Question

Calculate coefficient of quartile deviation from the following data:

X (Less than)	200	300	400	500	600
Frequency	8	20	40	46	50

Answer 1

Converting less than cumulative frequency into normal frequency distribution:

X	c.f.	Frequency
100−200	8	8
200−300	20	20−8=12
300−400	40	40−20=20
400−500	46	46−40=6
500−600	50	50−46=4
		N=50

Q₁ = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 12.5^th item

12.5^th item lies in group 200−300 and falls within 20^th c.f. of the series.

$Q_1=l_1+\frac{{\displaystyle \frac{N}{4}}-c.f.}{f}\times i$

Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval

$$\begin{gathered} Thus, \\ {Q}_1=200+\frac{12.5-8}{12}\times 100 \\ or,Q_1=200+\frac{450}{12} \\ or,Q_1=200+37.5 \\ \Rightarrow Q_1=237.5 \end{gathered}$$

Likewise,
Q₃ = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of $3\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 37.5th item

37.5^th item lies in group 300−400 and fall within the 40th c.f. of the series

$$\begin{gathered} Q_3=l_1+\frac{{\displaystyle \frac{3N}{4}}-c.f.}{f}\times i \\ or,Q_3=300+\frac{37.5-20}{20}\times 100 \\ or,Q_3=300+\frac{1750}{20} \\ \Rightarrow Q_3=300+87.5=387.5 \end{gathered}$$

$\mathit{C}\mathit{o}\mathit{e}\mathit{f}\mathit{f}\mathit{i}\mathit{c}\mathit{i}\mathit{e}\mathit{n}\mathit{t}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{Q}\mathbf{.}\mathit{D}\mathbf{.}=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{387.5-237.5}{387.5+237.5}=\frac{150}{625}=0.24$