Consider any set of observations x1, x2, x3, ldots, x101 It is given that x1

Question

Consider any set of observations x1, x2, x3, ldots, x101 It is given that x1< x2< x3< ldots< 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+ dots +x101/101) (D) x50

Tardigrade Admin · Accepted Answer

Mean deviation is minimum when it is considered about an item equidistant from the beginning and the end, i.e., the median. In this case, the median is (101+1/2) th, i.e., 51 st item, i.e., x51

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

$x_{1}$

$x_{51}$

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

$x_{50}$

Mean deviation is minimum when it is considered about an item equidistant from the beginning and the end, i.e., the median. In this case, the median is $\frac{101 + 1}{2}$ th, i.e., 51 st item, i.e., $x_{51}$

Q. Consider any set of observations x1​,x2​,x3​,…,x101​ It is 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​