3. Calculate Percentile deviation, Coefficient of P.D., Interpercentile range 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`

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`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`

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InterPercentile range `=P_90 - P_10=8-2.6667=5.3333`

Percentile deviation `=(P_90 - P_10)/2=(8-2.6667)/2=5.3333/2=2.6666` (Semi-InterPercentile range)

Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(8-2.6667)/(8+2.6667)=5.3333/10.6667=0.5`

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