Question

Write the variance of first n natural numbers.

 

Answer 1

​Sum of first n natural numbers 

\[= \frac{n\left( n + 1 \right)}{2}\]

Mean, \[\bar{X} = \frac{\text{ Sum of all the observations} }{\text{ Total number of observations } }\]

\[= \frac{\frac{n\left( n + 1 \right)}{2}}{n} = \frac{n + 1}{2}\]

\[\therefore \sigma^2 = \frac{\sum \left( x_i - \bar{X} \right)^2}{n} = \frac{\sum \left( x_i - \frac{n + 1}{2} \right)^2}{n}\]

\[ = \frac{1}{n}\sum \left[ {x_i}^2 - x_i \left( n + 1 \right) + \left( \frac{n + 1}{2} \right)^2 \right]\]

\[ = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6n} - \left[ \frac{n\left( n + 1 \right)}{2} \right]\left( \frac{n + 1}{n} \right) + \frac{\left( n + 1 \right)^2}{4n} \times n\]

\[ = \frac{\left( n + 1 \right)\left( 2n + 1 \right)}{6} - \frac{\left( n + 1 \right)^2}{2} + \frac{\left( n + 1 \right)^2}{4}\]

\[ = \frac{\left( n + 1 \right)\left( n - 1 \right)}{12} = \frac{\left( n^2 - 1 \right)}{12}\]