Question

The standard deviation $\sigma $ of the first N natural numbers can be obtained using which of the following,
A. $\sigma = \dfrac{{{N^2} - 1}}{{12}}$
B. $\sigma = \sqrt {\dfrac{{{N^2} - 1}}{{12}}} $
C. $\sigma = \sqrt {\dfrac{{N - 1}}{{12}}} $
D. $\sigma = \sqrt {\dfrac{{{N^2} - 1}}{{6N}}} $

Answer 1

Hint: Use the definition of standard deviation as the square root of the variance and $\operatorname{var} \left( X \right) = E\left( {{{\left( {X - \mu } \right)}^2}} \right)$ where $\mu = E\left( X \right)$. Use linearity of expression to prove that $\operatorname{var} \left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$.
Using formula form the sum of squares of first n natural numbers and the sum of first n natural number to find the individual terms in the expression for var(X), $\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ and $\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$.

Complete step by step solution:
We know that
$\operatorname{var} \left( X \right) = E\left( {{{\left( {X - \mu } \right)}^2}} \right)$
Using the formula, ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get,
$ \Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2} + {\mu ^2} - 2X\mu } \right)$
Now we know that
$E\left( {X + Y} \right) = E\left( X \right) + E\left( Y \right)$
Using the above formula, we get
$ \Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) + E\left( {{\mu ^2}} \right) + E\left( { - 2X\mu } \right)$
We know that $E\left( {aX} \right) = aE\left( X \right)$ and $E\left( a \right) = a$, we get
$ \Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) + {\mu ^2} - 2\mu E\left( X \right)$
Substitute $\mu = E\left( X \right)$ in the above equation,
$ \Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) + {\left( {E\left( X \right)} \right)^2} - 2E\left( X \right)E\left( X \right)$
Simplify the terms,
$ \Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$ ….. (1)
We know that,
$E\left( {f\left( X \right)} \right) = \sum\limits_{r \in S} {P\left( {X = r} \right)f\left( r \right)} $
Substitute $X$ in place of \[f\left( X \right)\],
$ \Rightarrow E\left( X \right) = \sum\limits_{r = 1}^n {P\left( {X = r} \right)r} $
Simplify the term,
$ \Rightarrow E\left( X \right) = \sum\limits_{r = 1}^n {\dfrac{1}{n} \times r} $
Take constant part out of the summation,
$ \Rightarrow E\left( X \right) = \dfrac{1}{n}\sum\limits_{r = 1}^n r $
Use $\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$, we get
$ \Rightarrow E\left( X \right) = \dfrac{1}{n} \times \dfrac{{n\left( {n + 1} \right)}}{2}$
Cancel out the common factors,
$ \Rightarrow E\left( X \right) = \dfrac{{\left( {n + 1} \right)}}{2}$ ….. (2)
Now, substitute ${X^2}$ in place of \[f\left( X \right)\],
$ \Rightarrow E\left( {{X^2}} \right) = \sum\limits_{r = 1}^n {P\left( {X = r} \right){r^2}} $
Simplify the term,
$ \Rightarrow E\left( {{X^2}} \right) = \sum\limits_{r = 1}^n {\dfrac{1}{n} \times {r^2}} $
Take constant part out of the summation,
$ \Rightarrow E\left( {{X^2}} \right) = \dfrac{1}{n}\sum\limits_{r = 1}^n {{r^2}} $
Use $\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$, we get
$ \Rightarrow E\left( {{X^2}} \right) = \dfrac{1}{n} \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Cancel out the common factors,
$ \Rightarrow E\left( {{X^2}} \right) = \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ ….. (3)
Substitute the values from equation (2) and (3) in equation (1),
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - {\left( {\dfrac{{n + 1}}{2}} \right)^2}$
Simplify the terms,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}$
Take LCM of the terms,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{2\left( {n + 1} \right)\left( {2n + 1} \right) - 3{{\left( {n + 1} \right)}^2}}}{{12}}$
Take $\dfrac{{n + 1}}{{12}}$ common from both terms,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)}}{{12}}\left[ {2\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]$
Simplify the terms in the bracket,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)}}{{12}}\left[ {4n + 2 - 3n - 3} \right]$
Subtract the like terms,
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{{12}}$
Using $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get
$ \Rightarrow \operatorname{var} \left( X \right) = \dfrac{{{n^2} - 1}}{{12}}$
We know that,
$\sigma \left( X \right) = \sqrt {\operatorname{var} \left( X \right)} $
Substitute the value,
$\therefore \sigma \left( X \right) = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $
Now we are given natural numbers as N so we will replace n with N.

So, the correct answer is “Option B”.

Note: Standard deviation measures the distribution of a dataset relative to its mean and is calculated as the square root of the variance and variance is the average of the squared differences of the values from the mean.