A) \[\frac{n\,M-a}{4}\]
B) \[\frac{n\,M+a}{2}\]
C) \[\frac{n\,M-A}{2}\]
D) n M + a
Correct Answer: A
Solution :
Let the mean of the remaining 4 observations be \[{{\bar{x}}_{1}}\]. Then, \[M=\frac{a+4{{{\bar{x}}}_{1}}}{(n-4)+4}\]Þ \[\overline{{{x}_{1}}}=\frac{nM-a}{4}\].
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