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Question

# Find out quartile deviation and coefficient of quartile deviation from the following data: Class Interval 0−10 10−20 20−30 30−40 40−50 50−60 Frequency 4 8 5 4 9 10

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Solution

## Class interval Frequency (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 4 8 5 4 9 10 4 4 + 8 = 12 12 + 5 = 17 17 + 4 = 21 21 + 9 = 30 30 + 10 = 40 $\Sigma f$=N = 40 Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item = Size of $\left(\frac{40}{4}\right)\mathrm{th}$ item = Size of 10th item 10th item lies in group 10 − 20 and fall within 12th c.f. of the series. ${\mathrm{Q}}_{1}={l}_{1}+\frac{\frac{\mathrm{N}}{4}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}$ 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 $Thus,\phantom{\rule{0ex}{0ex}}{Q}_{1}=10+\frac{10-4}{8}×10\phantom{\rule{0ex}{0ex}}or,{Q}_{1}=10+\frac{60}{8}\phantom{\rule{0ex}{0ex}}or,{Q}_{1}=10+7.5\phantom{\rule{0ex}{0ex}}\mathbf{⇒}\mathbf{}{\mathbit{Q}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{5}$ Likewise, Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item = Size of $3\left(\frac{40}{4}\right)\mathrm{th}$ item = Size of 30th item 30th item lies in group 40-50 and fall within the 30th c.f. of the series $\mathrm{Thus},\phantom{\rule{0ex}{0ex}}{Q}_{3}={l}_{1}+\frac{3\left(\frac{N}{4}\right)-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=40+\frac{30-21}{9}×10\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=40+\frac{9}{9}×10\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=40+10\phantom{\rule{0ex}{0ex}}\mathbf{⇒}\mathbf{}\mathbf{}{\mathbit{Q}}_{\mathbf{3}}\mathbf{=}\mathbf{50}\phantom{\rule{0ex}{0ex}}\mathrm{Quartile}\mathrm{deviation}\left(QD\right)=\frac{{\mathit{Q}}_{\mathit{3}}\mathit{-}{\mathit{Q}}_{\mathit{1}}}{\mathit{2}}=\frac{50-17.5}{2}=\frac{32.5}{2}=16.25\phantom{\rule{0ex}{0ex}}\mathrm{Coefficient}\mathrm{of}\mathrm{Q}.\mathrm{D}=\frac{{\mathit{Q}}_{\mathit{3}}\mathit{-}{\mathit{Q}}_{\mathit{1}}}{{\mathit{Q}}_{\mathit{3}}\mathit{+}{\mathit{Q}}_{\mathit{1}}}=\frac{50-17.5}{50+17.5}=\frac{32.5}{67.5}=0.48$

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