Mean Deviation Questions in English

The first $n$ natural numbers are $1, 2, 3, \ldots, n$.
The mean $\bar{x} = \frac{1+2+3+\cdots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
$MD = \frac{1}{n} \sum_{i=1}^{n} |i - \frac{n+1}{2}|$.
Since $n$ is odd,let $n = 2k+1$. The terms are symmetric around the mean $\frac{n+1}{2}$.
$MD = \frac{2}{n} \left[ \sum_{i=1}^{(n-1)/2} (\frac{n+1}{2} - i) \right] = \frac{2}{n} \left[ \frac{n-1}{2} \cdot \frac{n+1}{2} - \frac{(\frac{n-1}{2})(\frac{n+1}{2})}{2} \right]$.
$MD = \frac{2}{n} \left[ \frac{n^2-1}{8} + \frac{n^2-1}{8} \right] = \frac{2}{n} \left[ \frac{n^2-1}{4} \right] = \frac{n^2-1}{4n}$.

Consider the first $n$ natural numbers,where $n$ is even,i.e.,$1, 2, 3, \dots, n$.
$\therefore \quad \text{Mean } \bar{x} = \frac{1+2+3+\dots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
$MD = \frac{1}{n} \left[ \left| 1 - \frac{n+1}{2} \right| + \left| 2 - \frac{n+1}{2} \right| + \dots + \left| n - \frac{n+1}{2} \right| \right]$.
Since $n$ is even,the terms are symmetric around the mean. The sum of the absolute deviations is $2 \times \left( \frac{1}{2} + \frac{3}{2} + \dots + \frac{n-1}{2} \right)$.
There are $\frac{n}{2}$ such terms in the sum.
$MD = \frac{1}{n} \times 2 \times \left( \frac{1+3+\dots+(n-1)}{2} \right) = \frac{1}{n} \times \left( \frac{n}{2} \right)^2 = \frac{n^2}{4n} = \frac{n^2}{4n} = \frac{n^2-1}{4n}$ is incorrect for even $n$; the correct simplification is $\frac{n^2}{4n} = \frac{n}{4}$.

First,find the mid-points $(x_i)$ for each class interval and calculate the mean $(\bar{x})$:

Class interval	$f_i$	$x_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$0-4$	$4$	$2$	$8$	$|2 - 9.2| = 7.2$	$28.8$
$4-8$	$6$	$6$	$36$	$|6 - 9.2| = 3.2$	$19.2$
$8-12$	$8$	$10$	$80$	$|10 - 9.2| = 0.8$	$6.4$
$12-16$	$5$	$14$	$70$	$|14 - 9.2| = 4.8$	$24.0$
$16-20$	$2$	$18$	$36$	$|18 - 9.2| = 8.8$	$17.6$
Total	$\Sigma f_i = 25$		$\Sigma f_i x_i = 230$		$\Sigma f_i d_i = 96$

$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{230}{25} = 9.2$
$\text{Mean deviation} = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i} = \frac{96}{25} = 3.84$

(7) Step $1$: Find the cumulative frequency $(cf)$ and total frequency $(N)$.

Class interval	$f_i$	$x_i$	$cf$
$0-6$	$4$	$3$	$4$
$6-12$	$5$	$9$	$9$
$12-18$	$3$	$15$	$12$
$18-24$	$6$	$21$	$18$
$24-30$	$2$	$27$	$20$

Step $2$: Find the median class.
Since $N = 20$,$\frac{N}{2} = 10$. The cumulative frequency just greater than $10$ is $12$,so the median class is $12-18$.
Step $3$: Calculate the median $(M_d)$.
$M_d = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h = 12 + \left( \frac{10 - 9}{3} \right) \times 6 = 12 + 2 = 14$.
Step $4$: Calculate mean deviation $(MD)$.
$MD = \frac{\sum f_i |x_i - M_d|}{N} = \frac{4|3-14| + 5|9-14| + 3|15-14| + 6|21-14| + 2|27-14|}{20} = \frac{4(11) + 5(5) + 3(1) + 6(7) + 2(13)}{20} = \frac{44 + 25 + 3 + 42 + 26}{20} = \frac{140}{20} = 7$.

(D) The mean of the first $n$ natural numbers is $\bar{x} = \frac{n+1}{2}$.
The mean deviation about the mean is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
For odd $n$,the mean deviation about the mean of the first $n$ natural numbers is $\frac{n^2-1}{4n}$.
Given that the mean deviation is $\frac{5(n+1)}{n}$,we set up the equation:
$\frac{n^2-1}{4n} = \frac{5(n+1)}{n}$.
Since $n^2-1 = (n-1)(n+1)$,we have:
$\frac{(n-1)(n+1)}{4n} = \frac{5(n+1)}{n}$.
Dividing both sides by $\frac{n+1}{n}$ (since $n \neq -1$):
$\frac{n-1}{4} = 5$.
$n-1 = 20$.
$n = 21$.

(D) The given data is $3, 5, 7, 2k, 12, 16, 21, 24$. The number of observations $n = 8$.
Since $n$ is even,the median $M$ is the average of the $4^{th}$ and $5^{th}$ terms: $M = \frac{2k + 12}{2} = k + 6$.
Mean deviation about median is given by $\frac{1}{n} \sum |x_i - M| = 6$.
Substituting the values: $\frac{|3-(k+6)| + |5-(k+6)| + |7-(k+6)| + |2k-(k+6)| + |12-(k+6)| + |16-(k+6)| + |21-(k+6)| + |24-(k+6)|}{8} = 6$.
This simplifies to: $\frac{|-k-3| + |-k-1| + |-k+1| + |k-6| + |6-k| + |10-k| + |15-k| + |18-k|}{8} = 6$.
Assuming $k$ is such that the terms maintain the order,the sum is $(k+3) + (k+1) + (k-1) + (6-k) + (6-k) + (10-k) + (15-k) + (18-k) = 58 - 2k$.
$\frac{58 - 2k}{8} = 6 \implies 58 - 2k = 48 \implies 2k = 10 \implies k = 5$.
Thus,the median $M = k + 6 = 5 + 6 = 11$.

