4. Calculate Quartile deviation, Coefficient of Q.D., Interquartile 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:
Quartile 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`


`Q_1` class :

Class with `(n/4)^(th)` value of the observation in `cf` column

`=(88/4)^(th)` value of the observation in `cf` column

`=(22)^(th)` value of the observation in `cf` column

and it lies in the class `20 - 30`.

`:. Q_1` class : `20 - 30`

The lower boundary point of `20-30` is `20`.

`:. L=20`

`Q_1=L+(( n)/4 - cf)/f * c`

`=20+(22-15)/25*10`

`=20+(7)/25*10`

`=20+2.8`

`=22.8`

---

`Q_3` class :

Class with `((3n)/4)^(th)` value of the observation in `cf` column

`=((3*88)/4)^(th)` value of the observation in `cf` column

`=(66)^(th)` value of the observation in `cf` column

and it lies in the class `40 - 50`.

`:. Q_3` class : `40 - 50`

The lower boundary point of `40-50` is `40`.

`:. L=40`

`Q_3=L+((3 n)/4 - cf)/f * c`

`=40+(66-60)/12*10`

`=40+(6)/12*10`

`=40+5`

`=45`

---

InterQuartile range `=Q_3 - Q_1=45-22.8=22.2`

Quartile deviation `=(Q_3 - Q_1)/2=(45-22.8)/2=22.2/2=11.1` (Semi-InterQuartile range)

Coefficient of Quartile deviation `=(Q_3 - Q_1)/(Q_3 + Q_1)=(45-22.8)/(45+22.8)=22.2/67.8=0.3274`

---