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

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. }(\bar{{}x}) & =\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, $M=\dfrac{(\dfrac{12}{2})^{t / h} \text{ observation }+(\dfrac{12}{2}+1)^{t / h} \text{ observation }}{2}$

$ \begin{aligned} & =\dfrac{6^{\text{th }} \text{ observation }+7^{\text{th }} \text{ observation }}{2} \\ & =\dfrac{13+14}{2}=\dfrac{27}{2}=13.5 \end{aligned} $

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. }(M) & =\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{(\dfrac{10}{2})^{t h} \text{ observation }+(\dfrac{10}{2}+1)^{t / h} \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. }(M) & =\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 :

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

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

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

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

Answer :

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

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

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

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 $=(\dfrac{29+1}{2}) _{\text{observation }=15^{\text{th }} \text{ observation }}^{\text{th }}$ on

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. }(M)=\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, $\quad N=\sum _{i=1}^{8} f_i=50, \sum _{i=1}^{8} f_i x_i=17900$

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

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

Answer :

The following table is formed.

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

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

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

Answer :

$ \begin{array}{|c|c|c|c|c|c|} \hline \text { Marks } & \begin{array}{|c|c|c|c|c|c|} \text { Mid values } \\ x_i \end{array} & f_i & \text { e.f. } & \left|x_i-27.86\right| & f_i\left|x_i-27.86\right| \\ \hline 0-10 & 5 & 6 & 6 & 22.86 & 137.16 \\ 10-20 & 15 & 8 & 14 & 12.86 & 102.88 \\ 20-30 & 25 & 14 & 28 & 2.86 & 40.04 \\ 30-40 & 35 & 16 & 44 & 7.14 & 114.24 \\ 40-50 & 45 & 4 & 48 & 17.14 & 68.56 \\ 50-60 & 55 & 2 & 50 & 27.14 & 54.28 \\ \hline & & 50 & & & 517.16 \\ \hline \end{array} $

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

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

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

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, $I=35.5, C=37, f=26, h=5$, 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,

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