A) \[n\overline{x}\]
B) \[\frac{\overline{x}}{n}\]
C) 0
D) 1
Correct Answer: C
Solution :
[c] \[\because \overline{x}\]is mean of \[{{x}_{1}},{{x}_{2}}.{{x}_{3}},......{{x}_{n}}.\]. \[\therefore {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......{{x}_{n}}=n.\overline{x}\] ......(1) Now, \[({{x}_{1}}-\overline{x})+({{x}_{2}}-\overline{x})+({{x}_{3}}-\overline{x})+.......({{x}_{n}}-\overline{x})\] \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......{{x}_{n}}-n.\overline{x}\] [from (1)] \[=n\overline{x}-n\overline{x}=0\]
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