Consider any set of observations x1, x2, x3, ...., x101; it being given that x1

Question

Consider any set of observations x1, x2, x3, ...., x101; it being given that x1 < x2 < x3 < ... < x100 < x101 ; then the mean deviation of this set of observations about a point k is minimum when k equals (A) x1 (B) x51 (C) (x1+x2+...+x101/101) (D) x50

Tardigrade Admin · Accepted Answer

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+1/2) th i.e. 51st item i.e., x51.

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

$x_{1}$

$x_{51}$

$\frac{x _{1} + x _{2} + ... + x _{101}}{101}$

$x_{50}$

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 $\frac{101 + 1}{2} t h$ i.e. 51st item i.e., $x_{51}$ .

Q. Consider any set of observations x1​,x2​,x3​,....,x101​; it being given that x1​<x2​<x3​<...<x100​<x101​; then the mean deviation of this set of observations about a point k is minimum when k equals

x1​

x51​

101x1​+x2​+...+x101​​

x50​