A) \[M-{{x}_{n}}+{x}'\]
B) \[\frac{nM-{{x}_{n}}+{x}'}{n}\]
C) \[\frac{(n-1)M+{x}'}{n}\]
D) \[\frac{M-{{x}_{n}}+{x}'}{n}\]
Correct Answer: B
Solution :
\[M=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}......{{x}_{n}}}{n}\] i.e., \[\underset{n}{\mathop{\underline{\begin{matrix} nM \\ nM-{{x}_{n}} \\ nM-{{x}_{n}}+{x}' \\ \end{matrix}}}}\,\begin{matrix} = \\ = \\ = \\ \end{matrix}\underset{n}{\mathop{\underline{\begin{matrix} {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......{{x}_{n-1}}+{{x}_{n}} \\ {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......{{x}_{n-1}}\ \ \ \ \ \\ {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......{{x}_{n-1}}+{x}' \\ \end{matrix}}}}\,\] \ New average\[=\frac{nM-{{x}_{n}}+{x}'}{n}\].
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