1. Formula & Example
Formula
|
1. Mean `bar x = (sum fx)/n`
|
|
2. Median `M = L + (n/2 - cf)/f * c`
|
|
3. Mode `Z = 3M - 2 bar x`
|
Examples
1. Calculate Mean, Median, Mode from the following mixed data
| Class | Frequency | | 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:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `f*x`
`(4)=(2)xx(3)` | `cf`
`(6)` | | 1 | 3 | 1 `1=1` | 3 `3=3xx1`
`(4)=(2)xx(3)` | 3 `3=0+3`
`(6)=`Previous `(6)+(2)` | | 2 | 4 | 2 `2=2` | 8 `8=4xx2`
`(4)=(2)xx(3)` | 7 `7=3+4`
`(6)=`Previous `(6)+(2)` | | 5 | 10 | 5 `5=5` | 50 `50=10xx5`
`(4)=(2)xx(3)` | 17 `17=7+10`
`(6)=`Previous `(6)+(2)` | | 6 - 10 | 23 | 8 `8=(6+10)/2` | 184 `184=23xx8`
`(4)=(2)xx(3)` | 40 `40=17+23`
`(6)=`Previous `(6)+(2)` | | 10 - 20 | 20 | 15 `15=(10+20)/2` | 300 `300=20xx15`
`(4)=(2)xx(3)` | 60 `60=40+20`
`(6)=`Previous `(6)+(2)` | | 20 - 30 | 20 | 25 `25=(20+30)/2` | 500 `500=20xx25`
`(4)=(2)xx(3)` | 80 `80=60+20`
`(6)=`Previous `(6)+(2)` | | 30 - 50 | 15 | 40 `40=(30+50)/2` | 600 `600=15xx40`
`(4)=(2)xx(3)` | 95 `95=80+15`
`(6)=`Previous `(6)+(2)` | | 50 - 70 | 3 | 60 `60=(50+70)/2` | 180 `180=3xx60`
`(4)=(2)xx(3)` | 98 `98=95+3`
`(6)=`Previous `(6)+(2)` | | 70 - 100 | 2 | 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`
2. Calculate Mean, Median, Mode from the following mixed data
| Class | Frequency | | 2 | 1 | | 3 | 2 | | 4 | 2 | | 5 - 9 | 8 | | 10 - 14 | 15 | | 15 - 19 | 8 | | 20 - 29 | 4 |
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `f*x`
`(4)=(2)xx(3)` | `cf`
`(6)` | | 2 | 1 | 2 | 2 | 1 | | 3 | 2 | 3 | 6 | 3 | | 4 | 2 | 4 | 8 | 5 | | 5 - 9 | 8 | 7 | 56 | 13 | | 10 - 14 | 15 | 12 | 180 | 28 | | 15 - 19 | 8 | 17 | 136 | 36 | | 20 - 29 | 4 | 24.5 | 98 | 40 | | --- | --- | --- | --- | --- | | `n = 40` | ----- | `sum f*x=486` | ----- |
Mean `bar x = (sum fx)/n`
`=486/40`
`=12.15`
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(40/2)^(th)` observation
= value of `20^(th)` observation
From the column of cumulative frequency `cf`, we find that the `20^(th)` observation lies in the class `10 - 14`.
`:.` The median class is `9.5 - 14.5`.
Now,
`:. L = `lower boundary point of median class `=9.5`
`:. n = `Total frequency `=40`
`:. cf = `Cumulative frequency of the class preceding the median class `=13`
`:. f = `Frequency of the median class `=15`
`:. c = `class length of median class `=5`
Median `M = L + (n/2 - cf)/f * c`
`=9.5 + (20 - 13)/15 * 5`
`=9.5 + (7)/15 * 5`
`=9.5 + 2.3333`
`=11.8333`
Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`
`=3 * 11.8333 - 2 * 12.15`
`=35.5 - 24.3`
`=11.2`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
