1. Calculate Percentile deviation from the following grouped data
Solution:
Percentile deviation :
`x` | Frequency `f` | `cf` |
0 | 1 | 1 |
1 | 5 | 6 |
2 | 10 | 16 |
3 | 6 | 22 |
4 | 3 | 25 |
--- | --- | --- |
| n = 25 | -- |
Here, `n = 25`
`P_10 = ((10(n+1))/100)^(th)` value of the observation
`=((10*26)/100)^(th)` value of the observation
`=(2.6)^(th)` value of the observation
`=1`
`P_90 = ((90(n+1))/100)^(th)` value of the observation
`=((90*26)/100)^(th)` value of the observation
`=(23.4)^(th)` value of the observation
`=4`
Percentile deviation `=(P_90 - P_10)/2=(4-1)/2=3/2=1.5`
Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(4-1)/(4+1)=3/5=0.6`
2. Calculate Percentile deviation from the following grouped data
X | Frequency |
10 | 3 |
11 | 12 |
12 | 18 |
13 | 12 |
14 | 3 |
Solution:
Percentile deviation :
`x` | Frequency `f` | `cf` |
10 | 3 | 3 |
11 | 12 | 15 |
12 | 18 | 33 |
13 | 12 | 45 |
14 | 3 | 48 |
--- | --- | --- |
| n = 48 | -- |
Here, `n = 48`
`P_10 = ((10(n+1))/100)^(th)` value of the observation
`=((10*49)/100)^(th)` value of the observation
`=(4.9)^(th)` value of the observation
`=11`
`P_90 = ((90(n+1))/100)^(th)` value of the observation
`=((90*49)/100)^(th)` value of the observation
`=(44.1)^(th)` value of the observation
`=13`
Percentile deviation `=(P_90 - P_10)/2=(13-11)/2=2/2=1`
Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(13-11)/(13+11)=2/24=0.0833`
3. Calculate Percentile deviation from the following grouped data
Class | Frequency |
2 - 4 | 3 |
4 - 6 | 4 |
6 - 8 | 2 |
8 - 10 | 1 |
Solution:
Percentile deviation :
Class | Frequency `f` | `cf` |
2 - 4 | 3 | 3 |
4 - 6 | 4 | 7 |
6 - 8 | 2 | 9 |
8 - 10 | 1 | 10 |
--- | --- | --- |
| n = 10 | -- |
Here, `n = 10`
`P_10` class :
Class with `((10n)/100)^(th)` value of the observation in `cf` column
`=((10*10)/100)^(th)` value of the observation in `cf` column
`=(1)^(th)` value of the observation in `cf` column
and it lies in the class `2 - 4`.
`:. P_10` class : `2 - 4`
The lower boundary point of `2 - 4` is `2`.
`:. L = 2`
`P_10 = L + ((10 n)/100 - cf)/f * c`
`=2 + (1 - 0)/3 * 2`
`=2 + (1)/3 * 2`
`=2 + 0.6667`
`=2.6667`
`P_90` class :
Class with `((90n)/100)^(th)` value of the observation in `cf` column
`=((90*10)/100)^(th)` value of the observation in `cf` column
`=(9)^(th)` value of the observation in `cf` column
and it lies in the class `6 - 8`.
`:. P_90` class : `6 - 8`
The lower boundary point of `6 - 8` is `6`.
`:. L = 6`
`P_90 = L + ((90 n)/100 - cf)/f * c`
`=6 + (9 - 7)/2 * 2`
`=6 + (2)/2 * 2`
`=6 + 2`
`=8`
Percentile deviation `=(P_90 - P_10)/2=(8-2.6667)/2=5.3333/2=2.6666`
Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(8-2.6667)/(8+2.6667)=5.3333/10.6667=0.5`
4. Calculate Percentile deviation from the following grouped data
Class | Frequency |
0 - 2 | 5 |
2 - 4 | 16 |
4 - 6 | 13 |
6 - 8 | 7 |
8 - 10 | 5 |
10 - 12 | 4 |
Solution:
Percentile deviation :
Class | Frequency `f` | `cf` |
0 - 2 | 5 | 5 |
2 - 4 | 16 | 21 |
4 - 6 | 13 | 34 |
6 - 8 | 7 | 41 |
8 - 10 | 5 | 46 |
10 - 12 | 4 | 50 |
--- | --- | --- |
| n = 50 | -- |
Here, `n = 50`
`P_10` class :
Class with `((10n)/100)^(th)` value of the observation in `cf` column
`=((10*50)/100)^(th)` value of the observation in `cf` column
`=(5)^(th)` value of the observation in `cf` column
and it lies in the class `0 - 2`.
`:. P_10` class : `0 - 2`
The lower boundary point of `0 - 2` is `0`.
`:. L = 0`
`P_10 = L + ((10 n)/100 - cf)/f * c`
`=0 + (5 - 0)/5 * 2`
`=0 + (5)/5 * 2`
`=0 + 2`
`=2`
`P_90` class :
Class with `((90n)/100)^(th)` value of the observation in `cf` column
`=((90*50)/100)^(th)` value of the observation in `cf` column
`=(45)^(th)` value of the observation in `cf` column
and it lies in the class `8 - 10`.
`:. P_90` class : `8 - 10`
The lower boundary point of `8 - 10` is `8`.
`:. L = 8`
`P_90 = L + ((90 n)/100 - cf)/f * c`
`=8 + (45 - 41)/5 * 2`
`=8 + (4)/5 * 2`
`=8 + 1.6`
`=9.6`
Percentile deviation `=(P_90 - P_10)/2=(9.6-2)/2=7.6/2=3.8`
Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(9.6-2)/(9.6+2)=7.6/11.6=0.6552`
This material is intended as a summary. Use your textbook for detail explanation.
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