Find the mean deviation about the median for the data.

    \[36,72,46,42,60,45,53,46,51,49\]

Find the mean deviation about the median for the data.

    \[36,72,46,42,60,45,53,46,51,49\]

Solution:-

To find the median arrange the given observations in ascending order,

    \[36,42,45,46,46,49,51,53,60,72\]

Total number of observations =

    \[10\]

Then,

Median = (

    \[{{(10/2)}^{th}}\]

observation +

    \[{{((10/2)+1)}^{th}}\]

observation)/

    \[2\]

    \[{{(10/2)}^{th}}\]

observation =

    \[46\]

    \[{{((10/2)+1)}^{th}}\]

observation =

    \[49\]

Median =

    \[(46+49)/2\]

=

    \[95/2\]

=

    \[47.5\]

finding the absolute values of the respective deviations from the median, i.e.,

    \[\left| {{x}_{i}}-M \right|\]

are

    \[11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5\]

    \[\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-M \right|}=70\]

Finding Mean Deviation,

M.D (M)=

    \[\frac{1}{10}\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-M \right|}\]

=

    \[(1/10)\times 70\]

=

    \[7\]

Therefore, the mean deviation about the median for the given data is

    \[7\]

.