If $\sum\limits_{i = 1}^n {({x_i} - a) = {\rm{n}}} $ and $\sum\limits_{i = 1}^n {{{({x_i} - a)}^2} = {\rm{n}}} a,{\rm{ }}(n,a > 1),$ then the standard deviation of n observations ${x_1},{x_2},......,.{x_n}$ is

$\sqrt {n\left( {a - 1} \right)} $

JEE MATHEMATICS MCQ: STATISTICS

1 Easy

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Application +4 | -1

\sum\limits_{i = 1}^n {({x_i} - a) = \mathrm{n}}

and

\sum\limits_{i = 1}^n {{{({x_i} - a)}^2} = \mathrm{n}} a,\mathrm{ }(n,a > 1),

then the standard deviation of n observations

{x_1},{x_2},......,.{x_n}

View answer and explanation

Correct option: D

\sqrt {a - 1}

Option D (

\sqrt {a - 1}

) is correct because it represents the standard deviation calculated from the given sums, using the definition of variance and standard deviation.

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