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Question

# Calculate Quartile Deviation and its Coefficient from the following data: Size 5−10 10−15 15−20 20−25 25−30 30−35 35−40 40−45 45−50 Frequency 6 10 18 30 15 12 10 6 4

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Solution

## Size Frequency Cumulative Frequency 5−10 6 6 10−15 10 16 15−20 18 34 20−25 30 64 25−30 15 79 30−35 12 91 35−40 10 101 40−45 6 107 45−50 4 111 Σf=N = 111 Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item = Size of $\left(\frac{111}{4}\right)\mathrm{th}$ item = Size of 27.75th item 27.75 item lies in group 15−20 and falls within 34th c.f. of the series ${Q}_{1}={l}_{1}+\frac{\frac{N}{4}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}\mathrm{Here,}\phantom{\rule{0ex}{0ex}}l₁=Lower\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{class}interval\phantom{\rule{0ex}{0ex}}N= Sum\mathrm{total}\mathrm{of}\mathrm{the}frequencies\phantom{\rule{0ex}{0ex}}c.f.=Cumulative\mathrm{frequency}ofthe\mathrm{class}\mathrm{preceding}\mathrm{the}\mathrm{first}\mathrm{quartile}class\phantom{\rule{0ex}{0ex}}f= Frequency\mathrm{of}\mathrm{the}\mathrm{quartile}class\phantom{\rule{0ex}{0ex}}i= Classinterval\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}{Q}_{1}=15+\frac{27.75-16}{18}×5\phantom{\rule{0ex}{0ex}}or,{Q}_{1}=15+\frac{11.75×5}{18}\phantom{\rule{0ex}{0ex}}or,{Q}_{1}=15+\frac{58.75}{18}\phantom{\rule{0ex}{0ex}}or,{Q}_{1}=15+3.26\phantom{\rule{0ex}{0ex}}\mathbf{⇒}{\mathbit{Q}}_{\mathbf{1}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{26}$ Like wise, Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item = Size of 3$\left(\frac{111}{4}\right)\mathrm{th}\mathrm{item}$ = Size of 83.25th item 83.25th item lies in the group 30-35 within the 33rd c.f. of the series. $\mathrm{Thus},\phantom{\rule{0ex}{0ex}}{Q}_{3}={l}_{1}+\frac{\frac{3N}{4}-c.f}{f}×i\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=30+\frac{83.25-79}{12}×5\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=30+\frac{4.25×5}{12}\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=30+\frac{21.25}{12}\phantom{\rule{0ex}{0ex}}or,{Q}_{3}=30+1.77\phantom{\rule{0ex}{0ex}}⇒{Q}_{3}=31.77$ $\mathbit{Q}\mathbit{u}\mathbit{a}\mathbit{r}\mathbit{t}\mathbit{i}\mathbit{l}\mathbit{e}\mathbf{}\mathbit{D}\mathbit{e}\mathbit{v}\mathbit{i}\mathbit{a}\mathbit{t}\mathbit{i}\mathbit{o}\mathbit{n}=\frac{{Q}_{3}-{Q}_{1}}{2}=\frac{31.77-18.26}{2}=6.75$ $\mathbit{C}\mathbit{o}\mathbit{e}\mathbit{f}\mathbit{f}\mathbit{i}\mathbit{c}\mathbit{i}\mathbit{e}\mathbit{n}\mathbit{t}\mathbf{}\mathbit{o}\mathbit{f}\mathbf{}\mathbit{Q}\mathbf{.}\mathbit{D}\mathbf{.}=\frac{{Q}_{3}-{Q}_{1}}{{Q}_{3}+{Q}_{1}}=\frac{31.77-18.26}{31.77+18.26}=\frac{13.51}{50.03}=0.27$

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