\displaystyle \textbf{Question 1: }\text{Compute the mean deviation from the median of the following distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&0-10&10-20&20-30&30-40&40-50\\\hline\text{Frequency}&5&10&20&5&10\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Calculation of mean deviation from the median:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-25|&f_i|d_i|\\\hline0-10&5&5&5&20&100\\10-20&15&10&15&10&100\\20-30&25&20&35&0&0\\30-40&35&5&40&10&50\\40-50&45&10&50&20&200\\\hline&&N=\sum f_i=50&&&\sum_{i=1}^{5}f_i|d_i|=450\\\hline\end{array}
\displaystyle N=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than }25\text{ is }35.
\displaystyle \text{Therefore, the median class is }20-30.
\displaystyle \therefore l=20,\qquad f=20,\qquad F=15,\qquad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =20+\frac{25-15}{20}\times10
\displaystyle =20+5=25
\displaystyle \therefore M=25
\displaystyle \mathrm{M.D.}=\frac{1}{N}\sum_{i=1}^{5}f_i|d_i|
\displaystyle =\frac{450}{50}=9
\displaystyle \therefore \text{The mean deviation from the median is }9.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the mean deviation from the mean from the following data:}
\displaystyle \text{(i) } \\ \begin{array}{|c|cccccccc|}\hline\text{Classes}&0-100&100-200&200-300&300-400&400-500&500-600&600-700&700-800\\\hline\text{Frequency}&4&8&9&10&7&5&4&3\\\hline\end{array}
\displaystyle \text{(ii) } \\ \begin{array}{|c|cccccc|}\hline\text{Classes}&95-105&105-115&115-125&125-135&135-145&145-155\\\hline\text{Frequency}&9&13&16&26&30&12\\\hline\end{array}
\displaystyle \text{(iii) } \\ \begin{array}{|c|cccccc|}\hline\text{Classes}&0-10&10-20&20-30&30-40&40-50&50-60\\\hline\text{Frequency}&6&8&14&16&4&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Calculation of mean deviation from the mean:}
\displaystyle \text{Let the assumed mean be }a=450\text{ and the common factor be }h=100.
\displaystyle d_i=\frac{x_i-a}{h}=\frac{x_i-450}{100}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline\text{Class}&f_i&\text{Mid-values} \  x_i&d_i=\frac{x_i-450}{100}&f_id_i&|x_i-\overline X|=|x_i-358|&f_i|x_i-\overline X|\\\hline0-100&4&50&-4&-16&308&1232\\100-200&8&150&-3&-24&208&1664\\200-300&9&250&-2&-18&108&972\\300-400&10&350&-1&-10&8&80\\400-500&7&450&0&0&92&644\\500-600&5&550&1&5&192&960\\600-700&4&650&2&8&292&1168\\700-800&3&750&3&9&392&1176\\\hline&N=\sum f_i=50&&&\sum f_id_i=-46&&\sum f_i|x_i-\overline X|=7896\\\hline\end{array}
\displaystyle \text{Clearly, }a=450,\ h=100,\ N=50\text{ and }\sum f_id_i=-46
\displaystyle \therefore \overline X=a+h\left(\frac{1}{N}\sum f_id_i\right)
\displaystyle =450+100\left(\frac{-46}{50}\right)=358
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=7896\text{ and }N=\sum f_i=50
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|=\frac{7896}{50}=157.92
\displaystyle \\

\displaystyle \text{(ii) Calculation of mean deviation from the mean:}
\displaystyle \text{Let the assumed mean be }a=130\text{ and the common factor be }h=10.
