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Find mean deviation about median {{10-20,20-30,30-40,40-50,50-60},{15,25,20,12,8}}

Solution:
Your problem -> mean deviation about median {{10-20,20-30,30-40,40-50,50-60},{15,25,20,12,8}}

 Class(1) f(2) cf(3) Mid value (x)(4) |x-M|=|x-30|(5) f*|x-M|(6)=(2)xx(5) 10 - 20 15 15 15=0+15(3)=Previous (3)+(2) 15 15=(10+20)/2 15 |15-30|=15|x - 30| 225 225=15xx15(6)=(2)xx(5) 20 - 30 25 40 40=15+25(3)=Previous (3)+(2) 25 25=(20+30)/2 5 |25-30|=5|x - 30| 125 125=25xx5(6)=(2)xx(5) 30 - 40 20 60 60=40+20(3)=Previous (3)+(2) 35 35=(30+40)/2 5 |35-30|=5|x - 30| 100 100=20xx5(6)=(2)xx(5) 40 - 50 12 72 72=60+12(3)=Previous (3)+(2) 45 45=(40+50)/2 15 |45-30|=15|x - 30| 180 180=12xx15(6)=(2)xx(5) 50 - 60 8 80 80=72+8(3)=Previous (3)+(2) 55 55=(50+60)/2 25 |55-30|=25|x - 30| 200 200=8xx25(6)=(2)xx(5) --- --- --- --- --- --- -- n=80 -- -- -- sum f*|x-M|=830

To find Median Class
= value of (n/2)^(th) observation

= value of (80/2)^(th) observation

= value of 40^(th) observation

From the column of cumulative frequency cf, we find that the 40^(th) observation lies in the class 30 - 40.

:. The median class is 30 - 40.

Now,
:. L = lower boundary point of median class =30

:. n = Total frequency =80

:. cf = Cumulative frequency of the class preceding the median class =40

:. f = Frequency of the median class =20

:. c = class length of median class =10

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