\displaystyle \textbf{Question 1: }\text{Calculate the mean deviation about the median of the following}
\displaystyle \text{observations:}
\displaystyle \text{(i) }3011,2780,3020,2354,3541,4150,5000
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{Here, }d_i=x_i-M,\text{ where }M=\text{Median.}
\displaystyle \text{Arranging the data in ascending order: }2354,2780,3011,3020,3541,4150,5000
\displaystyle M=3020,\quad n=7
\displaystyle \begin{array}{|c|c|}\hline x_i&|x_i-3020|\\\hline3011&9\\2780&240\\3020&0\\2354&666\\3541&521\\4150&1130\\5000&1980\\\hline\text{Total}&4546\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{4546}{7}=649.42\text{ (approx.)}
\displaystyle \\

\displaystyle \text{(ii) }38,70,48,34,42,55,63,46,54,44
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{Here, }d_i=x_i-M,\text{ where }M=\text{Median.}
\displaystyle \text{Arranging the data in ascending order: }34,38,42,44,46,48,54,55,63,70
\displaystyle M=\frac{46+48}{2}=47,\quad n=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|x_i-47|\\\hline38&9\\70&23\\48&1\\34&13\\42&5\\55&8\\63&16\\46&1\\54&7\\44&3\\\hline\text{Total}&86\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{86}{10}=8.6
\displaystyle \\

\displaystyle \text{(iii) }34,66,30,38,44,50,40,60,42,51
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{Here, }d_i=x_i-M,\text{ where }M=\text{Median.}
\displaystyle \text{Arranging the data in ascending order: }30,34,38,40,42,44,50,51,60,66
\displaystyle M=\frac{42+44}{2}=43,\quad n=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|x_i-43|\\\hline34&9\\66&23\\30&13\\38&5\\44&1\\50&7\\40&3\\60&17\\42&1\\51&8\\\hline\text{Total}&87\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{87}{10}=8.7
\displaystyle \\

\displaystyle \text{(iv) }22,24,30,27,29,31,25,28,41,42
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{Here, }d_i=x_i-M,\text{ where }M=\text{Median.}
\displaystyle \text{Arranging the data in ascending order: }22,24,25,27,28,29,30,31,41,42
\displaystyle M=\frac{28+29}{2}=28.5,\quad n=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|x_i-28.5|\\\hline22&6.5\\24&4.5\\30&1.5\\27&1.5\\29&0.5\\31&2.5\\25&3.5\\28&0.5\\41&12.5\\42&13.5\\\hline\text{Total}&47\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{47}{10}=4.7
\displaystyle \\

\displaystyle \text{(v) }38,70,48,34,63,42,55,44,53,47
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{Here, }d_i=x_i-M,\text{ where }M=\text{Median.}
\displaystyle \text{Arranging the data in ascending order: }34,38,42,44,47,48,53,55,63,70
\displaystyle M=\frac{47+48}{2}=47.5,\qquad n=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-47.5|\\\hline38&9.5\\70&22.5\\48&0.5\\34&13.5\\63&15.5\\42&5.5\\55&7.5\\44&3.5\\53&5.5\\47&0.5\\\hline\text{Total}&84\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{84}{10}=8.4
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Calculate the mean deviation about the mean of the following}
\displaystyle \text{observations:}
\displaystyle \text{(i) }4,7,8,9,10,12,13,17
\displaystyle \text{(ii) }13,17,16,14,11,13,10,16,11,18,12,17
\displaystyle \text{(iii) }38,70,48,40,42,55,63,46,54,44
\displaystyle \text{(iv) }36,72,46,42,60,45,53,46,51,49
\displaystyle \text{(v) }57,64,43,67,49,59,44,47,61,59
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|,\text{ where }d_i=x_i-\overline{x}.
\displaystyle \text{(i) }4,7,8,9,10,12,13,17
\displaystyle \overline{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-10|\\\hline4&6\\7&3\\8&2\\9&1\\10&0\\12&2\\13&3\\17&7\\\hline\text{Total}&24\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{8}\sum_{i=1}^{8}|d_i|=\frac{24}{8}=3
\displaystyle \\

