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

6   Find the mean deviation about the mean for the data$\\$ $\begin{array}{|c|c|c|c|c|c|} \hline x_i & 10 & 30 & 50 & 70 & 90 \\ \hline f_i & 4 & 24 & 28 & 16 & 8 \\ \hline \end{array}$

Solution :

$\begin{array}{|c|c|c|c|c|c|} \hline x_i & f_i & f_i x_i & |x_i-\bar{x}|& f_i|x_i-\bar{x}| \\ \hline 10& 4 & 40 & 40& 160 \\ \hline 30 & 24 & 720 & 20 & 480 \\ \hline 50 & 28 & 1400 & 0 & 0 \\ \hline 70 & 16 & 1120 & 20 & 320 \\ \hline 90 & 8 & 720 & 40 & 320 \\ \hline & 80 & 4000 & & 1280 \\ \hline \end{array}$ $ N=\displaystyle\sum_{i=1}^{5} f_i =80,\displaystyle\sum_{i=1}^{5} f_i x_i=4000\\ \therefore \bar{x}=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i x_i=\dfrac{1}{80}*4000=50\\ \therefore MD(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i |x_i-\bar{x}|=\dfrac{1}{80}*1280=16$

7   Find the mean deviation about the median for the data.$\\$ $\begin{array}{|c|c|c|c|c|c|c|} \hline x_i & 5 & 7 & 9 & 10 & 12 & 15 \\ \hline f_i & 8 & 6 & 2 & 2 & 2 & 6\\ \hline \end{array}$

Solution :

The given observations are already in ascending order.$\\$ Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.$\\$ $\begin{array}{|c|c|c|} \hline x_i & f_i & c.f. \\ \hline 5 & 8 & 8 \\ \hline 7 & 6 & 14 \\ \hline 9 & 2 & 16 \\ \hline 10 & 2 & 18 \\ \hline 12 & 2 & 18 \\ \hline 15 & 6 & 26 \\ \hline \end{array}$ Here, $N = 26$, which is even.$\\$ Median is the mean of $13^{th}$ and $14^{th}$ observations. Both of these observations lie in the cumulative frequency $14$, for which the corresponding observation is $7.$$\\$ $\therefore \text{Median } =\dfrac{13^{th}\text{observation} + 14^{th} \text{oservation}}{2}\\ =\dfrac{7+7}{2}=7$$\\$ The absolute values of the deviations from median, i.e.$ |x _i - M |$, are $\\$ $ \begin{array}{|c|c|c|c|c|c|c|} \hline |x_i-M|& 2 & 0 & 2 & 3 & 5 & 8\\ \hline f_i & 8 & 6 & 2 & 2 & 2 & 6 \\ \hline f_i|x_i-M| & 16 & 0 & 4 & 6 & 10 & 48 \\ \hline \end{array}$ $\displaystyle\sum_{i=1}^{6} f_i =26 \text{and} \displaystyle\sum_{i=1}^{6} f_i|x_i-M|=84\\ M.D.(M)=\dfrac{1}{N}\displaystyle\sum_{i=1}^{6} f_i|x_i-M|=\dfrac{1}{26}* 84 =3.23$

8   Find the mean deviation about the median for the data$\\$ $\begin{array}{|c|c|c|c|} \hline x_i & 15 & 21 & 27 & 30 & 35 \\ \hline f_i & 3 & 5 & 6 & 7 & 8 \\ \hline \end{array}$

Solution :

The given observations are already in ascending order.$\\$ Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.$\\$ $\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & c.f. \\ \hline 15 & 3 & 3 \\ \hline 21 & 5 & 8\\ \hline 27 & 6 & 14 \\ \hline 30 & 7 & 21\\ \hline 35 & 8 & 29 \\ \hline \end{array}$ $\\$ Here,$ N=29 , $ which is odd $\\$ $\therefore $ Median$ =(\dfrac{7+7}{2})^{th} $ observation $ =15 ^{th}$ observation $\\$ This observation lies in the cumulative frequency $21$, for which the corresponding observation is $30.$$\\$ $\therefore \text{Median} =30 $$\\$ The absolute values of the deviations from median, i.e. $|x_ i -M|$ , are $\\$ $\begin{array}{|c|c|c|c|c|} \hline |x_i-M| & 15 & 9 & 3 & 0 & 5\\ \hline f_i & 3 & 5 & 6 & 7 & 8 \\ \hline f_i|x_i-M| & 45 & 45 & 18 & 0 & 40 \\ \hline \end{array}$$\\$ $\displaystyle\sum_{i=1}^{5} f_i =29, \displaystyle\sum_{i=1}^{5} f_i |x_i-M|=148\\ \therefore M.D.(M)=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i |x_i-M|=\dfrac{1}{29}*148=5.1$