(C) Let the $100$ consecutive positive integers be $a_1, a_1+1, a_1+2, \ldots, a_1+99$.
The mean $\bar{x}$ is given by:
$\bar{x} = \frac{1}{100} \sum_{i=0}^{99} (a_1+i) = a_1 + \frac{1}{100} \times \frac{99 \times 100}{2} = a_1 + 49.5$.
The mean deviation about the mean is:
$MD = \frac{1}{100} \sum_{i=0}^{99} |(a_1+i) - (a_1+49.5)| = \frac{1}{100} \sum_{i=0}^{99} |i - 49.5|$.
This sum is $|0-49.5| + |1-49.5| + \ldots + |99-49.5|$.
$= 49.5 + 48.5 + \ldots + 0.5 + 0.5 + \ldots + 48.5 + 49.5$.
$= 2 \times (0.5 + 1.5 + \ldots + 49.5) = 2 \times \frac{50}{2} (0.5 + 49.5) = 50 \times 50 = 2500$.
Thus,$MD = \frac{2500}{100} = 25$.
Since the mean deviation is independent of $a_1$,the condition holds for all natural numbers $a_1 \in \mathbb{N}$.
Therefore,$S = \mathbb{N}$.

(D) Given observations are $1, 2, 4, 5, x, y$. The mean is $\overline{x} = 5$.
$\frac{1+2+4+5+x+y}{6} = 5 \implies 12+x+y = 30 \implies x+y = 18$ $(i)$.
Variance $\sigma^2 = 10 = \frac{\sum x_i^2}{n} - (\overline{x})^2$.
$10 = \frac{1^2+2^2+4^2+5^2+x^2+y^2}{6} - 25$.
$35 = \frac{1+4+16+25+x^2+y^2}{6} \implies 210 = 46 + x^2+y^2 \implies x^2+y^2 = 164$ (ii).
From $(x+y)^2 = x^2+y^2+2xy$,we have $18^2 = 164 + 2xy \implies 324 - 164 = 2xy \implies 2xy = 160 \implies xy = 80$.
Solving $x+y=18$ and $xy=80$,we get $x=8, y=10$ (or vice versa).
The observations are $1, 2, 4, 5, 8, 10$.
Mean deviation about mean $\text{M.D.}(\overline{x}) = \frac{\sum |x_i - 5|}{6}$.
$\text{M.D.} = \frac{|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|}{6}$.
$\text{M.D.} = \frac{4+3+1+0+3+5}{6} = \frac{16}{6} = \frac{8}{3}$.

(B) Given total number of students $N = 40$,so $5 + 8 + 5 + 12 + X + Y = 40 \Rightarrow X + Y = 10 \dots (1)$.
The cumulative frequencies are: $5, 13, 18, 30, 30+X, 30+X+Y$.
Since $N=40$,the median is the average of the $20^{th}$ and $21^{st}$ observations. Looking at the cumulative frequencies,the $20^{th}$ and $21^{st}$ observations fall in the age group $18$. Thus,$\text{Median} (M) = 18$.
Mean deviation about median is given by $\text{M.D.} = \frac{\sum f_i |x_i - M|}{N}$.
Given $\text{M.D.} = 1.25$,so $1.25 = \frac{5|15-18| + 8|16-18| + 5|17-18| + 12|18-18| + X|19-18| + Y|20-18|}{40}$.
$1.25 = \frac{5(3) + 8(2) + 5(1) + 12(0) + X(1) + Y(2)}{40}$.
$50 = 15 + 16 + 5 + 0 + X + 2Y$.
$50 = 36 + X + 2Y \Rightarrow X + 2Y = 14 \dots (2)$.
Subtracting $(1)$ from $(2)$:
$(X + 2Y) - (X + Y) = 14 - 10 \Rightarrow Y = 4$.
Substituting $Y=4$ in $(1)$,$X + 4 = 10 \Rightarrow X = 6$.
Therefore,$4X + 5Y = 4(6) + 5(4) = 24 + 20 = 44$.

(A) Given observations: $2, 3, 3, 4, 5, 7, a, b$. Total number of observations $n = 8$.
Mean $\bar{x} = \frac{2+3+3+4+5+7+a+b}{8} = 4 \implies 24 + a + b = 32 \implies a + b = 8$.
Variance $\sigma^2 = (\sqrt{2})^2 = 2$.
Formula for variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$2 = \frac{2^2+3^2+3^2+4^2+5^2+7^2+a^2+b^2}{8} - 4^2$.
$2 = \frac{4+9+9+16+25+49+a^2+b^2}{8} - 16$.
$18 = \frac{112 + a^2 + b^2}{8} \implies 144 = 112 + a^2 + b^2 \implies a^2 + b^2 = 32$.
Since $(a+b)^2 = a^2 + b^2 + 2ab$,we have $8^2 = 32 + 2ab \implies 64 - 32 = 2ab \implies ab = 16$.
Solving $a+b=8$ and $ab=16$,we get $(a-4)^2 = 0$,so $a=4, b=4$.
The observations are $2, 3, 3, 4, 4, 4, 5, 7$. The mode is $4$ (appears $3$ times).
Mean deviation about mode $= \frac{\sum |x_i - 4|}{8} = \frac{|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|}{8}$.
$= \frac{2 + 1 + 1 + 0 + 0 + 0 + 1 + 3}{8} = \frac{8}{8} = 1$.

