4. Calculate Percentile deviation, Coefficient of P.D., Interpercentile range from the following grouped data

Class	Frequency
10 - 20	15
20 - 30	25
30 - 40	20
40 - 50	12
50 - 60	8
60 - 70	5
70 - 80	3

Solution:
Percentile deviation :

Class	Frequency `f`	`cf`
10 - 20	15	15
20 - 30	25	40
30 - 40	20	60
40 - 50	12	72
50 - 60	8	80
60 - 70	5	85
70 - 80	3	88
---	---	---
	`n = 88`	--

Here, `n = 88`


`P_10` class :

Class with `((10n)/100)^(th)` value of the observation in `cf` column

`=((10*88)/100)^(th)` value of the observation in `cf` column

`=(8.8)^(th)` value of the observation in `cf` column

and it lies in the class `10 - 20`.

`:. P_10` class : `10 - 20`

The lower boundary point of `10-20` is `10`.

`:. L=10`

`P_10=L+((10 n)/100 - cf)/f * c`

`=10+(8.8-0)/15*10`

`=10+(8.8)/15*10`

`=10+5.8667`

`=15.8667`

---

`P_90` class :

Class with `((90n)/100)^(th)` value of the observation in `cf` column

`=((90*88)/100)^(th)` value of the observation in `cf` column

`=(79.2)^(th)` value of the observation in `cf` column

and it lies in the class `50 - 60`.

`:. P_90` class : `50 - 60`

The lower boundary point of `50-60` is `50`.

`:. L=50`

`P_90=L+((90 n)/100 - cf)/f * c`

`=50+(79.2-72)/8*10`

`=50+(7.2)/8*10`

`=50+9`

`=59`

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InterPercentile range `=P_90 - P_10=59-15.8667=43.1333`

Percentile deviation `=(P_90 - P_10)/2=(59-15.8667)/2=43.1333/2=21.5666` (Semi-InterPercentile range)

Coefficient of Percentile deviation `=(P_90 - P_10)/(P_90 + P_10)=(59-15.8667)/(59+15.8667)=43.1333/74.8667=0.5761`

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