\displaystyle \textbf{Question 1: }\text{Calculate the mean deviation from the median of the following frequency}
\displaystyle \text{distribution:}
\displaystyle \begin{array}{|c|ccccccccc|}\hline\text{Height in inches}&58&59&60&61&62&63&64&65&66\\\hline\text{Number of students}&15&20&32&35&35&22&20&10&8\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{We first calculate the median of the given frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-61|&f_i|d_i|\\\hline58&15&15&3&45\\59&20&35&2&40\\60&32&67&1&32\\61&35&102&0&0\\62&35&137&1&35\\63&22&159&2&44\\64&20&179&3&60\\65&10&189&4&40\\66&8&197&5&40\\\hline&N=\sum f_i=197&&&\sum f_i|d_i|=336\\\hline\end{array}
\displaystyle N=197\quad\Rightarrow\quad\frac{N}{2}=\frac{197}{2}=98.5
\displaystyle \text{The cumulative frequency just greater than }98.5\text{ is }102.
\displaystyle \text{The corresponding value of }x_i\text{ is }61.
\displaystyle \therefore M=61
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|
\displaystyle =\frac{336}{197}=1.7055\ldots
\displaystyle \therefore \text{The mean deviation from the median is }1.71\text{ inches (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The number of telephone calls received at an exchange in }245
\displaystyle \text{successive one-minute intervals is shown in the following frequency distribution:}
\displaystyle \begin{array}{|c|cccccccc|}\hline\text{Number of calls}&0&1&2&3&4&5&6&7\\\hline\text{Frequency}&14&21&25&43&51&40&39&12\\\hline\end{array}
\displaystyle \text{Compute the mean deviation about the median.}
\displaystyle \text{Answer:}
\displaystyle \text{We first calculate the median of the given frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-4|&f_i|d_i|\\\hline0&14&14&4&56\\1&21&35&3&63\\2&25&60&2&50\\3&43&103&1&43\\4&51&154&0&0\\5&40&194&1&40\\6&39&233&2&78\\7&12&245&3&36\\\hline&N=\sum f_i=245&&&\sum f_i|d_i|=366\\\hline\end{array}
\displaystyle N=245\quad\Rightarrow\quad\frac{N}{2}=\frac{245}{2}=122.5
\displaystyle \text{The cumulative frequency just greater than }122.5\text{ is }154.
\displaystyle \text{The corresponding value of }x_i\text{ is }4.
\displaystyle \therefore M=4
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|
\displaystyle =\frac{366}{245}=1.493877\ldots
\displaystyle \therefore \text{The mean deviation about the median is }1.49\text{ calls (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate the mean deviation from the median of the following}
\displaystyle \text{frequency distribution:}
\displaystyle \begin{array}{|c|ccccccc|}\hline x_i&5&7&9&11&13&15&17\\\hline f_i&2&4&6&8&10&12&8\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{We first calculate the median of the given frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-13|&f_i|d_i|\\\hline5&2&2&8&16\\7&4&6&6&24\\9&6&12&4&24\\11&8&20&2&16\\13&10&30&0&0\\15&12&42&2&24\\17&8&50&4&32\\\hline&N=\sum f_i=50&&&\sum f_i|d_i|=136\\\hline\end{array}
\displaystyle N=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than or equal to }25\text{ is }30.
\displaystyle \text{The corresponding value of }x_i\text{ is }13.
\displaystyle \therefore M=13
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|
\displaystyle =\frac{136}{50}=2.72
\displaystyle \therefore \text{The mean deviation from the median is }2.72.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the mean deviation from the mean for the following data:}
\displaystyle \text{(i) }\begin{array}{|c|cccccc|}\hline x_i&5&7&9&10&12&15\\\hline f_i&8&6&2&2&2&6\\\hline\end{array}
\displaystyle \text{(ii) }\begin{array}{|c|ccccc|}\hline x_i&5&10&15&20&25\\\hline f_i&7&4&6&3&5\\\hline\end{array}
\displaystyle \text{(iii) }\begin{array}{|c|ccccc|}\hline x_i&10&30&50&70&90\\\hline f_i&4&24&28&16&8\\\hline\end{array}
\displaystyle \text{(iv) }\begin{array}{|c|ccccc|}\hline x_i&20&21&22&23&24\\\hline f_i&6&4&5&1&4\\\hline\end{array}
\displaystyle \text{(v) }\begin{array}{|c|cccccccc|}\hline x_i&1&3&5&7&9&11&13&15\\\hline f_i&3&3&4&14&7&4&3&4\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\overline{x}=\frac{\sum f_ix_i}{N},\qquad \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|x_i-\overline{x}|
\displaystyle \text{where }N=\sum f_i.
\displaystyle \text{(i) Calculation of mean deviation about the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&|x_i-\overline{x}|&f_i|x_i-9|\\\hline5&8&40&4&32\\7&6&42&2&12\\9&2&18&0&0\\10&2&20&1&2\\12&2&24&3&6\\15&6&90&6&36\\\hline&N=\sum f_i=26&\sum f_ix_i=234&&\sum f_i|x_i-9|=88\\\hline\end{array}
\displaystyle \overline{x}=\frac{\sum f_ix_i}{N}=\frac{234}{26}=9
\displaystyle \mathrm{M.D.}=\frac{1}{26}\sum_{i=1}^{n}f_i|x_i-9|=\frac{88}{26}
\displaystyle =3.38\text{ (approx.)}
\displaystyle \\

