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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 - 2015 15 `15=(10+20)/2` 11.6667 `|15-26.6667|=11.6667`
`|x - 26.6667|`
 175 `175=15xx11.6667`
`(5)=(2)xx(4)`
20 - 3025 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 - 4020 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 - 5012 45 `45=(40+50)/2` 18.3333 `|45-26.6667|=18.3333`
`|x - 26.6667|`
 220 `220=12xx18.3333`
`(5)=(2)xx(4)`
50 - 608 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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