If sum limits_{i=1}^{n}left(x_{i}-aright)=n a | vedclass

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Question

$S.D =\sqrt{\frac{\sum_{i=1}^{n}\left( x _{ i }- a \right)}{ n }-\left(\frac{\sum_{i=1}^{ n }\left( x _{ i }- a \right)}{ n }\right)^{2}}$

$=\sqrt{

Vedclass · Accepted Answer

$\sqrt{a-1}$

Vedclass · Answer

$S.D =\sqrt{\frac{\sum_{i=1}^{n}\left( x _{ i }- a \right)}{ n }-\left(\frac{\sum_{i=1}^{ n }\left( x _{ i }- a \right)}{ n }\right)^{2}}$

$=\sqrt{\frac{ na }{ n }-\left(\frac{ n }{ n }\right)^{2}}$

$\left\{\right.$ Given $\left.\sum_{i=1}^{n}\left(x_{i}-a\right)=n \sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\right\}$

$=\sqrt{a-1}$

Subject	Mathematics	Physics	Chemistty
Mean	$42$	$32$	$40.9$
Standard deviation	$12$	$15$	$20$

If $\sum \limits_{i=1}^{n}\left(x_{i}-a\right)=n$ and $\sum \limits_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a,(n, a>1)$ then the standard deviation of $n$ observations $x _{1}, x _{2}, \ldots, x _{ n }$ is

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