If $\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$ and $\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$ (n, a > 1) then the standard deviation of n observations x1 , x2 , ..., xn is :

Question

If $\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$ and $\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$
(n, a > 1) then the standard deviation of n
observations x₁ , x₂ , ..., x_n is :

$a$ – 1

$n\sqrt {a - 1} $

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

$\sqrt {a - 1} $

Solution

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