A) \[\bar{X}-{{x}_{2}}+\lambda \]
B) \[\frac{\bar{X}-{{x}_{2}}-\lambda }{n}\]
C) \[\frac{\bar{X}-{{x}_{2}}+\lambda }{n}\]
D) \[\frac{n\bar{X}-{{x}_{2}}+\lambda }{n}\]
Correct Answer: D
Solution :
[d] Mean of series \[({{x}_{1}},{{x}_{2}},{{x}_{3}}...{{x}_{n}})\] \[\bar{x}=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....{{x}_{n}}}{n}\] \[\Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...{{x}_{n}}=n\bar{x}\] Now we will replace \[{{x}_{2}}\] by so no. of elements in series will not change. New series will include \[\lambda \] and exclude \[{{x}_{2}}\] Hence new series sum: \[({{x}_{1}}+{{x}_{2}}+...{{x}_{n}})-{{x}_{2}}+\lambda =n\bar{x}+\lambda -{{x}_{2}}\] Now new mean \[=\frac{n\bar{x}+\lambda -{{x}_{2}}}{n}=\frac{n\bar{x}-{{x}_{2}}+\lambda }{n}\]
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