If any discrete series (when all the value are not same) the relationship between M.D about mean and S.D is
$
(a){\text{ M}}{\text{.D = S}}{\text{.D}} \\
(b){\text{ M}}{\text{.D > S}}{\text{.D}} \\
(c){\text{ M}}{\text{.D < S}}{\text{.D}} \\
(d){\text{ M}}{\text{.D}} \leqslant {\text{S}}{\text{.D}} \\
$
Answer
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Hint – In this question assume a discrete series of number as${x_1},{x_2},{x_3},...............{x_n}$, so use the direct formula for the mean deviation that is $M.D = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} }}{n}$, where ${x_i}$ is any ${i^{th}}$ element. Then use the direct formula for standard deviation that is $S.D = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - x} \right)}^2}} }}{n}} $. Try to figure out the relationship between the mean deviation and the standard deviation. This will help get the right answer.
Complete step-by-step answer:
Let the series of discrete numbers is
${x_1},{x_2},{x_3},...............{x_n}$
Let ${x_i}$ be the any general term where, i = 1, 2, 3 ...n
Let $\bar x$ be the mean of the above series having n terms.
So the mean deviation of the above series is given as
$M.D = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} }}{n}$ (Standard formula)
And standard deviation (S.D)
$S.D = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - x} \right)}^2}} }}{n}} $ (Standard formula)
$ \Rightarrow S.D = \dfrac{{\sum\limits_{i = 1}^n {\left( {{x_i} - x} \right)} }}{{\sqrt n }}$
Now as we know that $n > \sqrt n $
$ \Rightarrow \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} }}{n} < \dfrac{{\sum\limits_{i = 1}^n {\left( {{x_i} - x} \right)} }}{{\sqrt n }}$
$ \Rightarrow M.D < S.D$
So this is the required answer.
Hence option (C) is the correct answer.
Note – The physical significance of the mean, mean deviation and the standard deviation is very helpful in solving problems of these kinds. Mean deviation is the absolute value of the numerical difference between the number of a set and their mean. Standard deviation is basically the measure of the dispersion of data of a specific set from its mean.
Complete step-by-step answer:
Let the series of discrete numbers is
${x_1},{x_2},{x_3},...............{x_n}$
Let ${x_i}$ be the any general term where, i = 1, 2, 3 ...n
Let $\bar x$ be the mean of the above series having n terms.
So the mean deviation of the above series is given as
$M.D = \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} }}{n}$ (Standard formula)
And standard deviation (S.D)
$S.D = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - x} \right)}^2}} }}{n}} $ (Standard formula)
$ \Rightarrow S.D = \dfrac{{\sum\limits_{i = 1}^n {\left( {{x_i} - x} \right)} }}{{\sqrt n }}$
Now as we know that $n > \sqrt n $
$ \Rightarrow \dfrac{{\sum\limits_{i = 1}^n {\left| {{x_i} - \bar x} \right|} }}{n} < \dfrac{{\sum\limits_{i = 1}^n {\left( {{x_i} - x} \right)} }}{{\sqrt n }}$
$ \Rightarrow M.D < S.D$
So this is the required answer.
Hence option (C) is the correct answer.
Note – The physical significance of the mean, mean deviation and the standard deviation is very helpful in solving problems of these kinds. Mean deviation is the absolute value of the numerical difference between the number of a set and their mean. Standard deviation is basically the measure of the dispersion of data of a specific set from its mean.
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