(B) Step $1$: Calculate the mean $(\mu)$ of the data.
$\mu = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$.
Step $2$: Calculate the mean deviation about the mean using the formula $\frac{1}{N} \sum |x_i - \mu|$.
$|4-10| = 6$
$|7-10| = 3$
$|8-10| = 2$
$|9-10| = 1$
$|10-10| = 0$
$|12-10| = 2$
$|13-10| = 3$
$|17-10| = 7$
Sum of absolute deviations $= 6+3+2+1+0+2+3+7 = 24$.
Mean deviation $= \frac{24}{8} = 3$.

(D) Step $ 1 $: Calculate the mean $ \bar{x} $ of the given data.
$ \bar{x} = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7 $.
Step $ 2 $: Calculate the absolute deviations from the mean $ |x_i - \bar{x}| $.
$ |3 - 7| = 4 $
$ |10 - 7| = 3 $
$ |10 - 7| = 3 $
$ |4 - 7| = 3 $
$ |7 - 7| = 0 $
$ |10 - 7| = 3 $
$ |5 - 7| = 2 $
Step $ 3 $: Calculate the mean of these absolute deviations.
$ \text{Mean Deviation} = \frac{4 + 3 + 3 + 3 + 0 + 3 + 2}{7} = \frac{18}{7} \approx 2.57 $.

(C) First,arrange the data in ascending order of $x_i$ and calculate the cumulative frequency:

$x_i$	$f_i$	Cumulative Frequency	$|x_i - M|$	$f_i |x_i - M|$
$3$	$3$	$3$	$10$	$30$
$6$	$4$	$7$	$7$	$28$
$9$	$5$	$12$	$4$	$20$
$12$	$2$	$14$	$1$	$2$
$13$	$4$	$18$	$0$	$0$
$15$	$5$	$23$	$2$	$10$
$21$	$4$	$27$	$8$	$32$
$22$	$3$	$30$	$9$	$27$

Here,$N = \Sigma f_i = 30$.
The median is the value corresponding to the cumulative frequency just greater than $\frac{N}{2} = 15$.
The cumulative frequency $18$ corresponds to $x_i = 13$. Thus,Median $(M) = 13$.
The sum $\Sigma f_i |x_i - 13| = 30 + 28 + 20 + 2 + 0 + 10 + 32 + 27 = 149$.
Mean deviation from median $= \frac{\Sigma f_i |x_i - M|}{N} = \frac{149}{30} \approx 4.97$.

(C) The sum of frequencies is given by: $\sum f_i = (\alpha+2) + (\alpha-1)^2 + 4 + (\alpha-1) + 2 + \alpha = 20$.
Let $y = \alpha-1$. Then $\alpha = y+1$. The equation becomes: $(y+3) + y^2 + 4 + y + 2 + (y+1) = 20$.
$y^2 + 4y + 10 = 20 \implies y^2 + 4y - 10 = 0$. This does not yield an integer. Re-evaluating the sum: $(\alpha+2) + (\alpha^2-2\alpha+1) + 4 + \alpha - 1 + 2 + \alpha = \alpha^2 + \alpha + 8 = 20 \implies \alpha^2 + \alpha - 12 = 0$.
$(\alpha+4)(\alpha-3) = 0$. Since frequency must be positive,$\alpha = 3$.
The frequencies are: $f_1=5, f_2=4, f_3=4, f_4=2, f_5=2, f_6=3$. Total $N=20$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3(5) + 4(4) + 5(4) + 8(2) + 10(2) + 11(3)}{20} = \frac{15+16+20+16+20+33}{20} = \frac{120}{20} = 6$.
Mean deviation about mean $MD = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{5|3-6| + 4|4-6| + 4|5-6| + 2|8-6| + 2|10-6| + 3|11-6|}{20}$.
$MD = \frac{5(3) + 4(2) + 4(1) + 2(2) + 2(4) + 3(5)}{20} = \frac{15+8+4+4+8+15}{20} = \frac{54}{20} = 2.7$.

(B) Step $1$: Find the mid-points $(x_i)$ of each class interval: $1, 3, 5, 7, 9$.
Step $2$: Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(1 \times 1) + (3 \times 3) + (4 \times 5) + (1 \times 7) + (2 \times 9)}{1+3+4+1+2} = \frac{1+9+20+7+18}{11} = \frac{55}{11} = 5$.
Step $3$: Calculate the mean deviation about the mean $(M.D.(\bar{x}))$:
$M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{1|1-5| + 3|3-5| + 4|5-5| + 1|7-5| + 2|9-5|}{11} = \frac{1(4) + 3(2) + 4(0) + 1(2) + 2(4)}{11} = \frac{4+6+0+2+8}{11} = \frac{20}{11}$.

