A) \[\bar{x}+n\]
B) \[\bar{x}+\frac{n}{2}\]
C) \[\bar{x}+\frac{n+1}{2}\]
D) None of these
Correct Answer: C
Solution :
[c] Let \[{{x}_{1}},{{x}_{2}}....{{x}_{n}}\] be n items. Then \[\bar{x}=\frac{1}{n}\Sigma {{x}_{i}}\] Let \[{{y}_{1}}={{x}_{1}}+1,{{y}_{2}}={{x}_{2}}+2,{{y}_{3}}={{x}_{3}}+3,..{{y}_{n}}={{x}_{n}}+n\] Then the mean of the new series is \[\frac{1}{n}\Sigma {{y}_{i}}=\frac{1}{n}\sum\limits_{i=1}^{n}{({{x}_{i}}+\frac{1}{i})}\] \[=\frac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}+\frac{1}{n}(1+2+3+...+n)}\] \[=\bar{x}+\frac{1}{n}.\frac{n(n+1)}{2}=\bar{x}+\frac{n+1}{2}.\]
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