9   Find the mean deviation about the mean for the data.$\\$ $\begin{array}{|c|c|c|c|c|} \hline \text{Income per day} & \text{Number of persons} \\ \hline 0-100 & 4 \\ \hline 100 - 200 & 8 \\ \hline 200 - 300 & 9 \\ \hline 300 -400 & 10 \\ \hline 400 -500 & 7 \\ \hline 500 -600 & 5 \\ \hline 600-700 & 4 \\ \hline 700 -800 &3 \\ \hline \end{array}$

Solution :

The following table is formed.$\\$ $ \begin{array}{|c|c|c|c|} \hline \text{Income per day} & \text{Number of persons f_i} & Mid-point x_i & f_ix_i & |x_i-\bar{x}| & f_i|x_i-\bar{x}| \\ \hline 0-100 & 4 & 50 & 200 & 308 & 1232 \\ \hline 100-200 & 8 & 150 & 1200 & 208 & 1664 \\ \hline 200 -300 & 9 & 250 & 2250 & 108 7 972 \\ \hline 300-400 & 10 & 350 & 3500 & 8 & 80 \\ \hline 400 -500 & 7 & 450 & 3150 & 92 & 644 \\ \hline 500-600 & 5 & 550 & 2750 & 192 & 960 \\ \hline 600-700 & 4 & 650 & 2600 & 292 & 1168 \\ \hline 700 -800 & 3 7 750 & 2250 & 392 & 1176 \\ \hline & 50 & 17900 & & 7896 \\ \hline \end{array}$ $\\$ Here, $ N=\displaystyle\sum_{i=1}^{8} f_i =50,\displaystyle\sum_{i=1}^{8} f_ix_i=17900\\ \therefore \bar{x}=\dfrac{1}{N}\displaystyle\sum_{i=1}^{8} f_ix_i=\dfrac{1}{50}*17900=358\\ M.D.(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{8} f_i|x_i-\bar{x}|=\dfrac{1}{50}*7896=157.92$

10   Find the mean deviation about the mean for the data$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Height in cms} & \text{Number of boys}\\ \hline 95-105 & 9 \\ \hline 105-115 & 13 \\ \hline 115-125 & 26 \\ \hline 125- 135 & 30 \\ \hline 135 - 145 & 12 \\ \hline 145 - 155 & 10 \\ \hline \end{array}$

Solution :

The following table is formed.$\\$ $\begin{array}{|c|c|c|c|c|c|} \hline \text{Height in cms} & \text{Number of boys f_i} & \text{Mid-point}x_i & f_i x_i & |x_i-\bar{x}|& f_i|x_i-\bar{x}|\\ \hline 95-105 & 9 & 100 & 900 & 25.3 & 227.7\\ \hline 105-115 &13 &110 &1430 & 15.3 &198.9 \\ \hline 115-125 & 26 &120 & 3120 & 5.3 & 137.8 \\ \hline 125-135 & 30 & 130 & 3900 & 4.7 & 141\\ \hline 135-145 & 12 & 140 & 1680 & 14.7 & 176.4 \\ \hline 145-155 7 10 & 150 & 1500 & 24.7 & 247\\ \hline \end{array}$ $\\$ Here,$N=\displaystyle\sum_{i=1}^{6} f_i =100, \displaystyle\sum_{i=1}^{6} f_ix_i=12530\\ \therefore \bar{x} =\dfrac{1}{N} \displaystyle\sum_{i=1}^{6} f_ix_i=\dfrac{1}{100}* 12530=125.3\\ M.D.(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{6} f_i|x_i-\bar{x}|=\dfrac{1}{100}*1128.8= 11.28 $