A) 2
B) 4
C) 3
D) 1
Correct Answer: B
Solution :
\[\sum\limits_{i=1}^{n}{{{({{x}_{i}}+1)}^{2}}\,\,=\,\,9n\,\,\Rightarrow \,\,\sum\limits_{i\,=\,1}^{n}{{{x}^{2}}_{i}+2\,\sum\limits_{i=1}^{n}{{{x}_{i}}+n}}}\] = 9 ?. (1) \[\sum\limits_{i=1}^{n}{{{({{x}_{i}}-1)}^{2}}=5n\Rightarrow \sum\limits_{i\,=\,1}^{n}{{{x}^{2}}_{i}-2\,\sum\limits_{i=1}^{n}{{{x}_{i}}+n}}}\] = 5n ...(2) Eq. (1) + (2) \[\Rightarrow \,\,\,2\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}\,+\,2\,n\,\,=\,\,14\,n}\] Eq. (1) \[-\] (2) \[\Rightarrow \,\,\,4\sum\limits_{i=1}^{n}{{{x}_{i}}\,=\,\,4\,n}\] \[\Rightarrow \,\,\,\sum\limits_{i=1}^{n}{{{x}_{i}}\,=\,\,n}\] \[S.D.\,\,\,\sigma \,\,=\,\,\frac{\sum{{{x}_{i}}^{2}}}{n}\,-\,{{(\overline{x})}^{2}}\] \[\sigma \,\,=\,\,\sqrt{\frac{6\,n}{n}\,-\,(1)}\] \[\sigma \,\,=\,\,\sqrt{5}\]
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