1   Find the mean deviation about the mean for the data $4, 7, 8, 9, 10, 12, 13, 17$

Solution :

The given data is $4, 7, 8, 9, 10, 12, 13, 17$$\\$ Mean of the data,$\bar{x}=\dfrac{4+7+8+9+10+12+13+17}{8}\\ =\dfrac{80}{8}=10$$\\$ The deviations of the respective observations from the mean $\bar{x}$, i. e.,$x_i-\bar{x},$ are $-6,-3,-2,-1,0,2,3,7$$\\$ The absolute values of the deviations, i.e. $|x _i - \bar{x}|,$ , are$6,3,2,1,0,2,3,7$$\\$ The required mean deviation about the mean is$\\$ $M.D.(\bar{x})=\dfrac{\displaystyle\sum_{i=1}^{8}|x_i-\bar{x}|}{8}\\ =\dfrac{6+3+2+1+0+2+3+7}{8}=\dfrac{24}{8}=3$

2   Find the mean deviation about the mean for the data $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$

Solution :

The given data is $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$ Mean of the given data,$\\$ $\bar{x}=\dfrac{38+70+48+40+55+63+46+54+44}{10}\\ =\dfrac{500}{10}=50$$\\$ The deviations of the respective observations from the mean $\bar{x}, $ i.e.,$x_i-\bar{x},$ are$-12,20,-2,-10,-8,5,13,-4,4,-6$$\\$ The absolute values of the deviations,i.e.,$|x_i-\bar{x}|,$ are$12,20,2,10,8,5,13,4,4,6$$\\$ The required mean deviation about the mean is $\\$ $M.D.(\bar{x})=\dfrac{\displaystyle\sum_{i=1}^{8}|x_i-\bar{x}|}{10}\\ =\dfrac{12+20+2+10+8+5+13+4+4+6}{10}\\ =\dfrac{84}{10}=8.4$

3   Find the mean deviation about the median for the data. $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

Solution :

The given data is $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$$\\$ Here, the numbers of observations are $12,$ which is even.$\\$ Arranging the data in ascending order, we obtain$\\$ $10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$$\\$ Median,$M=\\ \dfrac{(\dfrac{12}{2})^{th}\text{observation}+(\dfrac{12}{2}+1)^{th}\text{observation}}{2}\\ =\dfrac{6^{th}\text{observation} + 7^{th}\text{observation} }{2}\\ =\dfrac{13+14}{2}\\ =\dfrac{27}{2}=13.5$$\\$ The deviations of the respective observations from the median,i.e.,$x_i-M,$ are$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,0.5,2.5,2.5,3.5,3.5,4.5$$\\$ The absolute values of the deviations,$|x_i-M|$ are $\\$ $3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5$$\\$ The required mean deviation about the median is $\\$ $ M.D.(M)=\dfrac{\displaystyle\sum_{i=1}^{10}|x_i-M|}{12}\\ =\dfrac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}\\ =\dfrac{28}{12}=2.33$

4   Find the mean deviation about the median for the data $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

Solution :

The given data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$$\\$ Here, the number of observations is $10,$ which is even. Arranging the data in ascending order, we obtain $36, 42, 45, 46, 46, 49, 51, 53, 60, 72$$\\$ Median $M=\\ \dfrac{(\dfrac{10}{2})^{th}\text{observation} +(\dfrac{10}{2}+1)^{th}\text{observation}}{2}\\ =\dfrac{5^{th}\text{observation} + 6^{th} \text{observation}}{2}\\ =\dfrac{46+49}{2}\\ =\dfrac{95}{2}=47.5$$\\$ The deviations of the respective observations from the median, i.e. $x_i-M$ are $-11.5,-5.5,-2.5,-1.5,-1.5,1.5,3.5,5.5,12.5,24.5$$\\$ The absolute values of the deviations, $|x_i-M|,$ are$\\$ $11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5$$\\$ Thus, the required mean deviation about the median is$\\$ $M.D.(M)=\dfrac{ \displaystyle\sum_{i=1}^{10}|x_i-M|}{10}\\ =\dfrac{11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5}{10}\\ =\dfrac{70}{10}=7$

5   Find the mean deviation about the mean for the data.$\\$ $\begin{array}{|c|c|c|c|c|} \hline x_i&5&10&15&20&25 \\ \hline f_i&7&4&6&3&5 \\ \hline \end{array}$

Solution :

$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & f_ix_i & |x_i-\bar{x}| & f_i|x_i-\bar{x}| \\ \hline 5 & 7 & 35 & 9 & 63 \\ \hline 10 & 4 & 40 & 4 & 16 \\ \hline 15 & 6 & 90 & 1 & 6 \\ \hline 20 & 3 & 60 & 6 & 18 \\ \hline 25 & 5 & 125 & 11 & 55 \\ \hline & 25 & 350 & & 158\\ \hline \end{array}$$\\$ $N = \displaystyle\sum_{i = 1}^{5} f_i = 25 \\ \displaystyle\sum_{i = 1}^{5} f_i x_i = 350 \\ \therefore \bar{x} = \dfrac{1}{N}\displaystyle\sum_{i = 1}^{5}f_i x_i \\ = \dfrac{1}{25} * 350 = 14 \\ \therefore MD(\bar{x}) = \dfrac{1}{N} \displaystyle\sum_{i = 1}^{5} f_i |x_i - \bar{x}| \\ \dfrac{1}{25} * 158 = 6.32$

$\displaystyle\sum_{i = 1}^{5} f_i x_i = 350 \\ \therefore \bar{x} = \dfrac{1}{N}\displaystyle\sum_{i = 1}^{5}f_i x_i \\ = \dfrac{1}{25} * 350 = 14 \\ \therefore MD(\bar{x}) = \dfrac{1}{N} \displaystyle\sum_{i = 1}^{5} f_i |x_i - \bar{x}| \\ \dfrac{1}{25} * 158 = 6.32$