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Measures of dispersion — Statistics STD 11 Commerce — Question

Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsMeasures of dispersion5 Marks
Question
Find range, coefficient of range, quartile deviation, coefficient of quartile deviation, mean deviation and coefficient of mean deviation from the following data of number of emergency visits of $80$ doctors to their patients in a town :
No. of visits $3$ $5$ $8$ $12$ $17$ $20$ $24$ $30$ $35$
No. of doctors $1$ $3$ $7$ $15$ $20$ $13$ $10$ $7$ $4$
✓

Answer

No. of visits $x$ No. of doctors $f$ $f.x$ $|x-\overline{ x }|\bar{x}=18.05$ $f|x-\bar{x}|$ Cumulative frequency $cf$
$3$ $1$ $3$ $15.05$ $15.05$ $1$
$5$ $3$ $15$ $13.05$ $39.15$ $4$
$8$ $7$ $56$ $10.05$ $70.35$ $11$
$12$ $15$ $180$ $6.05$ $90.75$ $26$
$17$ $20$ $340$ $1.05$ $21.00$ $46$
$20$ $13$ $260$ $1.95$ $25.35$ $59$
$24$ $10$ $240$ $5.95$ $59.50$ $69$
$30$ $7$ $210$ $11.95$ $83.65$ $76$
$35$ $4$ $140$ $16.95$ $67.80$ $80$
Total $n = 80$ $\sum fx =1444$ - $\sum f | x -\overline{ x }|= 472.60$ -
Range :
$X_H=35, x_L=3$
Here, $R=X_H-x_L=35-3=32$ Visits
First quartile :
$Q _1=$ Value of $\left(\frac{ n +1}{4}\right)$ th observation
$=$ value of $\left(\frac{80+1}{4}\right)$ th observation
$=$ value of $20.25^{\text {th }}$ observation
Referring to column cf, $Q_1=12$
Quartile deviation :
$Q _{ d }=\frac{ Q ^3- Q 1}{2}=\frac{24-12}{2}=\frac{12}{2}=6$ visits
Mean :
$\overline{ x }=\frac{\sum fx }{ n }=\frac{1444}{80}=18.05 \text { visits }$
Mean deviation :
$MD =\frac{\sum f | x -\bar{x}|}{ n }=\frac{472.60}{80}=5.91 \text { visits }$
Coefficient of range :
Coefficient of range $=\frac{= xH - xL }{= xH + xL }=\frac{35-3}{35+3}=\frac{32}{38}=0.84$
Third quartile :
$Q _3=$ Value of $3\left(\frac{ n +1}{4}\right)$ th observation
$=$ Value of $3(20.25)$ th observation
$=$ Value of $60.75$ th observation
Referring to column cf, $Q_3=24$
Coefficient of Quartile deviation :
Coefficient of $Q _{ d }=\frac{ Q 3- Q 1}{ Q 3+ Q 1}=\frac{24-12}{24+12}=\frac{12}{36}=0.33$
Coefficient of mean deviation :
Coefficient of $MD =\frac{ MD }{ x }=\frac{5.91}{18.05}=0.33$

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