\displaystyle \text{(ii) }13,17,16,14,11,13,10,16,11,18,12,17
\displaystyle \overline{x}=\frac{13+17+16+14+11+13+10+16+11+18+12+17}{12}
\displaystyle =\frac{168}{12}=14
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-14|\\\hline13&1\\17&3\\16&2\\14&0\\11&3\\13&1\\10&4\\16&2\\11&3\\18&4\\12&2\\17&3\\\hline\text{Total}&28\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{12}\sum_{i=1}^{12}|d_i|=\frac{28}{12}=\frac{7}{3}
\displaystyle =2.33\text{ (approx.)}
\displaystyle \\

\displaystyle \text{(iii) }38,70,48,40,42,55,63,46,54,44
\displaystyle \overline{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}
\displaystyle =\frac{500}{10}=50
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-50|\\\hline38&12\\70&20\\48&2\\40&10\\42&8\\55&5\\63&13\\46&4\\54&4\\44&6\\\hline\text{Total}&84\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{84}{10}=8.4
\displaystyle \\

\displaystyle \text{(iv) }36,72,46,42,60,45,53,46,51,49
\displaystyle \overline{x}=\frac{36+72+46+42+60+45+53+46+51+49}{10}
\displaystyle =\frac{500}{10}=50
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-50|\\\hline36&14\\72&22\\46&4\\42&8\\60&10\\45&5\\53&3\\46&4\\51&1\\49&1\\\hline\text{Total}&72\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{72}{10}=7.2
\displaystyle \\

\displaystyle \text{(v) }57,64,43,67,49,59,44,47,61,59
\displaystyle \overline{x}=\frac{57+64+43+67+49+59+44+47+61+59}{10}
\displaystyle =\frac{550}{10}=55
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-55|\\\hline57&2\\64&9\\43&12\\67&12\\49&6\\59&4\\44&11\\47&8\\61&6\\59&4\\\hline\text{Total}&74\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{74}{10}=7.4
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate the mean deviation of the following income groups of five}
\displaystyle \text{and seven members from their medians:}
\displaystyle \begin{array}{|c|c|}\hline\text{I: Income in Rs.}&\text{II: Income in Rs.}\\\hline4000&3800\\4200&4000\\4400&4200\\4600&4400\\4800&4600\\-&4800\\-&5800\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{where }d_i=x_i-M\text{ and }M=\text{Median.}
\displaystyle \text{For income group I:}
\displaystyle \text{The observations in ascending order are }4000,4200,4400,4600,4800.
\displaystyle M=4400,\qquad n=5
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-4400|\\\hline4000&400\\4200&200\\4400&0\\4600&200\\4800&400\\\hline\text{Total}&1200\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{5}\sum_{i=1}^{5}|d_i|=\frac{1200}{5}=240
\displaystyle \text{For income group II:}
\displaystyle \text{The observations in ascending order are }3800,4000,4200,4400,4600,4800,5800.
\displaystyle M=4400,\qquad n=7
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-4400|\\\hline3800&600\\4000&400\\4200&200\\4400&0\\4600&200\\4800&400\\5800&1400\\\hline\text{Total}&3200\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{7}\sum_{i=1}^{7}|d_i|=\frac{3200}{7}
\displaystyle =457.14\text{ (approx.)}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The lengths, in cm, of }10\text{ rods in a shop are given below:}
\displaystyle 40.0,52.3,55.2,72.9,52.8,79.0,32.5,15.2,27.9,30.2
\displaystyle \text{(i) Find the mean deviation from the median.}
\displaystyle \text{(ii) Find the mean deviation from the mean.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Mean deviation about the median}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{where }d_i=x_i-M\text{ and }M=\text{Median.}
\displaystyle \text{Arranging the observations in ascending order:}
\displaystyle 15.2,27.9,30.2,32.5,40.0,52.3,52.8,55.2,72.9,79.0
\displaystyle M=\frac{40.0+52.3}{2}=46.15,\qquad n=10
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-46.15|\\\hline40.0&6.15\\52.3&6.15\\55.2&9.05\\72.9&26.75\\52.8&6.65\\79.0&32.85\\32.5&13.65\\15.2&30.95\\27.9&18.25\\30.2&15.95\\\hline\text{Total}&166.40\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{166.4}{10}=16.64\text{ cm}
\displaystyle \text{(ii) Mean deviation about the mean}
\displaystyle \overline{x}=\frac{40.0+52.3+55.2+72.9+52.8+79.0+32.5+15.2+27.9+30.2}{10}
\displaystyle =\frac{458}{10}=45.8
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-45.8|\\\hline40.0&5.8\\52.3&6.5\\55.2&9.4\\72.9&27.1\\52.8&7.0\\79.0&33.2\\32.5&13.3\\15.2&30.6\\27.9&17.9\\30.2&15.6\\\hline\text{Total}&166.4\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{1}{10}\sum_{i=1}^{10}|d_i|=\frac{166.4}{10}=16.64\text{ cm}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In Question 1 (iii), (iv) and (v), find the number of observations lying}
\displaystyle \text{between }\overline{X}-\mathrm{M.D.}\text{ and }\overline{X}+\mathrm{M.D.},\text{ where M.D. is the mean}
\displaystyle \text{deviation from the mean.}
\displaystyle \text{Answer:}
\displaystyle \text{Formula used: }\mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|d_i|
\displaystyle \text{where }d_i=x_i-\overline{x}\text{ and }\overline{x}=\text{Mean.}
\displaystyle \text{(i) For the observations }34,66,30,38,44,50,40,60,42,51
\displaystyle \overline{x}=\frac{34+66+30+38+44+50+40+60+42+51}{10}
\displaystyle =\frac{455}{10}=45.5
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-45.5|\\\hline34&11.5\\66&20.5\\30&15.5\\38&7.5\\44&1.5\\50&4.5\\40&5.5\\60&14.5\\42&3.5\\51&5.5\\\hline\text{Total}&90\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{90}{10}=9
\displaystyle \overline{x}-\mathrm{M.D.}=45.5-9=36.5
\displaystyle \overline{x}+\mathrm{M.D.}=45.5+9=54.5
\displaystyle \text{The observations lying between }36.5\text{ and }54.5\text{ are }38,40,42,44,50,51.
\displaystyle \therefore \text{The required number of observations is }6.
\displaystyle \\

