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Question: Consider any set of observations \(x_{1},x_{2},.x_{3},....,x_{101}\); it being given that \(x_{1} < ...

Consider any set of observations x1,x2,.x3,....,x101x_{1},x_{2},.x_{3},....,x_{101}; it being given that x1<x2<x3<....<x100<x101x_{1} < x_{2} < x_{3} < .... < x_{100} < x_{101}; then the mean deviation of this set of observations about a point k is minimum when k equals

A

x1x_{1}

B

x51x_{51}

C

x1+x2+...+x101101\frac{x_{1} + x_{2} + ... + x_{101}}{101}

D

x50x_{50}

Answer

x51x_{51}

Explanation

Solution

Mean deviation is minimum when it is considered about the item, equidistant from the beginning and the end i.e., the median. In this case median is 101+12th\frac{101 + 1}{2}th

i.e., 51st item i.e., x51x_{51}.