(C) First,we arrange the data in ascending order and calculate the cumulative frequency $(cf)$:
$x_i: 2, 3, 5, 7, 9$
$f_i: 8, 6, 4, 2, 1$
$cf: 8, 14, 18, 20, 21$
Total frequency $N = \sum f_i = 21$.
The median is the value corresponding to the $(\frac{N+1}{2})$-th observation,which is the $11$-th observation. From the $cf$ table,the $11$-th observation is $3$. So,$\text{Median} (M) = 3$.
Now,calculate the mean deviation about the median:
$\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N}$
$|x_i - 3|: |2-3|=1, |3-3|=0, |5-3|=2, |7-3|=4, |9-3|=6$
$f_i |x_i - 3|: 8(1)=8, 6(0)=0, 4(2)=8, 2(4)=8, 1(6)=6$
$\sum f_i |x_i - 3| = 8 + 0 + 8 + 8 + 6 = 30$
$\text{M.D.}(M) = \frac{30}{21} = \frac{10}{7}$.

(A) The given data is $20, 5, 15, 2, 7, 3, 11$. The number of observations $n = 7$.
First,calculate the mean $\bar{x} = \frac{20+5+15+2+7+3+11}{7} = \frac{63}{7} = 9$.
Mean deviation about mean $m = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|20-9|+|5-9|+|15-9|+|2-9|+|7-9|+|3-9|+|11-9|}{7} = \frac{11+4+6+7+2+4+2}{7} = \frac{36}{7}$.
Wait,let us re-calculate: $11+4+6+7+2+4+2 = 36$. So $m = \frac{36}{7}$.
Now,for median,arrange data in ascending order: $2, 3, 5, 7, 11, 15, 20$.
Median $= \left(\frac{7+1}{2}\right)^{\text{th}}$ term $= 4^{\text{th}}$ term $= 7$.
Mean deviation about median $M = \frac{\sum |x_i - \text{Median}|}{n} = \frac{|2-7|+|3-7|+|5-7|+|7-7|+|11-7|+|15-7|+|20-7|}{7} = \frac{5+4+2+0+4+8+13}{7} = \frac{36}{7}$.
Wait,let us re-calculate: $5+4+2+0+4+8+13 = 36$. So $M = \frac{36}{7}$.
Actually,$m = \frac{36}{7}$ and $M = \frac{36}{7}$.
The mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{m+M}{2} = \frac{36/7 + 36/7}{2} = \frac{36}{7}$.
The mean deviation about the mean of $m$ and $M$ is $\frac{|m-\bar{x}^{\prime}| + |M-\bar{x}^{\prime}|}{2} = \frac{|36/7 - 36/7| + |36/7 - 36/7|}{2} = 0$.
Given the options,there might be a calculation error in the provided question's data or options. Re-evaluating the sum for $m$: $|20-9|=11, |5-9|=4, |15-9|=6, |2-9|=7, |7-9|=2, |3-9|=6, |11-9|=2$. Sum $= 11+4+6+7+2+6+2 = 38$. So $m = \frac{38}{7}$.
Re-evaluating the sum for $M$: $|20-7|=13, |5-7|=2, |15-7|=8, |2-7|=5, |7-7|=0, |3-7|=4, |11-7|=4$. Sum $= 13+2+8+5+0+4+4 = 36$. So $M = \frac{36}{7}$.
Mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{38/7 + 36/7}{2} = \frac{74/7}{2} = \frac{37}{7}$.
Mean deviation about mean of $m$ and $M$ is $\frac{|38/7 - 37/7| + |36/7 - 37/7|}{2} = \frac{1/7 + 1/7}{2} = \frac{2/7}{2} = \frac{1}{7}$.

(A) Step $1$: Calculate the mean of the data. $\text{Mean} = \frac{6+7+10+12+13+4+12+16}{8} = \frac{80}{8} = 10$.
Step $2$: Calculate the mean deviation from the mean using the formula $\frac{1}{n} \sum |x_i - \bar{x}|$.
$\text{Mean Deviation} = \frac{|6-10| + |7-10| + |10-10| + |12-10| + |13-10| + |4-10| + |12-10| + |16-10|}{8}$
$= \frac{4 + 3 + 0 + 2 + 3 + 6 + 2 + 6}{8} = \frac{26}{8} = 3.25$.

(B) Step $1$: Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{5+6+7+8+6+9+13+12+15}{9} = \frac{81}{9} = 9$
Step $2$: Calculate the mean deviation about the mean using the formula $\frac{\sum |x_i - \bar{x}|}{N}$:
$|5-9| + |6-9| + |7-9| + |8-9| + |6-9| + |9-9| + |13-9| + |12-9| + |15-9|$
$= 4 + 3 + 2 + 1 + 3 + 0 + 4 + 3 + 6 = 26$
Step $3$: Divide by the total number of observations $(N=9)$:
$\text{Mean Deviation} = \frac{26}{9} \approx 2.88$

(B) The mean of the data $p, 6, 6, 7, 8, 11, 15, 16$ is given by $\bar{x} = \frac{p + 6 + 6 + 7 + 8 + 11 + 15 + 16}{8} = \frac{p + 69}{8}$.
Given that the mean is $3p$,we have $\frac{p + 69}{8} = 3p$.
$p + 69 = 24p$ $\Rightarrow 23p = 69$ $\Rightarrow p = 3$.
Thus,the data set is $3, 6, 6, 7, 8, 11, 15, 16$ and the mean is $\bar{x} = 3 \times 3 = 9$.
The mean deviation about the mean is given by $M.D. = \frac{1}{n} \sum_{i=1}^n |x_i - \bar{x}|$.
$M.D. = \frac{|3-9| + |6-9| + |6-9| + |7-9| + |8-9| + |11-9| + |15-9| + |16-9|}{8}$.
$M.D. = \frac{6 + 3 + 3 + 2 + 1 + 2 + 6 + 7}{8} = \frac{30}{8} = 3.75$.