\displaystyle d_i=\frac{x_i-a}{h}=\frac{x_i-130}{10}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline\text{Class}&f_i&\text{Mid-values} \  x_i&d_i=\frac{x_i-130}{10}&f_id_i&|x_i-\overline X|=|x_i-128.58|&f_i|x_i-\overline X|\\\hline95-105&9&100&-3&-27&28.58&257.22\\105-115&13&110&-2&-26&18.58&241.54\\115-125&16&120&-1&-16&8.58&137.28\\125-135&26&130&0&0&1.42&36.92\\135-145&30&140&1&30&11.42&342.60\\145-155&12&150&2&24&21.42&257.04\\\hline&N=\sum f_i=106&&&\sum f_id_i=-15&&\sum f_i|x_i-\overline X|=1272.60\\\hline\end{array}
\displaystyle \text{Clearly, }a=130,\ h=10,\ N=106\text{ and }\sum f_id_i=-15
\displaystyle \therefore \overline X=a+h\left(\frac{1}{N}\sum f_id_i\right)
\displaystyle =130+10\left(\frac{-15}{106}\right)=128.58\text{ (approx.)}
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=1272.60\text{ and }N=\sum f_i=106
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|
\displaystyle =\frac{1272.60}{106}=12.005\text{ (approx.)}
\displaystyle \\

\displaystyle \text{(iii) Calculation of mean deviation from the mean:}
\displaystyle \text{Let the assumed mean be }a=25\text{ and the common factor be }h=10.
\displaystyle d_i=\frac{x_i-a}{h}=\frac{x_i-25}{10}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline\text{Class}&f_i&\text{Mid-values} \  x_i&d_i=\frac{x_i-25}{10}&f_id_i&|x_i-\overline X|=|x_i-27|&f_i|x_i-\overline X|\\\hline0-10&6&5&-2&-12&22&132\\10-20&8&15&-1&-8&12&96\\20-30&14&25&0&0&2&28\\30-40&16&35&1&16&8&128\\40-50&4&45&2&8&18&72\\50-60&2&55&3&6&28&56\\\hline&N=\sum f_i=50&&&\sum f_id_i=10&&\sum f_i|x_i-\overline X|=512\\\hline\end{array}
\displaystyle \text{Clearly, }a=25,\ h=10,\ N=50\text{ and }\sum f_id_i=10
\displaystyle \therefore \overline X=a+h\left(\frac{1}{N}\sum f_id_i\right)
\displaystyle =25+10\left(\frac{10}{50}\right)=27
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=512\text{ and }N=\sum f_i=50
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|=\frac{512}{50}=10.24
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Compute the mean deviation from the mean of the following distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline\text{Class}&10-20&20-30&30-40&40-50&50-60&60-70&70-80&80-90\\\hline\text{Number of students}&8&10&15&25&20&18&9&5\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Calculation of mean deviation from the mean:}
\displaystyle \text{Let the assumed mean be }a=45\text{ and the common factor be }h=10.
\displaystyle d_i=\frac{x_i-a}{h}=\frac{x_i-45}{10}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline\text{Class}&f_i&x_i&d_i=\frac{x_i-45}{10}&f_id_i&|x_i-\overline X|=|x_i-49|&f_i|x_i-\overline X|\\\hline10-20&8&15&-3&-24&34&272\\20-30&10&25&-2&-20&24&240\\30-40&15&35&-1&-15&14&210\\40-50&25&45&0&0&4&100\\50-60&20&55&1&20&6&120\\60-70&18&65&2&36&16&288\\70-80&9&75&3&27&26&234\\80-90&5&85&4&20&36&180\\\hline&N=\sum f_i=110&&&\sum f_id_i=44&&\sum f_i|x_i-\overline X|=1644\\\hline\end{array}
\displaystyle \text{Clearly, }N=110,\ a=45,\ h=10\text{ and }\sum f_id_i=44
\displaystyle \therefore \overline X=a+h\left(\frac{1}{N}\sum f_id_i\right)
\displaystyle =45+10\left(\frac{44}{110}\right)=45+4=49
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=1644\text{ and }N=\sum f_i=110
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|=\frac{1644}{110}
\displaystyle =14.9454\ldots=14.95\text{ (approx.)}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The age distribution of the life insurance policy holders is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline\text{Age (on nearest birthday)}&17-19.5&20-25.5&26-35.5&36-40.5&41-50.5&51-55.5&56-60.5&61-70.5\\\hline\text{No. of persons}&5&16&12&26&14&12&6&5\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{To convert the classes into continuous classes, subtract }0.25\text{ from each lower limit}
\displaystyle \text{and add }0.25\text{ to each upper limit.}
\displaystyle \text{Calculation of mean deviation from the median:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-38.63|&f_i|d_i|\\\hline16.75-19.75&18.25&5&5&20.38&101.90\\19.75-25.75&22.75&16&21&15.88&254.08\\25.75-35.75&30.75&12&33&7.88&94.56\\35.75-40.75&38.25&26&59&0.38&9.88\\40.75-50.75&45.75&14&73&7.12&99.68\\50.75-55.75&53.25&12&85&14.62&175.44\\55.75-60.75&58.25&6&91&19.62&117.72\\60.75-70.75&65.75&5&96&27.12&135.60\\\hline&&N=\sum f_i=96&&&\sum_{i=1}^{8}f_i|d_i|=988.86\\\hline\end{array}
\displaystyle \text{Clearly, }N=96\Rightarrow\frac{N}{2}=48
\displaystyle \text{The cumulative frequency just greater than }48\text{ is }59.
