Answer :

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. $\left(\bar{{}x}\right)=\dfrac{\sum _{i=1}^{8}|x_i-\bar{{}x}|}{8}=\dfrac{6+3+2+1+0+2+3+7}{8}=\dfrac{24}{8}=3$

Answer :

The given data is $38,70,48,40,42,55,63,46,54,44$

Mean of the given data, $\bar{{}x}=\dfrac{38+70+48+40+42+55+63+46+54+44}{10}=\dfrac{500}{10}=50$

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

$ \begin{aligned} \text{ M.D. }\left(\bar{{}x}\right) & =\dfrac{\sum _{i=1}^{10}|x_i-\bar{{}x}|}{10} \\ \\ & =\dfrac{12+20+2+10+8+5+13+4+4+6}{10} \\ \\ & =\dfrac{84}{10} \\ \\ & =8.4 \end{aligned} $

Answer :

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, $M=\dfrac{\left(\dfrac{12}{2}\right)^{t h} \text{ observation }+\left(\dfrac{12}{2}+1\right)^{th} \text{ observation }}{2}$

$ \qquad\qquad \ =\dfrac{6^{\text{th }} \text{ observation }+7^{\text{th }} \text{ observation }}{2}$

$\qquad\qquad \ =\dfrac{13+14}{2}=\dfrac{27}{2}=13.5$

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

$ \begin{aligned} \text{ M.D. }\left(M\right) & =\dfrac{\sum _{i=1}^{12}|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 \end{aligned} $

Answer :

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 $36,42,45,46,46,49,51,53,60,72$

$ \begin{aligned} \text{ Median } M & =\dfrac{\left(\dfrac{10}{2}\right)^{t h} \text{ observation }+\left(\dfrac{10}{2}+1\right)^{th} \text{ observation }}{2} \\ \\ & =\dfrac{5^{\text{th }} \text{ observation }+6^{\text{th }} \text{ observation }}{2} \\ \\ & =\dfrac{46+49}{2}=\dfrac{95}{2}=47.5 \end{aligned} $

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

$ \begin{aligned} \text{ M.D. }\left(M\right) & =\dfrac{\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 \end{aligned} $

Answer :

$N=\sum _{i=1}^{5} f_i=25$

$\sum _{i=1}^{5} f_i x_i=350$

$\therefore \ \ \ \overline{x}=\dfrac{1}{N} \sum _{i=1}^{5} f_i x_i=\dfrac{1}{25} \times 350=14$

$\therefore \ \ \ MD\left(\overline{x}\right)=\dfrac{1}{N} \sum _{i=1}^{5} f_i|x_i-\overline{x}|=\dfrac{1}{25} \times 158=6.32$

Answer :

$N=\sum _{i=1}^{5} f_i=80, \sum _{i=1}^{5} f_i x_i=4000$

$\therefore \ \ \ \overline{x}=\dfrac{1}{N} \sum _{i=1}^{5} f_i x_i=\dfrac{1}{80} \times 4000=50$

$MD\left(\overline{x}\right) =\dfrac{1}{N} \sum _{i=1}^{5} f_i|x_i-\overline{x}|=\dfrac{1}{80} \times 1280=16$

Answer :

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

Here, $N=26$, which is even.

Median is the mean of $13^{\text{th }}$ and $14^{\text{th }}$ observations. Both of these observations lie in the cumulative frequency $14 ,$ for which the corresponding observation is $7 .$

$\therefore \ \ \ $ Median $=\dfrac{13^{\text{is }} \text{ observation }+14^{\text{th }} \text{ observation }}{2}=\dfrac{7+7}{2}=7$

The absolute values of the deviations from median, i.e. $|x_i-M|$, are

$ \begin{aligned} & \sum _{i=1}^{6} f_i=26 \quad \sum _{i=1}^{6} f_i|x_i-M|=84 \\ \\ & \text{ M.D.(M) }=\dfrac{1}{N} \sum _{i=1}^{6} f_i|x_i-M|=\dfrac{1}{26} \times 84=3.23 \end{aligned} $

Answer :

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

Here, $N=29$, which is odd.

$\therefore \ \ \ $ Median $=\left(\dfrac{29+1}{2}\right) {\text{observation }=15^{\text{th }} \text{ observation }}^{\text{th }}$

This observation lies in the cumulative frequency $21 ,$ for which the corresponding observation is $30 .$

$\therefore \ \ \ $ Median $=30$

The absolute values of the deviations from median, i.e. $|x_i-M|$, are

$ \begin{aligned} & \sum _{i=1}^{5} f_i=29, \\ \\ & \sum _{i=1}^{5} f_i|x_i-M|=148 \\ \\ & \quad \text{ M.D. }\left(M\right)=\dfrac{1}{N} \sum _{i=1}^{5} f_i|x_i-M|=\dfrac{1}{29} \times 148=5.1 \end{aligned} $

Answer :

The following table is formed.

Here, $\quad N=\sum _{i=1}^{8} f_i=50, \sum _{i=1}^{8} f_i x_i=17900$

$\therefore \ \ \ \overline{x}=\dfrac{1}{N} \sum _{i=1}^{8} f_i x_i=\dfrac{1}{50} \times 17900=358$

$\text{M.D.} \left(\overline{x}\right)=\dfrac{1}{N} \sum _{i=1}^{8} f_i|x_i-\overline{x}|=\dfrac{1}{50} \times 7896=157.92$

Answer :

The following table is formed.

Here, $\quad N=\sum _{i=1}^{6} f_i=100, \sum _{i=1}^{6} f_i x_i=12530$

$\therefore \ \ \ \overline{x}=\dfrac{1}{N} \sum _{i=1}^{6} f_i x_i=\dfrac{1}{100} \times 12530=125.3$

$\text{M.D.}\left(\overline{x}\right)=\dfrac{1}{N} \sum _{i=1}^{6} f_i|x_i-\overline{x}|=\dfrac{1}{100} \times 1128.8=11.28$

Answer :

$ \dfrac{\mathrm{N}}{2}=\dfrac{5 \mathrm{0}}{2}=25 $

$\therefore \ \ \ $ Median class is $20-30 $

$\therefore \ \ \ $ Median $=20+\dfrac{25-14}{14} \times 10=20+7.86=27.86$

$\text{M.D.}$ about median $=\dfrac{1}{N} \sum _{i=1}^n f i\left|x_i-M\right|=\dfrac{1}{50} \times 517.16=10.34$

Answer :

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval.

The table is formed as follows.

The class interval containing the $\dfrac{N^{t h}}{2}$ or $50^{\text{th }}$ item is $35.5 - 40.5$.

Therefore, $35.5 - 40.5$ is the median class.

It is known that,

Median $=L+\dfrac{\dfrac{N}{2}-C}{f} \times h$

Here, $L=35.5, C=37, f=26, h=5$, and $N=100$

$\therefore \ \ \ $ Median $=35.5+\dfrac{50-37}{26} \times 5=35.5+\dfrac{13 \times 5}{26}=35.5+2.5=38$

Thus, mean deviation about the median is given by,

$\text{M.D.(M)}$ $=\dfrac{1}{N} \sum _{i=1}^{8} f_i|x_i-M|=\dfrac{1}{100} \times 735=7.35$