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Find mode {{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 `->` mode {{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)`
Mid value `(x)`
`(3)`
`f*x`
`(4)=(2)xx(3)`
`cf`
`(6)`
13 1 `1=1` 3 `3=3xx1`
`(4)=(2)xx(3)`
 3 `3=0+3`
`(6)=`Previous `(6)+(2)`
24 2 `2=2` 8 `8=4xx2`
`(4)=(2)xx(3)`
 7 `7=3+4`
`(6)=`Previous `(6)+(2)`
510 5 `5=5` 50 `50=10xx5`
`(4)=(2)xx(3)`
 17 `17=7+10`
`(6)=`Previous `(6)+(2)`
6 - 1023 8 `8=(6+10)/2` 184 `184=23xx8`
`(4)=(2)xx(3)`
 40 `40=17+23`
`(6)=`Previous `(6)+(2)`
10 - 2020 15 `15=(10+20)/2` 300 `300=20xx15`
`(4)=(2)xx(3)`
 60 `60=40+20`
`(6)=`Previous `(6)+(2)`
20 - 3020 25 `25=(20+30)/2` 500 `500=20xx25`
`(4)=(2)xx(3)`
 80 `80=60+20`
`(6)=`Previous `(6)+(2)`
30 - 5015 40 `40=(30+50)/2` 600 `600=15xx40`
`(4)=(2)xx(3)`
 95 `95=80+15`
`(6)=`Previous `(6)+(2)`
50 - 703 60 `60=(50+70)/2` 180 `180=3xx60`
`(4)=(2)xx(3)`
 98 `98=95+3`
`(6)=`Previous `(6)+(2)`
70 - 1002 85 `85=(70+100)/2` 170 `170=2xx85`
`(4)=(2)xx(3)`
 100 `100=98+2`
`(6)=`Previous `(6)+(2)`
---------------
`n = 100`-----`sum f*x=1995`-----


Mean `bar x = (sum fx)/n`

`=1995/100`

`=19.95`



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`



Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`

`=3 * 15 - 2 * 19.95`

`=45 - 39.9`

`=5.1`








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