"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 , ..., xn is :"

Accepted Answer

"S.D = $$\sqrt {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n} - {{\left( {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n}} \right)}^2}} $$

= $\sqrt {{{na} \over n} - {{\left( {{n \over n}} \right)}^2}}$

= $\sqrt {a - 1}$"

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 :

Solution

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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 :

Solution

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More Statistics questions

Drill 25 more like these. Every day. Free.

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 :