\displaystyle \text{Therefore, the median class is }35.75-40.75.
\displaystyle \therefore l=35.75,\qquad f=26,\qquad F=33,\qquad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =35.75+\frac{48-33}{26}\times5
\displaystyle =38.63\text{ (approx.)}
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\text{Median}|=988.86\text{ and }N=\sum f_i=96
\displaystyle \therefore \mathrm{M.D.}=\frac{1}{N}\sum f_i|x_i-\text{Median}|=\frac{988.86}{96}
\displaystyle =10.30\text{ (approx.)}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the mean deviation from the mean and from the median of the following}
\displaystyle \text{distribution:}
\displaystyle \begin{array}{|c|ccccc|}\hline\text{Marks}&0-10&10-20&20-30&30-40&40-50\\\hline\text{Number of students}&5&8&15&16&6\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Calculation of mean deviation from the median:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-28|&f_i|d_i|\\\hline0-10&5&5&5&23&115\\10-20&15&8&13&13&104\\20-30&25&15&28&3&45\\30-40&35&16&44&7&112\\40-50&45&6&50&17&102\\\hline&&N=\sum f_i=50&&&\sum_{i=1}^{5}f_i|d_i|=478\\\hline\end{array}
\displaystyle \text{Clearly, }N=50\Rightarrow\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than }25\text{ is }28.
\displaystyle \text{The corresponding class is }20-30.
\displaystyle \text{Therefore, the median class is }20-30.
\displaystyle \therefore l=20,\qquad f=15,\qquad F=13,\qquad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =20+\frac{25-13}{15}\times10
\displaystyle =20+\frac{12}{15}\times10=28
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\text{Median}|=478\text{ and }N=\sum f_i=50
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\text{Median}|=\frac{478}{50}=9.56
\displaystyle \\

\displaystyle \text{Calculation of mean deviation from the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&f_ix_i&|x_i-\overline X|=|x_i-27|&f_i|x_i-\overline X|\\\hline0-10&5&5&25&22&110\\10-20&15&8&120&12&96\\20-30&25&15&375&2&30\\30-40&35&16&560&8&128\\40-50&45&6&270&18&108\\\hline&&N=\sum f_i=50&\sum_{i=1}^{5}f_ix_i=1350&&\sum_{i=1}^{5}f_i|x_i-\overline X|=472\\\hline\end{array}
\displaystyle \text{Clearly, }N=50\text{ and }\sum f_ix_i=1350
\displaystyle \therefore \overline X=\frac{\sum f_ix_i}{N}=\frac{1350}{50}=27
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=472\text{ and }N=\sum f_i=50
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|=\frac{472}{50}=9.44
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Calculate the mean deviation about the median age for the age distribution}
\displaystyle \text{of }100\text{ persons given below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline\text{Age}&16-20&21-25&26-30&31-35&36-40&41-45&46-50&51-55\\\hline\text{Number of persons}&5&6&12&14&26&12&16&9\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Since the class intervals are not continuous, subtract }0.5\text{ from each lower limit}
\displaystyle \text{and add }0.5\text{ to each upper limit to obtain continuous class intervals.}
\displaystyle \text{Calculation of mean deviation from the median:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-38|&f_i|d_i|\\\hline15.5-20.5&18&5&5&20&100\\20.5-25.5&23&6&11&15&90\\25.5-30.5&28&12&23&10&120\\30.5-35.5&33&14&37&5&70\\35.5-40.5&38&26&63&0&0\\40.5-45.5&43&12&75&5&60\\45.5-50.5&48&16&91&10&160\\50.5-55.5&53&9&100&15&135\\\hline&&N=\sum f_i=100&&&\sum_{i=1}^{8}f_i|d_i|=735\\\hline\end{array}
\displaystyle \text{Clearly, }N=100\Rightarrow\frac{N}{2}=\frac{100}{2}=50
\displaystyle \text{The cumulative frequency just greater than }50\text{ is }63.
