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Find mean deviation about median {{10,3},{11,12},{12,18},{13,12},{14,3}}

Solution:
Your problem `->` mean deviation about median {{10,3},{11,12},{12,18},{13,12},{14,3}}


`x`
`(1)`
`f`
`(2)`
`cf`
`(3)`
`|x-M|=|x-12|`
`(4)`
`f*|x-M|`
`(5)=(2)xx(4)`
103 3 `3=0+3`
`(3)=`Previous `(3)+(2)`
 2 `|x - 12|=|10-12|=2` 6 `6=3xx2`
`(5)=(2)xx(4)`
1112 15 `15=3+12`
`(3)=`Previous `(3)+(2)`
 1 `|x - 12|=|11-12|=1` 12 `12=12xx1`
`(5)=(2)xx(4)`
1218 33 `33=15+18`
`(3)=`Previous `(3)+(2)`
 0 `|x - 12|=|12-12|=0` 0 `0=18xx0`
`(5)=(2)xx(4)`
1312 45 `45=33+12`
`(3)=`Previous `(3)+(2)`
 1 `|x - 12|=|13-12|=1` 12 `12=12xx1`
`(5)=(2)xx(4)`
143 48 `48=45+3`
`(3)=`Previous `(3)+(2)`
 2 `|x - 12|=|14-12|=2` 6 `6=3xx2`
`(5)=(2)xx(4)`
---------------
--`n=48`----`sum f*|x-M|=36`


Median :
M = value of `(n/2)^(th)` observation

= value of `(48/2)^(th)` observation

= value of `24^(th)` observation

From the column of cumulative frequency `cf`, we find that the `24^(th)` observation is `12`.

Hence, the median of the data is `12`.

Mean deviation of Median





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