Question

The variance of first \[n\] natural numbers
A.\[\dfrac{[n(n+1)]}{2}\]
B.\[\dfrac{[(n+1)(n+5)]}{12}\]
C.\[\dfrac{[(n+1)(n-5)]}{12}\]
D.\[\dfrac{({{n}^{2}}-1)}{12}\]

Answer 1

Hint: Firstly we will find out the mean that is \[\overline{x}\] and then we will find out the summation that is \[\sum{{{x}_{i}}}\] and then put the value of \[\sum{{{x}_{i}}}\] in the mean \[\overline{x}\] after that find out the variance by putting the obtained values and after simplifying it check which option is correct in the given options.

Complete answer:
Statistics is the study of the collection, organisation and analysis of data and it includes the planning of data collection in terms of design of experiments.
Variance measures the dispersion of a set of data points around their mean and dispersion is the measure of a variation and it is non-negative where values do not cancel out. Commonly there are four types of dispersion; they are range, quartile deviation, mean deviation and standard deviation. Note that the sum of deviation through the statistical mean is always zero.
Mean deviation of a distribution is the arithmetic mean of the absolute deviations of the terms of the distribution from its statistical mean, that is arithmetic mean, median, mode.We must also know that mean deviation of a distribution is the arithmetic mean of absolute deviation of terms.
Standard deviation is the positive square root of the arithmetic mean of the squared deviations from the mean of the distribution and it is also a form of average deviation from the mean.
 Variance is the average squared deviation from the mean of the data and is also the square of standard deviation.
Now according to the question:
Firstly find out the mean:
\[\overline{x}=\dfrac{\sum{{{x}_{i}}}}{n}\]
And we know that \[\sum{{{x}_{i}}}=1,2,3,4........n\]
\[\Rightarrow \sum{{{x}_{i}}}=\dfrac{n(n+1)}{2}\]
Putting the value of \[\sum{{{x}_{i}}}\] in the mean we get:
\[\Rightarrow \overline{x}=\dfrac{\dfrac{n(n+1)}{2}}{n}\]
\[\Rightarrow \overline{x}=\dfrac{n(n+1)}{2n}\]
\[\Rightarrow \overline{x}=\dfrac{(n+1)}{2}\]
Now we can find out the variance as:
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{n}\sum{({{x}_{i}}-}\overline{x}{{)}^{2}}\]
Now applying \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] in \[{{({{x}_{i}}-\overline{x})}^{2}}\] we will get:
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{n}\sum{({{x}_{i}}^{2}-2{{x}_{i}}\overline{x}+(}\overline{x}{{)}^{2}})\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{n}(\sum{{{x}_{i}}^{2}-2\overline{x}\sum{{{x}_{i}}}+}{{(\overline{x})}^{2}}\sum{1})\]
Putting the values of \[{{x}_{i}}\] , \[\overline{x}\] and \[\sum{1}=n\] we get:
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{n}\left( \dfrac{n(n+1)(2n+1)}{6}-2\times \dfrac{(n+1)}{2}\times \dfrac{n(n+1)}{2}+{{(\dfrac{(n+1)}{2})}^{2}}\times n \right)\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{1}{n}\left( \dfrac{n(n+1)(2n+1)}{6}-(n+1)\times \dfrac{n(n+1)}{2}+{{(\dfrac{(n+1)}{2})}^{2}}\times n \right)\]
Now taking \[\dfrac{1}{n}\] inside the bracket:
\[\Rightarrow {{\sigma }^{2}}=\left( \dfrac{n(n+1)(2n+1)}{6n}-\dfrac{(n+1)}{n}\times \dfrac{n(n+1)}{2}+{{(\dfrac{(n+1)}{2})}^{2}}\times \dfrac{n}{n} \right)\]
\[\Rightarrow {{\sigma }^{2}}=\left( \dfrac{(n+1)(2n+1)}{6}-(n+1)\times \dfrac{(n+1)}{2}+{{(\dfrac{(n+1)}{2})}^{2}} \right)\]
\[\Rightarrow {{\sigma }^{2}}=\left( \dfrac{(n+1)(2n+1)}{6}-\dfrac{{{(n+1)}^{2}}}{2}+\dfrac{{{(n+1)}^{2}}}{4} \right)\]
Now taking \[\dfrac{(n+1)}{2}\] in common we get:
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)}{2}\left( \dfrac{(2n+1)}{3}-(n+1)+\dfrac{(n+1)}{2} \right)\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)}{2}\left( \dfrac{(2n+1)}{3}-\dfrac{(n+1)}{2} \right)\]
Taking LCM:
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)}{2}\left( \dfrac{4n+2-(3n+1)}{6} \right)\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)}{2}\left( \dfrac{4n+2-3n-1}{6} \right)\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)}{2}\times \dfrac{(n-1)}{6}\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{(n+1)(n-1)}{12}\]
 As we know that \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{({{n}^{2}}-{{1}^{2}})}{12}\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{({{n}^{2}}-1)}{12}\]
Hence option D is correct as the variance is \[\dfrac{({{n}^{2}}-1)}{12}\].

Note:
You know that if two distribution have some arithmetic mean but different coefficient of variance, then the distribution with less coefficient of variation is less deviated in comparison to the distribution having greater coefficient of variation and coefficient of variance is an absolute quantity, it does not have any unit.