A) \[\frac{\sqrt{{{n}^{2}}+1}}{12}\]
B) \[\frac{{{n}^{2}}-1}{12}\]
C) \[\sqrt{\frac{{{n}^{2}}-1}{12}}\]
D) \[\frac{{{n}^{2}}+1}{12}\]
Correct Answer: C
Solution :
Since, \[\Sigma n=\frac{n(n+1)}{2}\] and \[\Sigma {{n}^{2}}=\frac{n(n+1)\,(2n+1)}{6}\] \[\therefore \] \[SD=\sqrt{\frac{\Sigma {{x}^{2}}}{n}-{{\left( \Sigma \frac{x}{n} \right)}^{2}}}\] \[=\sqrt{\frac{\Sigma {{n}^{2}}}{n}-{{\left( \frac{\Sigma n}{n} \right)}^{2}}}\] \[=\sqrt{\frac{n(n+1)(2n+1)}{6n}-{{\left( \frac{n+(n+1)}{2n} \right)}^{2}}}\] \[=\sqrt{\frac{(n+1)(2n+1)}{6}-\frac{{{(n+1)}^{2}}}{4}}\] \[=\sqrt{\frac{{{n}^{2}}-1}{12}}\]
You need to login to perform this action.
You will be redirected in
3 sec
