1. Formula & Example
Formula
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1. Mean `bar x = (sum fx)/n`
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2. Median `M = L + (n/2 - cf)/f * c`
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3. Mode `Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
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Examples
1. Calculate Mean, Median, Mode from the following grouped data
| Class | Frequency | | 2 - 4 | 3 | | 4 - 6 | 4 | | 6 - 8 | 2 | | 8 - 10 | 1 |
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | Mid value `(x)`
`(3)` | `f*x`
`(4)=(2)xx(3)` | `cf`
`(6)` | | 2-4 | 3 | 3 `3=(2+4)/2` | 9 `9=3xx3`
`(4)=(2)xx(3)` | 3 `3=0+3`
`(6)=`Previous `(6)+(2)` | | 4-6 | 4 | 5 `5=(4+6)/2` | 20 `20=4xx5`
`(4)=(2)xx(3)` | 7 `7=3+4`
`(6)=`Previous `(6)+(2)` | | 6-8 | 2 | 7 `7=(6+8)/2` | 14 `14=2xx7`
`(4)=(2)xx(3)` | 9 `9=7+2`
`(6)=`Previous `(6)+(2)` | | 8-10 | 1 | 9 `9=(8+10)/2` | 9 `9=1xx9`
`(4)=(2)xx(3)` | 10 `10=9+1`
`(6)=`Previous `(6)+(2)` | | --- | --- | --- | --- | --- | | -- | `n = 10` | -- | `sum f*x=52` | -- |
Mean `bar x = (sum fx)/n`
`=52/10`
`=5.2`
To find Median Class
= value of `(n/2)^(th)` observation
= value of `(10/2)^(th)` observation
= value of `5^(th)` observation
From the column of cumulative frequency `cf`, we find that the `5^(th)` observation lies in the class `4 - 6`.
`:.` The median class is `4 - 6`.
Now,
`:. L = `lower boundary point of median class `=4`
`:. n = `Total frequency `=10`
`:. cf = `Cumulative frequency of the class preceding the median class `=3`
`:. f = `Frequency of the median class `=4`
`:. c = `class length of median class `=2`
Median `M = L + (n/2 - cf)/f * c`
`=4 + (5 - 3)/4 * 2`
`=4 + (2)/4 * 2`
`=4 + 1`
`=5`
To find Mode Class
Here, maximum frequency is `4`.
`:.` The mode class is `4 - 6`.
`:. L = `lower boundary point of mode class `=4`
`:. f_1 = ` frequency of the mode class `=4`
`:. f_0 = ` frequency of the preceding class `=3`
`:. f_2 = ` frequency of the succedding class `=2`
`:. c = ` class length of mode class `=2`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`=4 + ((4 - 3)/(2*4 - 3 - 2)) * 2`
`=4 + (1/3) * 2`
`=4 + 0.6667`
`=4.6667`
2. Calculate Mean, Median, Mode from the following grouped data
Solution:
`x`
`(1)` | Frequency `(f)`
`(2)` | `f*x`
`(3)=(2)xx(1)` | `cf`
`(5)` | | 0 | 1 | 0 `0=1xx0`
`(3)=(2)xx(1)` | 1 `1=0+1`
`(5)=`Previous `(5)+(2)` | | 1 | 5 | 5 `5=5xx1`
`(3)=(2)xx(1)` | 6 `6=1+5`
`(5)=`Previous `(5)+(2)` | | 2 | 10 | 20 `20=10xx2`
`(3)=(2)xx(1)` | 16 `16=6+10`
`(5)=`Previous `(5)+(2)` | | 3 | 6 | 18 `18=6xx3`
`(3)=(2)xx(1)` | 22 `22=16+6`
`(5)=`Previous `(5)+(2)` | | 4 | 3 | 12 `12=3xx4`
`(3)=(2)xx(1)` | 25 `25=22+3`
`(5)=`Previous `(5)+(2)` | | --- | --- | --- | --- | | `n=25` | `sum f*x=55` | -- |
Mean `bar x = (sum fx)/n`
`=55/25`
`=2.2`
Median :
M = value of `(n/2)^(th)` observation
= value of `(25/2)^(th)` observation
= value of `12^(th)` observation
From the column of cumulative frequency `cf`, we find that the `12^(th)` observation is `2`.
Hence, the median of the data is `2`.
Mode :
the frequency of observation `2` is maximum (`10`)
`:. Z = 2`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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