If _i=1^n(x_i-a)=n and _i=1^n(x_i-a)^2=na, (n,a1), then the standard deviation of n obser…

Question

If $\displaystyle\sum_{i=1}^n(x_i-a)=n$ and $\displaystyle\sum_{i=1}^n(x_i-a)^2=na$, $(n,a>1)$, then the standard deviation of n observations $x_1,x_2,\ldots,x_n$ is:

Accepted Answer

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

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

= $$\sqrt {a - 1} $$

Solutions · 1