1. Tardigrade
  2. Question
  3. Mathematics
  4. Standard deviation of the first 2n + 1 natural numbers is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Standard deviation of the first $2n + 1$ natural numbers is equal to

Statistics

Solution:

$\bar{x} =$ mean $=\frac{ 1+2+3+...+\left(2n+1\right)}{\left(2n+1\right)} $
$ = \frac{\sum\left(2n+1\right)}{\left(2n+1\right)} $
$ = \frac{\left(2n+1\right)\left(2n+2\right)}{2\left(2n+1\right)} $
$= n+1$
Varience $= \sigma^{2} = \frac{1}{2n+1} \sum\left(x_{i}-\bar{x}\right)^{2} $
$= \frac{2}{2n+1} \left[n^{2}+\left(n-1\right)^{2}+.........+\left(1\right)^{2}\right]$
$ = \frac{2}{2n+1} \sum n^{2} = \frac{2}{2n+1}\cdot\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
$ = \frac{n\left(n+1\right)}{3} $
$ S.D. = \sqrt{Varience}$
$ = \sqrt{\frac{n\left(n+1\right)}{3}}.$