Question

Using raw data given below derive interquartile range, semiquartile range, and coefficient of quartile deviation. Choose the correct choice/s given below. Data given is $X =20,12,18,28,32,10$

• A. Interquartile range $=15.25$
• B. Coefficient of quartile deviation $=0.42$
• C. Coefficient of quartile deviation $=0.39$
• D. Semiquartile range $=7.01$

Answer 1

Answer: (A), (C)

Arrange the data in ascending order $X=10,12,18,20,25,32$
Number of items $=6$
$Q _1=(\frac{n+1}{4})^{\frac{6}{n}}$ item $=(\frac{6+1}{4})^{\text {th }}$ item
$Q _1=(1.75)^{\text {th }}$ item
So, $Q _1= I ^{ st }$ item $+0.75(2^{\text {nd }}.$ item $-[^{\text {st }}.$ item $)$
$=10+0.75(12-10)$
$=10+0.75(2)$
$=10+1.50=11.50$
$Q_3=3(\frac{N+1}{4})^{\text {th }}$ item $=3(\frac{6+1}{4})^{\text {th }}$ item
$Q _3=(5.25)^{\text {th }}$ item
$Q _3=^{\text {th }}$ item $+0.25(6^{\text {th }}.$ item $-5^{\text {th }}$ item $)$
$=25+0.25(32-25)$
$Q_3=25+0.25(7)=26.75$
Inter-quartile range $= Q _3- Q _1$
$=26.75-11.50=15.25$
Semi-quartile range
$=\frac{Q_3-Q_1}{2}=\frac{15.25}{2}=7.625$
Coefficient of quartile deviation
$=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{26.75-11.50}{26.75+11.50} $
$=\frac{15.25}{38.25}=0.39$