\displaystyle \text{(ii) Calculation of mean deviation about the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&|x_i-\overline{x}|&f_i|x_i-14|\\\hline5&7&35&9&63\\10&4&40&4&16\\15&6&90&1&6\\20&3&60&6&18\\25&5&125&11&55\\\hline&N=\sum f_i=25&\sum f_ix_i=350&&\sum f_i|x_i-14|=158\\\hline\end{array}
\displaystyle \overline{x}=\frac{\sum f_ix_i}{N}=\frac{350}{25}=14
\displaystyle \mathrm{M.D.}=\frac{1}{25}\sum_{i=1}^{n}f_i|x_i-14|=\frac{158}{25}=6.32
\displaystyle \\

\displaystyle \text{(iii) Calculation of mean deviation about the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&|x_i-\overline{x}|&f_i|x_i-50|\\\hline10&4&40&40&160\\30&24&720&20&480\\50&28&1400&0&0\\70&16&1120&20&320\\90&8&720&40&320\\\hline&N=\sum f_i=80&\sum f_ix_i=4000&&\sum f_i|x_i-50|=1280\\\hline\end{array}
\displaystyle \overline{x}=\frac{\sum f_ix_i}{N}=\frac{4000}{80}=50
\displaystyle \mathrm{M.D.}=\frac{1}{80}\sum_{i=1}^{n}f_i|x_i-50|=\frac{1280}{80}=16
\displaystyle \\

\displaystyle \text{(iv) Calculation of mean deviation about the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&|x_i-\overline{x}|&f_i|x_i-21.65|\\\hline20&6&120&1.65&9.90\\21&4&84&0.65&2.60\\22&5&110&0.35&1.75\\23&1&23&1.35&1.35\\24&4&96&2.35&9.40\\\hline&N=\sum f_i=20&\sum f_ix_i=433&&\sum f_i|x_i-21.65|=25\\\hline\end{array}
\displaystyle \overline{x}=\frac{\sum f_ix_i}{N}=\frac{433}{20}=21.65
\displaystyle \mathrm{M.D.}=\frac{1}{20}\sum_{i=1}^{n}f_i|x_i-21.65|=\frac{25}{20}=1.25
\displaystyle \\

\displaystyle \text{(v) Calculation of mean deviation about the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&|x_i-\overline{x}|&f_i|x_i-8|\\\hline1&3&3&7&21\\3&3&9&5&15\\5&4&20&3&12\\7&14&98&1&14\\9&7&63&1&7\\11&4&44&3&12\\13&3&39&5&15\\15&4&60&7&28\\\hline&N=\sum f_i=42&\sum f_ix_i=336&&\sum f_i|x_i-8|=124\\\hline\end{array}
\displaystyle \overline{x}=\frac{\sum f_ix_i}{N}=\frac{336}{42}=8
\displaystyle \mathrm{M.D.}=\frac{1}{42}\sum_{i=1}^{n}f_i|x_i-8|=\frac{124}{42}
\displaystyle =2.95\text{ (approx.)}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the mean deviation from the median for the following data:}
\displaystyle \text{(i) }\begin{array}{|c|ccccc|}\hline x_i&15&21&27&30&35\\\hline f_i&3&5&6&7&8\\\hline\end{array}
\displaystyle \text{(ii) }\begin{array}{|c|ccccccc|}\hline x_i&74&89&42&54&91&94&35\\\hline f_i&20&12&2&4&5&3&4\\\hline\end{array}
\displaystyle \text{(iii) }\begin{array}{|c|ccccc|}\hline x_i&10&11&12&14&15\\\hline f_i&2&3&8&3&4\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) We first calculate the median of the given frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-30|&f_i|d_i|\\\hline15&3&3&15&45\\21&5&8&9&45\\27&6&14&3&18\\30&7&21&0&0\\35&8&29&5&40\\\hline&N=\sum f_i=29&&&\sum f_i|d_i|=148\\\hline\end{array}
\displaystyle N=29\quad\Rightarrow\quad\frac{N}{2}=\frac{29}{2}=14.5
\displaystyle \text{The cumulative frequency just greater than }14.5\text{ is }21.
\displaystyle \text{The corresponding value of }x_i\text{ is }30.
\displaystyle \therefore M=30
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|=\frac{148}{29}
\displaystyle =5.1034\ldots=5.10\text{ (approx.)}
\displaystyle \\

\displaystyle \text{(ii) We first arrange the values of }x_i\text{ in ascending order and calculate the median.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-74|&f_i|d_i|\\\hline35&4&4&39&156\\42&2&6&32&64\\54&4&10&20&80\\74&20&30&0&0\\89&12&42&15&180\\91&5&47&17&85\\94&3&50&20&60\\\hline&N=\sum f_i=50&&&\sum f_i|d_i|=625\\\hline\end{array}
\displaystyle N=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than or equal to }25\text{ is }30.
\displaystyle \text{The corresponding value of }x_i\text{ is }74.
\displaystyle \therefore M=74
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|=\frac{625}{50}=12.5
\displaystyle \\

\displaystyle \text{(iii) We first calculate the median of the given frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-12|&f_i|d_i|\\\hline10&2&2&2&4\\11&3&5&1&3\\12&8&13&0&0\\14&3&16&2&6\\15&4&20&3&12\\\hline&N=\sum f_i=20&&&\sum f_i|d_i|=25\\\hline\end{array}
\displaystyle N=20\quad\Rightarrow\quad\frac{N}{2}=\frac{20}{2}=10
\displaystyle \text{The cumulative frequency just greater than or equal to }10\text{ is }13.
\displaystyle \text{The corresponding value of }x_i\text{ is }12.
\displaystyle \therefore M=12
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{n}f_i|d_i|=\frac{25}{20}=1.25
\displaystyle \\

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