Question

In a discrete series (when all values are not same), the relationship between mean deviation (M.D) about mean and standard deviation (S.D) is
(a) $M.D=S.D$
(b) $M.D\ge S.D$
(c) $M.D > S.D$
(d) $M.D < S.D$

Answer 1

We solve this problem by using the formula of mean deviation about mean and standard deviation.
The formula for mean deviation about mean is given as
$M.D=\dfrac{\sum{\left| {{x}_{i}}-\bar{x} \right|}}{n}$
The formula for standard deviation is given as
$S.D=\sqrt{\dfrac{\sum{{{\left( \left| {{x}_{i}}-\bar{x} \right| \right)}^{2}}}}{n}}$
By using the above two formulas we try to find the relation between M.D and S.D.

Complete step by step answer:
We are asked to find the relation between M.D about mean and S.D
We know that the formula of mean deviation about mean is given as
$M.D=\dfrac{\sum{\left| {{x}_{i}}-\bar{x} \right|}}{n}$
We also know that the formula of standard deviation is given as
$S.D=\sqrt{\dfrac{\sum{{{\left( \left| {{x}_{i}}-\bar{x} \right| \right)}^{2}}}}{n}}$
Let us take the difference of square of S.D and M.D as
$\Rightarrow {{\left( S.D \right)}^{2}}-{{\left( M.D \right)}^{2}}=\dfrac{\sum{{{\left( \left| {{x}_{i}}-\bar{x} \right| \right)}^{2}}}}{n}-{{\left( \dfrac{\sum{\left| {{x}_{i}}-\bar{x} \right|}}{n} \right)}^{2}}$
Let us assume that the value of modulus that is $\left| {{x}_{i}}-\bar{x} \right|={{k}_{1}}$ then we get

& \Rightarrow {{\left( S.D \right)}^{2}}-{{\left( M.D \right)}^{2}}=\dfrac{\sum{{{k}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum{{{k}_{i}}}}{n} \right)}^{2}} \\\ & \Rightarrow {{\left( S.D \right)}^{2}}-{{\left( M.D \right)}^{2}}=p \\\ \end{aligned}$$ Here, we can see that the value of $$p$$ is always greater than 0 because we can see that the first term includes the sum of squares of terms while the second term includes squares of sum of terms. But we are given that all the terms of series are not same so we can say that $$p$$ not equal to 0 but greater than 0 By using the above conditions we can say that $$\begin{aligned} & \Rightarrow {{\left( S.D \right)}^{2}}-{{\left( M.D \right)}^{2}}>0 \\\ & \Rightarrow {{\left( S.D \right)}^{2}} > {{\left( M.D \right)}^{2}} \\\ \end{aligned}$$ We know that the mean deviation and standard deviation are always positive so, by applying the square root in above inequality we get $$\Rightarrow S.D > M.D$$ Therefore we can conclude that the relation between S.D and M.D is $$S.D > M.D$$ or $$M.D < S.D$$. **So, the correct answer is “Option d”.** **Note:** We have a shortcut for solving this problem. We have the relation between M.D and S.D for any series as $$\Rightarrow M.D=\dfrac{4}{5}S.D$$ From this relation we can say that M.D is less than S.D because the coefficient of S.D is less than 1 that is $$\Rightarrow M.D < S.D$$ Therefore we can conclude that the relation between S.D and M.D is $$S.D > M.D$$ or $$M.D < S.D$$ So, option (d) is the correct answer.