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Find mean deviation about mode {{10-20,15},{20-30,25},{30-40,20},{40-50,12},{50-60,8}} [ Calculator, Method and examples ]

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

 Class(1) f(2) Mid value (x)(3) |x-Z|=|x-26.6667|(4) f*|x-Z|(5)=(2)xx(4) 10 - 20 15 15 15=(10+20)/2 11.6667 |15-26.6667|=11.6667|x - 26.6667| 175 175=15xx11.6667(5)=(2)xx(4) 20 - 30 25 25 25=(20+30)/2 1.6667 |25-26.6667|=1.6667|x - 26.6667| 41.6667 41.6667=25xx1.6667(5)=(2)xx(4) 30 - 40 20 35 35=(30+40)/2 8.3333 |35-26.6667|=8.3333|x - 26.6667| 166.6667 166.6667=20xx8.3333(5)=(2)xx(4) 40 - 50 12 45 45=(40+50)/2 18.3333 |45-26.6667|=18.3333|x - 26.6667| 220 220=12xx18.3333(5)=(2)xx(4) 50 - 60 8 55 55=(50+60)/2 28.3333 |55-26.6667|=28.3333|x - 26.6667| 226.6667 226.6667=8xx28.3333(5)=(2)xx(4) --- --- --- --- --- -- n=80 -- -- sum f*|x-Z|=830

To find Mode Class
Here, maximum frequency is 25.

:. The mode class is 20 - 30.

:. L = lower boundary point of mode class =20

:. f_1 =  frequency of the mode class =25

:. f_0 =  frequency of the preceding class =15

:. f_2 =  frequency of the succedding class =20

:. c =  class length of mode class =10

Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c

=20 + ((25 - 15)/(2*25 - 15 - 20)) * 10

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