas of median is
$ \Rightarrow $$y = \left( {\dfrac{{n + 1}}{2}} \right)$th observations (as observations are odd)
Substituting the values in formula we get
Here n$ = 101$
$ \Rightarrow $$y = \left( {\dfrac{{101 + 1}}{2}} \right)$th observation
On solving the expression
$ \Rightarrow $$y = \left( {\dfrac{{102}}{2}} \right) = {51^{^{th}}}$observation
It means at the ${x_{51}}$I.e. at this value of median mean deviation will be minimum

Hence option (B) is the correct answer i.e.${x_{51}}$

Note: Many students cannot solve this problem because most of them don’t know when mean deviation will be minimum. It will be minimum when the observations are taken from median as median is the middle value.
Commit to memory:
Median=$\left( {\dfrac{{n + 1}}{2}} \right)$th observation (when n is odd)