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Find median {{1,3},{2,4},{5,10},{6-10,23},{10-20,20},{20-30,20},{30-50,15},{50-70,3},{70-100,2}}

Solution:
Your problem `->` median {{1,3},{2,4},{5,10},{6-10,23},{10-20,20},{20-30,20},{30-50,15},{50-70,3},{70-100,2}}


Class
`(1)`
Frequency `(f)`
`(2)`
`cf`
`(6)`
13 3 `3=0+3`
`(6)=`Previous `(6)+(2)`
24 7 `7=3+4`
`(6)=`Previous `(6)+(2)`
510 17 `17=7+10`
`(6)=`Previous `(6)+(2)`
6 - 1023 40 `40=17+23`
`(6)=`Previous `(6)+(2)`
10 - 2020 60 `60=40+20`
`(6)=`Previous `(6)+(2)`
20 - 3020 80 `80=60+20`
`(6)=`Previous `(6)+(2)`
30 - 5015 95 `95=80+15`
`(6)=`Previous `(6)+(2)`
50 - 703 98 `98=95+3`
`(6)=`Previous `(6)+(2)`
70 - 1002 100 `100=98+2`
`(6)=`Previous `(6)+(2)`
---------
`n = 100`-----


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

= value of `(100/2)^(th)` observation

= value of `50^(th)` observation

From the column of cumulative frequency `cf`, we find that the `50^(th)` observation lies in the class `10 - 20`.

`:.` The median class is `10 - 20`.

Now,
`:. L = `lower boundary point of median class `=10`

`:. n = `Total frequency `=100`

`:. 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`

Median `M = L + (n/2 - cf)/f * c`

`=10 + (50 - 40)/20 * 10`

`=10 + (10)/20 * 10`

`=10 + 5`

`=15`








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