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13.Statistics
medium
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
A
$n \sqrt{ a -1}$
B
$\sqrt{a-1}$
C
$a-1$
D
$\sqrt{n(a-1)}$
(JEE MAIN-2020)
Solution
$S.D =\sqrt{\frac{\sum_{i=1}^{n}\left( x _{ i }- a \right)}{ n }-\left(\frac{\sum_{i=1}^{ n }\left( x _{ i }- a \right)}{ n }\right)^{2}}$
$=\sqrt{\frac{ na }{ n }-\left(\frac{ n }{ n }\right)^{2}}$
$\left\{\right.$ Given $\left.\sum_{i=1}^{n}\left(x_{i}-a\right)=n \sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\right\}$
$=\sqrt{a-1}$
Standard 11
Mathematics
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