\displaystyle \text{(ii) For the observations }22,24,30,27,29,31,25,28,41,42
\displaystyle \overline{x}=\frac{22+24+30+27+29+31+25+28+41+42}{10}
\displaystyle =\frac{299}{10}=29.9
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-29.9|\\\hline22&7.9\\24&5.9\\30&0.1\\27&2.9\\29&0.9\\31&1.1\\25&4.9\\28&1.9\\41&11.1\\42&12.1\\\hline\text{Total}&48.8\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{48.8}{10}=4.88
\displaystyle \overline{x}-\mathrm{M.D.}=29.9-4.88=25.02
\displaystyle \overline{x}+\mathrm{M.D.}=29.9+4.88=34.78
\displaystyle \text{The observations lying between }25.02\text{ and }34.78\text{ are }27,28,29,30,31.
\displaystyle \therefore \text{The required number of observations is }5.
\displaystyle \\

\displaystyle \text{(iii) For the observations }38,70,48,34,63,42,55,44,53,47
\displaystyle \overline{x}=\frac{38+70+48+34+63+42+55+44+53+47}{10}
\displaystyle =\frac{494}{10}=49.4
\displaystyle \begin{array}{|c|c|}\hline x_i&|d_i|=|x_i-49.4|\\\hline38&11.4\\70&20.6\\48&1.4\\34&15.4\\63&13.6\\42&7.4\\55&5.6\\44&5.4\\53&3.6\\47&2.4\\\hline\text{Total}&86.8\\\hline\end{array}
\displaystyle \mathrm{M.D.}=\frac{86.8}{10}=8.68
\displaystyle \overline{x}-\mathrm{M.D.}=49.4-8.68=40.72
\displaystyle \overline{x}+\mathrm{M.D.}=49.4+8.68=58.08
\displaystyle \text{The observations lying between }40.72\text{ and }58.08\text{ are }42,44,47,48,53,55.
\displaystyle \therefore \text{The required number of observations is }6.
\displaystyle \\

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