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Question

# Find the mean deviation from the mean and from median of the following distribution: Marks 0-10 10-20 20-30 30-40 40-50 No. of students 5 8 15 16 6

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Solution

## Computation of mean distribution from the median: Marks Number of Students ${f}_{i}$ Cumulative Frequency Midpoints ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-28\right|$ ${f}_{\mathrm{i}}\left|{d}_{\mathit{i}}\right|$ ${f}_{i}{x}_{i}$ $\left|{x}_{i}-27\right|$ ${f}_{\mathrm{i}}\left|{x}_{\mathit{i}}-27\right|$ 0−10 5 5 5 23 115 25 22 110 10−20 8 13 15 13 104 120 12 96 20−30 15 28 25 3 45 375 2 30 30−40 16 44 35 7 112 560 8 128 40−50 6 50 45 17 102 270 18 108 $N=50$ $\sum _{i=1}^{5}{f}_{i}\left|{d}_{i}\right|=478$ 1350 $\sum _{i=1}^{5}{f}_{i}\left|{x}_{i}-27\right|=472\phantom{\rule{0ex}{0ex}}$ $N=50,\frac{N}{2}=25$ The cumulative frequency just greater than $\frac{N}{2}=25$ is 28 and the corresponding class is 20−30. Thus, the median class is 20−30. Using formula: $\therefore {}{l}{=}{20}{,}{}{F}{=}{13}{,}{}{f}{=}{15}{,}{}{h}{=}{10}{}{}\phantom{\rule{0ex}{0ex}}{}{\mathrm{Median}}{}{}{=}{l}{+}\frac{\frac{N}{2}-F}{f}{}{×}{}{h}{}{}{}{}{}{}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{Substituting}}{}{\mathrm{the}}{}{\mathrm{values}}{:}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{Median}}{}{=}{20}{+}\frac{25-13}{15}{×}{10}{}{}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{20}{}{+}{8}{}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{28}$ ${}{\mathrm{Mean}}{}{\mathrm{distribution}}{}{\mathrm{from}}{}{\mathrm{the}}{}{\mathrm{median}}{}{=}\frac{\sum _{i=1}^{5}{f}_{i}\left|{d}_{i}\right|}{N}{}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{}\frac{478}{50}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{9}{.}{56}\phantom{\rule{0ex}{0ex}}$ $\phantom{\rule{0ex}{0ex}}{\mathrm{Mean}}{}{\left(}\overline{)X}{\right)}{=}\frac{\sum _{i=1}^{5}{f}_{i}{x}_{i}}{N}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}\frac{1350}{50}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{27}\phantom{\rule{0ex}{0ex}}{\mathrm{Mean}}{}{\mathrm{deviation}}{}{\mathrm{from}}{}{\mathrm{the}}{}{\mathrm{mean}}{}{}{=}\frac{1}{50}{×}\sum _{\mathrm{i}=1}^{5}{{f}}_{i}\left|{x}_{\mathrm{i}}-27\right|\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}\frac{472}{50}\phantom{\rule{0ex}{0ex}}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{=}{9}{.}{44}$ Mean deviation from the median and the mean are 9.56 and 9.44, respectively.

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