Consider any set of 201 observations $x_{1}, x_{2}, \dots, x_{200}, x_{201}$ . It is given that $x_{1} < x_{2} < \dots < x_{200} < x_{201}$ . Then, the mean deviation of this set of observations about a point k is minimum, when k equals

Q. If both mean and the standard deviation of

50

observation

x_{1}, x_{2}, . . ., x_{50}

are equal to

16

, then the mean of

(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, . . ., (x_{50} - 4)^{2}

is :

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For the values x1,x2.....x101 of a distribution x1<x2<x3<.....<x100<x101. The mean deviation of this distribution with respect to a number k will be minimum when k is equal to

The correct option is C x51We know that mean deviation is minimum when it is taken by median .So, here K is median of given observations. Now, number of observations are 101.K=(n+12) th observation = 51th observation .∴K=x51

For the values $x_{1}, x_{2} . . . . . x_{101}$ of a distribution $x_{1} < x_{2} < x_{3} < . . . . . < x_{100} < x_{101} .$ The mean deviation of this distribution with respect to a number k will be minimum when k is equal to

The correct option is C $x_{51}$
We know that mean deviation is minimum when it is taken by median .
So, here K is median of given observations.
Now, number of observations are 101.
$K = (\frac{n + 1}{2})$ th observation = $51^{t h}$ observation .
$∴ K = x_{51}$