A) \[\frac{1}{n}\]
B) \[\sqrt{2}\]
C) \[2\]
D) \[\frac{\sqrt{2}}{n}\]
Correct Answer: C
Solution :
[c] \[\because \sigma =\sqrt{\frac{\Sigma x{{i}^{2}}}{N}-{{\left( \frac{\Sigma {{x}_{i}}}{N} \right)}^{2}}}\] \[\therefore 2=\sqrt{\frac{({{a}^{2}}+{{a}^{2}}...'2n'times)}{2n}-0}\] \[\Rightarrow 4=\frac{2n{{a}^{2}}}{2n}\Rightarrow {{a}^{2}}=4\Rightarrow \left| a \right|=2\]
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