Question
Find the quartile deviation of advertisement expenditure using following frequency distribution of advertising expenditure of $50$ companies :
Advertisement cost (thousand $Rs$. ) $0 – 5$ $5 – 15$ $15 – 30$ $30 – 40$ $40 – 60$ $60 -100$ Total
No. of companies $3$ $8$ $15$ $10$ $8$ $6$ $50$

Answer

Expenditure of advertisement (thousand $Rs.$ ) No. of companies $f$ Cumulative frequency $cf$
$0 – 5$ $3$ $3$
$5 – 15$ $8$ $11$
$15 – 30$ $15$ $26$
$30 – 40$ $10$ $36$
$40 – 60$ $8$ $44$
$60 – 100$ $6$ $50$
Total $n = 50$  
First quartile :
$Q _1$ class $=$ class that includes $\left(\frac{n}{4}\right)$ th observation
$=$ class that includes $\left(\frac{50}{4}\right)$
$=12.5$ th observation
Referring to column cf, $Q_1$ class $=15-30$
Now, $Q_1=L+\frac{\left(\frac{n}{4}\right)-c t}{f} X c$
Putting $L=15, \frac{n}{4}=12.5$, $cf$ $=11, f=15$ and $c=15$ in the formula, $Q _1=15+\frac{12.5-11}{15} \times 15$
$=15+1.5=$ $Rs 16.5$ thousand
Third quartile :
$Q _3$ Class $=$ Class that includes $3\left(\frac{n}{4}\right)$ th observation
$=$ Class that includes $3(12.5)$
$=37.5$ th observation
Referring to column cf, $Q_3$ class $=40-60$
Now, $Q _3=L+\frac{3\left(\frac{n}{4}\right)- cf }{f} \times c$
Putting, $L=40,3\left(\frac{n}{4}\right)=37.5, c f=36, f=8$ and $c=20$ in the formula, $Q _3=40+\frac{37.5-36}{8} \times 20=40+\frac{1.5 \times 20}{8}=40+\frac{30}{8}=40+3.75=$ $Rs. 43.75$ thousand
Quartile deviation :
$Q _{ d }=\frac{Q 3-Q 1}{2}=\frac{43.75-16.5}{2}=\frac{27.5}{2}=13.625 \approx 13.63 \text { thousand }$

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