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

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x_{1}

x_{51}

x_{50}

\frac{x_{1} - x_{2} - \dots \dots . . - x_{101}}{101}

answer is B.

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Detailed Solution

We know that mean deviation is minimum when it is taken by median.
So, here K is the median of given observations.
Now, the number of observations is 101.

k = (\frac{n + 1}{2})

th observation= 51th observation
Then, k =

x_{51}

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