Question

Calculate the quartile deviation for the following:

$C.I.$	$0-9$	$10-19$	$20-29$	$30-39$	$40-49$	$50-59$	$60-69$
$f$	$8$	$16$	$17$	$34$	$10$	$5$	$10$

Answer 1

We observe that the class intervals are in inclusive form. We have to convert them into exclusive form by adding the correction factor. You have learnt to add the correction factor in $8^{th}$ standard.

Class interval	$f$	$f_c$
$-0.5-9.5$	$8$	$8$
$9.5-19.5$	$16$	$24$
$19.5-29.5$	$17$	$41$	$Q_{1}Class$
$29.5-39.5$	$34$	$75$	$Q_{3}Class$
$39.5-49.5$	$10$	$85$
$49.5-59.5$	$5$	$90$
$59.5-69.5$	$10$	$100$

To find $Q_1$:
Here $N=100$. Hence $N/4=25$ and $25$ in $f_c$ column corresponds to $CI19.5-29.65$.
Hence
$LRL=19.5,f_c=24,f_m=17$ and $i=10$.
Thus we have
$Q_1=LRL+\left(\frac{\frac{N}{4}-f_c}{f_m}\right)\times i=19.5+\left(\frac{25-24}{17}\right)\times 10$
$=19.5+(0.058\times 10)=19.5+0.58=20.08$.
To find $Q_3$:
We have $3N/4=(3\times 100)/4=75$ and $75$ corresponds to $CI29.5-39.5$ in $f_c$ column.
Thus
$LRL=29.5,f_c=41,f_m=34$ and $i=10$.
Therefore
$Q_3=LRL+\left(\frac{\frac{3N}{4}-f_c}{f_m}\right)\times i=29.5+\left(\frac{75-41}{34}\right)\times 10$
$=29.5+(1\times 10)=29.5+10=39.5$.
We now compute quartile deviation:
quartile deviation $=\frac{Q_3-Q_1}{2}=\frac{39.5-20.08}{2}=\frac{19.42}{2}=9.71$.