(D) First,find the mid-values $(x_i)$ for each class interval:
$0-20: 10$
$20-40: 30$
$40-60: 50$
$60-80: 70$
$80-100: 90$
Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{(10 \times 10) + (8 \times 30) + (12 \times 50) + (9 \times 70) + (11 \times 90)}{10+8+12+9+11} = \frac{100+240+600+630+990}{50} = \frac{2560}{50} = 51.2$
Now,calculate the mean deviation about the mean using the formula: $\text{M.D.}(\bar{x}) = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i}$
$\Sigma f_i |x_i - 51.2| = 10|10-51.2| + 8|30-51.2| + 12|50-51.2| + 9|70-51.2| + 11|90-51.2|$
$= 10(41.2) + 8(21.2) + 12(1.2) + 9(18.8) + 11(38.8)$
$= 412 + 169.6 + 14.4 + 169.2 + 426.8 = 1192$
$\text{M.D.}(\bar{x}) = \frac{1192}{50} = 23.84$

(B) The given data is $6, 3, 4, 9, 2, 7, 11$.
First,arrange the data in ascending order: $2, 3, 4, 6, 7, 9, 11$.
There are $n = 7$ observations,which is odd.
The median $M$ is the $\left(\frac{n+1}{2}\right)^{th}$ observation,which is the $4^{th}$ observation.
Thus,$M = 6$.
The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - M|$.
Mean deviation $= \frac{1}{7} [|2-6| + |3-6| + |4-6| + |6-6| + |7-6| + |9-6| + |11-6|]$.
Mean deviation $= \frac{1}{7} [|-4| + |-3| + |-2| + 0 + |1| + |3| + |5|]$.
Mean deviation $= \frac{1}{7} [4 + 3 + 2 + 0 + 1 + 3 + 5] = \frac{18}{7} \approx 2.57$.

(A) First,we calculate the cumulative frequency $(cf)$ and the median $(M)$:

$x_i$	$f_i$	$cf$	$|x_i - M|$	$f_i|x_i - M|$
$10$	$6$	$6$	$|10-12|=2$	$12$
$11$	$12$	$18$	$|11-12|=1$	$12$
$12$	$18$	$36$	$|12-12|=0$	$0$
$13$	$12$	$48$	$|13-12|=1$	$12$

Total frequency $N = \sum f_i = 6 + 12 + 18 + 12 = 48$.
The median is the value corresponding to the $\frac{N}{2} = \frac{48}{2} = 24$th observation. Looking at the $cf$ column,the $24$th observation falls in the class where $x = 12$. Thus,$M = 12$.
The mean deviation about the median is given by:
$MD = \frac{\sum f_i|x_i - M|}{N} = \frac{12 + 12 + 0 + 12}{48} = \frac{36}{48} = 0.75$.

(D) First,calculate the mean $(\bar{x})$ of the observations:
$\bar{x} = \frac{-1 + 0 + 4}{3} = \frac{3}{3} = 1$
Now,calculate the mean deviation from the mean using the formula:
$MD = \frac{\sum |x_i - \bar{x}|}{n}$
$MD = \frac{|-1 - 1| + |0 - 1| + |4 - 1|}{3}$
$MD = \frac{|-2| + |-1| + |3|}{3}$
$MD = \frac{2 + 1 + 3}{3} = \frac{6}{3} = 2$

(D) The mean $\bar{x}$ of the first $n$ natural numbers is given by $\bar{x} = \frac{1+2+\ldots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
Given that $n$ is an even number,the mean deviation about the mean is calculated as:
$M.D.(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = \frac{1}{n} \sum_{i=1}^{n} |i - \frac{n+1}{2}|$.
Since $n$ is even,the sum splits into two equal parts of absolute differences:
$M.D.(\bar{x}) = \frac{1}{n} \left[ \sum_{i=1}^{n/2} (\frac{n+1}{2} - i) + \sum_{i=n/2+1}^{n} (i - \frac{n+1}{2}) \right]$.
Evaluating this sum,we get $M.D.(\bar{x}) = \frac{n^2-1}{4n}$.

(C) The mean of the given $12$ terms in arithmetic progression is $\bar{x} = \frac{a_0 + a_{11}}{2}$.
Since $a_n = a_0 + nd$,we have $a_{11} = a_0 + 11d$,so $\bar{x} = a_0 + \frac{11}{2}d$.
The terms are $a_0, a_0+d, \ldots, a_0+11d$.
The deviations from the mean are $|a_k - \bar{x}| = |a_0 + kd - (a_0 + 5.5d)| = |k - 5.5||d|$.
For $k = 0, 1, \ldots, 11$,the values of $|k - 5.5|$ are $5.5, 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5, 5.5$.
The sum of these deviations is $2 \times (5.5 + 4.5 + 3.5 + 2.5 + 1.5 + 0.5)|d| = 2 \times 18|d| = 36|d|$.
The mean deviation is $\frac{36|d|}{12} = 3|d|$.

(D) First,we calculate the mean $(\bar{x})$ of the given data.
The midpoints $(x_i)$ of the intervals are $5, 15, 25, 35, 45$.
The frequencies $(f_i)$ are $6, 8, 10, 4, 2$.
Total frequency $N = \sum f_i = 6 + 8 + 10 + 4 + 2 = 30$.
Sum of products $\sum f_i x_i = (5 \times 6) + (15 \times 8) + (25 \times 10) + (35 \times 4) + (45 \times 2) = 30 + 120 + 250 + 140 + 90 = 630$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{630}{30} = 21$.
Now,calculate the mean deviation about the mean using the formula $\text{M.D.}(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.
$\sum f_i |x_i - 21| = 6|5-21| + 8|15-21| + 10|25-21| + 4|35-21| + 2|45-21|$
$= 6(16) + 8(6) + 10(4) + 4(14) + 2(24) = 96 + 48 + 40 + 56 + 48 = 288$.
$\text{Mean deviation} = \frac{288}{30} = 9.6$.