\displaystyle \text{The corresponding class is }35.5-40.5.
\displaystyle \text{Therefore, the median class is }35.5-40.5.
\displaystyle \therefore l=35.5,\qquad f=26,\qquad F=37,\qquad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =35.5+\frac{50-37}{26}\times5
\displaystyle =35.5+\frac{13}{26}\times5
\displaystyle =35.5+2.5=38
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\text{Median}|=735\text{ and }N=\sum f_i=100
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\text{Median}|=\frac{735}{100}=7.35
\displaystyle \therefore \text{The mean deviation about the median age is }7.35\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Calculate the mean deviation about the mean for the following frequency}
\displaystyle \text{distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class interval}&0-4&4-8&8-12&12-16&16-20\\\hline\text{Frequency}&4&6&8&5&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{To avoid lengthy calculations, we compute }\overline X\text{ by the step-deviation method.}
\displaystyle \overline X=a+h\left(\frac{1}{N}\sum_{i=1}^{n}f_id_i\right),\qquad d_i=\frac{x_i-a}{h}
\displaystyle \text{where }a\text{ is the assumed mean and }h\text{ is the common factor.}
\displaystyle \text{Let }a=10\text{ and }h=4.
\displaystyle \text{Calculation of mean deviation from the mean:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&d_i=\frac{x_i-10}{4}&f_id_i&|x_i-\overline X|=|x_i-9.2|&f_i|x_i-\overline X|\\\hline0-4&2&4&-2&-8&7.2&28.8\\4-8&6&6&-1&-6&3.2&19.2\\8-12&10&8&0&0&0.8&6.4\\12-16&14&5&1&5&4.8&24\\16-20&18&2&2&4&8.8&17.6\\\hline&&N=\sum f_i=25&&\sum_{i=1}^{5}f_id_i=-5&&\sum_{i=1}^{5}f_i|x_i-\overline X|=96\\\hline\end{array}
\displaystyle \text{Clearly, }N=25\text{ and }\sum f_id_i=-5
\displaystyle \therefore \overline X=a+h\left(\frac{1}{N}\sum_{i=1}^{n}f_id_i\right)
\displaystyle =10+4\left(\frac{-5}{25}\right)
\displaystyle =10-\frac{20}{25}=10-0.8=9.2
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\overline X|=96\text{ and }N=\sum f_i=25
\displaystyle \therefore M.D.=\frac{1}{N}\sum f_i|x_i-\overline X|=\frac{96}{25}=3.84
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Calculate the mean deviation from the median of the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class interval}&0-6&6-12&12-18&18-24&24-30\\\hline\text{Frequency}&4&5&3&6&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Calculation of mean deviation from the median:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&x_i&f_i&\text{Cumulative frequency}&|d_i|=|x_i-14|&f_i|d_i|\\\hline0-6&3&4&4&11&44\\6-12&9&5&9&5&25\\12-18&15&3&12&1&3\\18-24&21&6&18&7&42\\24-30&27&2&20&13&26\\\hline&&N=\sum f_i=20&&&\sum_{i=1}^{5}f_i|d_i|=140\\\hline\end{array}
\displaystyle \text{Clearly, }N=20\Rightarrow\frac{N}{2}=\frac{20}{2}=10
\displaystyle \text{The cumulative frequency just greater than }10\text{ is }12.
\displaystyle \text{The corresponding class is }12-18.
\displaystyle \text{Therefore, the median class is }12-18.
\displaystyle \therefore l=12,\qquad f=3,\qquad F=9,\qquad h=6
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =12+\frac{10-9}{3}\times6
\displaystyle =12+\frac{1}{3}\times6=14
\displaystyle \text{Let }d_i=x_i-\text{Median}.\text{ Therefore, }|d_i|=|x_i-\text{Median}|.
\displaystyle \text{From the above table we get}
\displaystyle \sum f_i|x_i-\text{Median}|=140\text{ and }N=\sum f_i=20
\displaystyle \therefore \mathrm{M.D.}=\frac{1}{N}\sum f_i|x_i-\text{Median}|=\frac{140}{20}=7
\displaystyle \\

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