Question
Question: For the values of \[{x_1},\;{x_2},{\text{ }} \ldots \ldots \ldots .{x_{101}}\] of a distribution \({...
For the values of 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 –
a) x1
b) x51
c) x50
d) 101x1+x2+.......+x101
Solution
In this question, they give one distribution. From that distribution we have to choose which term is equal to k when the mean deviation of this distribution with respect to k will be minimum. We solve this problem by taking k as the median of the given observation.
Complete step-by-step answer:
As we know mean deviation is minimum when median has been taken.
Therefore, here k is taken as the median of the given observations.
Total number of observations in the question is 101.
Now we have to apply the formula of median which is 2n+1
Since k is the median and n=101 in this question therefore we can write-
⇒k=2n+1
By substituting the total number of observations,
⇒k=2101+1
Adding the terms we get,
⇒k=2102
Dividing the terms we get,
⇒k=51
Hence, k=51th observation
∴ Thus, k=x51
So, the correct answer is “Option b”.
Note: Mean, median and mode are the three kinds of averages which have wide application in statistics.
Median is generally used to find the middle value or centre of a set of data or information.
It can also be used for an open-end distribution and it is more useful than mean.
The main formula which is used for finding out the value of median for an odd discrete series is 2n+1 where n represents the total number of data or numbers available in a distribution and the formula for even discrete series is 21[2n+(2n+1)]
It is only applicable in quantitative data but not qualitative.
Value of the median is not dependent on all the values of the data in a data set and it does not depend on the individual value of a particular data of a set.