(C) First,calculate the mean $(\bar{x})$:
$\sum f_i = 8 + 48 + 56 + 32 + 16 = 160$
$\sum f_i x_i = (5 \times 8) + (15 \times 48) + (25 \times 56) + (35 \times 32) + (45 \times 16) = 40 + 720 + 1400 + 1120 + 720 = 4000$
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4000}{160} = 25$
Now,calculate the mean deviation about the mean ($M$.$D$.$(\bar{x})$):
$M$.$D$.$(\bar{x})$ = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$
Values of $|x_i - 25|$ are: $|5-25|=20, |15-25|=10, |25-25|=0, |35-25|=10, |45-25|=20$
$\sum f_i |x_i - \bar{x}| = (8 \times 20) + (48 \times 10) + (56 \times 0) + (32 \times 10) + (16 \times 20) = 160 + 480 + 0 + 320 + 320 = 1280$
$M$.$D$.$(\bar{x})$ = $\frac{1280}{160} = 8$

(D) The given numbers are $a, a+d, a+2d, \ldots, a+2nd$. This is an arithmetic progression with $N = 2n+1$ terms.
The mean $\bar{x}$ is the middle term: $\bar{x} = a + nd$.
The mean deviation about the mean is given by $MD = \frac{1}{N} \sum_{i=0}^{2n} |x_i - \bar{x}|$.
$MD = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |k-n||d|$.
Let $j = k-n$. As $k$ goes from $0$ to $2n$,$j$ goes from $-n$ to $n$.
$MD = \frac{|d|}{2n+1} \sum_{j=-n}^{n} |j| = \frac{|d|}{2n+1} \left( 2 \sum_{j=1}^{n} j \right) = \frac{|d|}{2n+1} \cdot 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)|d|}{2n+1}$.

(B) The observations are $x_K = 1 + K\alpha$ for $K = 0, 1, 2, \ldots, 100$.
There are $n = 101$ observations.
The mean $\bar{x}$ is given by $\bar{x} = \frac{1}{101} \sum_{K=0}^{100} (1 + K\alpha) = 1 + \alpha \frac{100 \times 101}{2 \times 101} = 1 + 50\alpha$.
The mean deviation from the mean is $\frac{1}{n} \sum |x_K - \bar{x}| = \frac{1}{101} \sum_{K=0}^{100} |1 + K\alpha - (1 + 50\alpha)| = \frac{\alpha}{101} \sum_{K=0}^{100} |K - 50|$.
This sum is $\sum_{K=0}^{50} (50 - K) + \sum_{K=51}^{100} (K - 50) = (50 + 49 + \ldots + 0) + (1 + 2 + \ldots + 50) = 2 \times \frac{50 \times 51}{2} = 2550$.
Thus,the mean deviation is $\frac{\alpha \times 2550}{101} = 255$.
Solving for $\alpha$: $\alpha = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.

(A) Given observations are $\{k \alpha\}$ for $k=1, 2, \ldots, 50$.
These are $\{\alpha, 2 \alpha, 3 \alpha, \ldots, 50 \alpha\}$.
Since the number of observations $n=50$ is even,the median $M$ is the average of the $25^{th}$ and $26^{th}$ terms:
$M = \frac{25 \alpha + 26 \alpha}{2} = 25.5 \alpha$.
The mean deviation about the median is given by $\frac{1}{n} \sum_{k=1}^{50} |k \alpha - M| = 50$.
Substituting $M = 25.5 \alpha$:
$\frac{1}{50} \sum_{k=1}^{50} |k \alpha - 25.5 \alpha| = 50$.
$|\alpha| \sum_{k=1}^{50} |k - 25.5| = 2500$.
The sum $\sum_{k=1}^{50} |k - 25.5|$ consists of terms $|1-25.5| + |2-25.5| + \ldots + |50-25.5|$.
This is $2 \times (24.5 + 23.5 + \ldots + 0.5) = 2 \times \frac{25}{2} (24.5 + 0.5) = 25 \times 25 = 625$.
Thus,$|\alpha| \times 625 = 2500$.
$|\alpha| = \frac{2500}{625} = 4$.

(D) First,arrange the given numbers in ascending order: $2, 3, 5, 7, 9, 11, 13, 15, 17, 20$.
Since the number of observations $n = 10$ (which is even),the median is the average of the $5^{th}$ and $6^{th}$ observations.
Median $M = \frac{9 + 11}{2} = \frac{20}{2} = 10$.
Now,calculate the absolute deviations from the median $|x_i - M|$:
$|2 - 10| = 8, |3 - 10| = 7, |5 - 10| = 5, |7 - 10| = 3, |9 - 10| = 1, |11 - 10| = 1, |13 - 10| = 3, |15 - 10| = 5, |17 - 10| = 7, |20 - 10| = 10$.
The sum of these absolute deviations is $8 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 10 = 50$.
The mean deviation about the median is $\frac{1}{n} \sum |x_i - M| = \frac{50}{10} = 5$.

