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STD 11 - 13. statistics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 13. statistics4 Marks
MCQ
The variance of the first $n$ natural numbers is
  • ✓
    $\frac{{{n^2} - 1}}{{12}}$
  • B
    $\frac{{{n^2} - 1}}{6}$
  • C
    $\frac{{{n^2} + 1}}{6}$
  • D
    $\frac{{{n^2} + 1}}{{12}}$
✓

Answer

Correct option: A.
$\frac{{{n^2} - 1}}{{12}}$
a
(a) Variance $ = {({\rm{S}}{\rm{.D}}{\rm{.}})^2}$$ = \frac{1}{n}\Sigma {x^2} - {\left( {\frac{{\Sigma x}}{n}} \right)^2}$,$\left( {\because \;\;\bar x = \frac{{\Sigma x}}{n}} \right)$

$ = \frac{{n(n + 1)\;(2n + 1)}}{{6n}} - {\left( {\frac{{n(n + 1)}}{{2n}}} \right)^2} $

$= \frac{{{n^2} - 1}}{{12}}$.

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