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Question: The standard deviation \(\sigma \) of the first N natural numbers can be obtained using which of the...

The standard deviation σ\sigma of the first N natural numbers can be obtained using which of the following,
A. σ=N2−112\sigma = \dfrac{{{N^2} - 1}}{{12}}
B. σ=N2−112\sigma = \sqrt {\dfrac{{{N^2} - 1}}{{12}}}
C. σ=N−112\sigma = \sqrt {\dfrac{{N - 1}}{{12}}}
D. σ=N2−16N\sigma = \sqrt {\dfrac{{{N^2} - 1}}{{6N}}}

Explanation

Solution

Use the definition of standard deviation as the square root of the variance and var⁡(X)=E((X−μ)2)\operatorname{var} \left( X \right) = E\left( {{{\left( {X - \mu } \right)}^2}} \right) where μ=E(X)\mu = E\left( X \right). Use linearity of expression to prove that var⁡(X)=E(X2)−[E(X)]2\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), ∑r=1nr2=n(n+1)(2n+1)6\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} and ∑r=1nr=n(n+1)2\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}.

Complete step by step solution:
We know that
var⁡(X)=E((X−μ)2)\operatorname{var} \left( X \right) = E\left( {{{\left( {X - \mu } \right)}^2}} \right)
Using the formula, (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we get,
⇒var⁡(X)=E(X2+μ2−2Xμ)\Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2} + {\mu ^2} - 2X\mu } \right)
Now we know that
E(X+Y)=E(X)+E(Y)E\left( {X + Y} \right) = E\left( X \right) + E\left( Y \right)
Using the above formula, we get
⇒var⁡(X)=E(X2)+E(μ2)+E(−2Xμ)\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(aX)=aE(X)E\left( {aX} \right) = aE\left( X \right) and E(a)=aE\left( a \right) = a, we get
⇒var⁡(X)=E(X2)+μ2−2μE(X)\Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) + {\mu ^2} - 2\mu E\left( X \right)
Substitute μ=E(X)\mu = E\left( X \right) in the above equation,
⇒var⁡(X)=E(X2)+(E(X))2−2E(X)E(X)\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,
⇒var⁡(X)=E(X2)−[E(X)]2\Rightarrow \operatorname{var} \left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} ….. (1)
We know that,
E(f(X))=∑r∈SP(X=r)f(r)E\left( {f\left( X \right)} \right) = \sum\limits_{r \in S} {P\left( {X = r} \right)f\left( r \right)}
Substitute XX in place of f(X)f\left( X \right),
⇒E(X)=∑r=1nP(X=r)r\Rightarrow E\left( X \right) = \sum\limits_{r = 1}^n {P\left( {X = r} \right)r}
Simplify the term,
⇒E(X)=∑r=1n1n×r\Rightarrow E\left( X \right) = \sum\limits_{r = 1}^n {\dfrac{1}{n} \times r}
Take constant part out of the summation,
⇒E(X)=1n∑r=1nr\Rightarrow E\left( X \right) = \dfrac{1}{n}\sum\limits_{r = 1}^n r
Use ∑r=1nr=n(n+1)2\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}, we get
⇒E(X)=1n×n(n+1)2\Rightarrow E\left( X \right) = \dfrac{1}{n} \times \dfrac{{n\left( {n + 1} \right)}}{2}
Cancel out the common factors,
⇒E(X)=(n+1)2\Rightarrow E\left( X \right) = \dfrac{{\left( {n + 1} \right)}}{2} ….. (2)
Now, substitute X2{X^2} in place of f(X)f\left( X \right),
⇒E(X2)=∑r=1nP(X=r)r2\Rightarrow E\left( {{X^2}} \right) = \sum\limits_{r = 1}^n {P\left( {X = r} \right){r^2}}
Simplify the term,
⇒E(X2)=∑r=1n1n×r2\Rightarrow E\left( {{X^2}} \right) = \sum\limits_{r = 1}^n {\dfrac{1}{n} \times {r^2}}
Take constant part out of the summation,
⇒E(X2)=1n∑r=1nr2\Rightarrow E\left( {{X^2}} \right) = \dfrac{1}{n}\sum\limits_{r = 1}^n {{r^2}}
Use ∑r=1nr2=n(n+1)(2n+1)6\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}, we get
⇒E(X2)=1n×n(n+1)(2n+1)6\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,
⇒E(X2)=(n+1)(2n+1)6\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),
⇒var⁡(X)=(n+1)(2n+1)6−(n+12)2\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,
⇒var⁡(X)=(n+1)(2n+1)6−(n+1)24\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,
⇒var⁡(X)=2(n+1)(2n+1)−3(n+1)212\Rightarrow \operatorname{var} \left( X \right) = \dfrac{{2\left( {n + 1} \right)\left( {2n + 1} \right) - 3{{\left( {n + 1} \right)}^2}}}{{12}}
Take n+112\dfrac{{n + 1}}{{12}} common from both terms,
⇒var⁡(X)=(n+1)12[2(2n+1)−3(n+1)]\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,
⇒var⁡(X)=(n+1)12[4n+2−3n−3]\Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)}}{{12}}\left[ {4n + 2 - 3n - 3} \right]
Subtract the like terms,
⇒var⁡(X)=(n+1)(n−1)12\Rightarrow \operatorname{var} \left( X \right) = \dfrac{{\left( {n + 1} \right)\left( {n - 1} \right)}}{{12}}
Using (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}, we get
⇒var⁡(X)=n2−112\Rightarrow \operatorname{var} \left( X \right) = \dfrac{{{n^2} - 1}}{{12}}
We know that,
σ(X)=var⁡(X)\sigma \left( X \right) = \sqrt {\operatorname{var} \left( X \right)}
Substitute the value,
∴σ(X)=n2−112\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.