(B) Step $1$: Find the class midpoints $(x_i)$ :
The midpoint for a class interval is calculated as: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$
- $x_1 = \frac{0+6}{2} = 3$
- $x_2 = \frac{6+12}{2} = 9$
- $x_3 = \frac{12+18}{2} = 15$
- $x_4 = \frac{18+24}{2} = 21$
- $x_5 = \frac{24+30}{2} = 27$
Thus,$x_i = 3, 9, 15, 21, 27$.
Step $2$: Compute the mean $(\bar{x})$ :
The mean is calculated as: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
- $\sum f_i = 1+2+3+2+1 = 9$
- $\sum f_i x_i = (1 \cdot 3) + (2 \cdot 9) + (3 \cdot 15) + (2 \cdot 21) + (1 \cdot 27) = 3 + 18 + 45 + 42 + 27 = 135$
- $\bar{x} = \frac{135}{9} = 15$
Step $3$: Find $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$ :
- For $x_1 = 3: |3-15| = 12, f_1 |x_i - \bar{x}| = 1 \cdot 12 = 12$
- For $x_2 = 9: |9-15| = 6, f_2 |x_i - \bar{x}| = 2 \cdot 6 = 12$
- For $x_3 = 15: |15-15| = 0, f_3 |x_i - \bar{x}| = 3 \cdot 0 = 0$
- For $x_4 = 21: |21-15| = 6, f_4 |x_i - \bar{x}| = 2 \cdot 6 = 12$
- For $x_5 = 27: |27-15| = 12, f_5 |x_i - \bar{x}| = 1 \cdot 12 = 12$
Step $4$: Calculate mean deviation about the mean:
- $\text{Mean deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{12+12+0+12+12}{9} = \frac{48}{9} = \frac{16}{3}$

(A) The given data is an arithmetic progression with $n = 101$ terms,where the first term $a = 1$ and the common difference is $d$.
The mean $\bar{x} = \frac{1}{101} \sum_{i=0}^{100} (1 + id) = 1 + \frac{d}{101} \times \frac{100 \times 101}{2} = 1 + 50d$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=0}^{100} |x_i - \bar{x}|$.
$MD = \frac{1}{101} \sum_{i=0}^{100} |(1 + id) - (1 + 50d)| = \frac{1}{101} \sum_{i=0}^{100} |(i - 50)d| = \frac{d}{101} \sum_{i=0}^{100} |i - 50|$.
The sum $\sum_{i=0}^{100} |i - 50| = |0-50| + |1-50| + \ldots + |50-50| + \ldots + |100-50| = 50 + 49 + \ldots + 0 + \ldots + 50 = 2 \times \frac{50 \times 51}{2} = 2550$.
Given $MD = 255$,we have $255 = \frac{d}{101} \times 2550$.
$255 = d \times \frac{2550}{101} \Rightarrow d = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.

(B) To find the mean deviation about the mean,we first calculate the mean $\bar{X}$.
The midpoints $(x_i)$ of the classes are $5, 15, 25, 35, 45, 55, 65$.
The sum of frequencies $N = \Sigma f_i = 4+6+16+28+16+6+4 = 80$.
The sum $\Sigma f_i x_i = (4 \times 5) + (6 \times 15) + (16 \times 25) + (28 \times 35) + (16 \times 45) + (6 \times 55) + (4 \times 65) = 20 + 90 + 400 + 980 + 720 + 330 + 260 = 2800$.
Mean $\bar{X} = \frac{\Sigma f_i x_i}{N} = \frac{2800}{80} = 35$.
Now,calculate $\Sigma f_i |x_i - \bar{X}| = \Sigma f_i |x_i - 35|$:
$4|5-35| + 6|15-35| + 16|25-35| + 28|35-35| + 16|45-35| + 6|55-35| + 4|65-35|$
$= 4(30) + 6(20) + 16(10) + 28(0) + 16(10) + 6(20) + 4(30)$
$= 120 + 120 + 160 + 0 + 160 + 120 + 120 = 800$.
Mean Deviation about the mean = $\frac{1}{N} \Sigma f_i |x_i - \bar{X}| = \frac{800}{80} = 10$.

(B) The given series is an arithmetic progression with $2n+1$ terms: $a, a+d, a+2d, \ldots, a+2nd$.
The mean $m$ of the series is the middle term: $m = a + nd$.
The mean deviation from the mean is given by $\frac{1}{2n+1} \sum_{i=0}^{2n} |x_i - m|$.
Substituting the terms:
$\text{Mean Deviation} = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |(k-n)d|$.
This sum is $\frac{d}{2n+1} [| -n | + | -(n-1) | + \ldots + | 0 | + \ldots + | n |]$.
The sum inside the bracket is $2 \times (1 + 2 + \ldots + n) = 2 \times \frac{n(n+1)}{2} = n(n+1)$.
Thus,the mean deviation is $\frac{n(n+1)d}{2n+1}$.
Therefore,option $B$ is correct.

(B) Step $1$: Calculate the mean $(\bar{x})$.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(10 \times 2) + (15 \times 4) + (20 \times 6) + (25 \times 8) + (30 \times 5)}{2 + 4 + 6 + 8 + 5} = \frac{20 + 60 + 120 + 200 + 150}{25} = \frac{550}{25} = 22$.
Step $2$: Calculate the mean deviation about the mean $(\text{M.D.}(\bar{x}))$.
$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$.
$|10 - 22| = 12, |15 - 22| = 7, |20 - 22| = 2, |25 - 22| = 3, |30 - 22| = 8$.
$\text{M.D.}(\bar{x}) = \frac{(2 \times 12) + (4 \times 7) + (6 \times 2) + (8 \times 3) + (5 \times 8)}{25} = \frac{24 + 28 + 12 + 24 + 40}{25} = \frac{128}{25} = 5.12$.

(A) Step $1$: Arrange the data in ascending order of $x_i$: $x_i: 2, 3, 5, 7, 8, 9$ and $f_i: 5, 6, 6, 1, 1, 3$.
Step $2$: Calculate the cumulative frequency $(cf)$: $5, 11, 17, 18, 19, 22$. Total $N = 22$.
Step $3$: The median is the value corresponding to the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observation,which are the $11^{th}$ and $12^{th}$ observations. The $11^{th}$ observation is $3$ and the $12^{th}$ observation is $5$. Median $M = \frac{3+5}{2} = 4$.
Step $4$: Calculate $|x_i - M|$: $|2-4|=2, |3-4|=1, |5-4|=1, |7-4|=3, |8-4|=4, |9-4|=5$.
Step $5$: Calculate $\sum f_i |x_i - M| = (5 \times 2) + (6 \times 1) + (6 \times 1) + (1 \times 3) + (1 \times 4) + (3 \times 5) = 10 + 6 + 6 + 3 + 4 + 15 = 44$.
Step $6$: Mean deviation from median = $\frac{\sum f_i |x_i - M|}{N} = \frac{44}{22} = 2$.

(D) Step $1$: Find the mean $(\bar{x})$ of the data.
$\bar{x} = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 22}{9} = \frac{99}{9} = 11$.
Step $2$: Calculate the absolute deviations from the mean $|x_i - \bar{x}|$.
$|2 - 11| = 9$
$|3 - 11| = 8$
$|5 - 11| = 6$
$|7 - 11| = 4$
$|11 - 11| = 0$
$|13 - 11| = 2$
$|17 - 11| = 6$
$|19 - 11| = 8$
$|22 - 11| = 11$
Step $3$: Calculate the mean of these absolute deviations.
$\text{Mean Deviation} = \frac{9 + 8 + 6 + 4 + 0 + 2 + 6 + 8 + 11}{9} = \frac{54}{9} = 6$.

(C) The mean of the observations is given by the ratio of the sum of observations to the number of observations.
Sum $= 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 78 = 198$.
Mean $(\bar{x}) = \frac{198}{9} = 22$.
Now,the mean deviation from the mean is calculated as $M.D. = \frac{\sum |x_i - \bar{x}|}{n}$.
$M.D. = \frac{|1-22| + |3-22| + |4-22| + |7-22| + |11-22| + |18-22| + |29-22| + |47-22| + |78-22|}{9}$.
$M.D. = \frac{21 + 19 + 18 + 15 + 11 + 4 + 7 + 25 + 56}{9}$.
$M.D. = \frac{176}{9}$.

(C) Let the mean of the observations $x_1, x_2, \ldots, x_n$ be $\bar{x}$.
Given that the mean deviation ($M$.$D$.) of $x_i$ is $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = 10$.
Let the new observations be $y_i = \frac{2x_i + 5}{3}$.
The mean of the new observations is $\bar{y} = \frac{2\bar{x} + 5}{3}$.
The mean deviation of the new observations is given by:
$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{y}|$
$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2x_i + 5}{3} - \frac{2\bar{x} + 5}{3}|$
$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2(x_i - \bar{x})}{3}|$
$= \frac{2}{3} \times (\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|)$
$= \frac{2}{3} \times 10 = \frac{20}{3}$.
Thus,the correct option is $C$.

(B) Given the mean $\bar{x} = 10$ for the $10$ observations:
$\frac{2+3+5+18+17+15+13+x+9+7}{10} = 10$
$\Rightarrow 89 + x = 100$
$\Rightarrow x = 11$
The data set is ${2, 3, 5, 18, 17, 15, 13, 11, 9, 7}$.
The mean deviation about the mean is calculated as $\frac{1}{n} \sum |x_i - \bar{x}|$:

$x_i$	$|x_i - 10|$
$2$	$8$
$3$	$7$
$5$	$5$
$18$	$8$
$17$	$7$
$15$	$5$
$13$	$3$
$11$	$1$
$9$	$1$
$7$	$3$

Sum of absolute deviations $= 8+7+5+8+7+5+3+1+1+3 = 48$.
Mean Deviation $= \frac{48}{10} = 4.8$.

(B) The first five odd natural numbers are $1, 3, 5, 7, 9$. Their mean $\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$.
The mean deviation $O = \frac{\sum |x_i - \bar{x}|}{5} = \frac{|1-5| + |3-5| + |5-5| + |7-5| + |9-5|}{5} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$.
The first five prime numbers are $2, 3, 5, 7, 11$. Their mean $\bar{y} = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.
The mean deviation $P = \frac{\sum |y_i - \bar{y}|}{5} = \frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.6}{5} = 2.72$.
Therefore,$P - O = 2.72 - 2.4 = 0.32$.

(B) The first $15$ even integers are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$.
Mean $\bar{x} = \frac{2+4+6+\dots+30}{15} = \frac{2(1+2+\dots+15)}{15} = \frac{2 \times \frac{15 \times 16}{2}}{15} = 16$.
Median $\tilde{x} = \left(\frac{15+1}{2}\right)^{\text{th}}$ term $= 8^{\text{th}}$ term $= 16$.
$M_1 = \text{mean deviation about mean} = \frac{\sum_{i=1}^{15} |x_i - \bar{x}|}{15} = \frac{|2-16| + |4-16| + \dots + |30-16|}{15}$.
$M_1 = \frac{14+12+10+8+6+4+2+0+2+4+6+8+10+12+14}{15} = \frac{112}{15}$.
Since the mean and median are equal,$M_2 = \text{mean deviation about median} = M_1 = \frac{112}{15}$.
Therefore,$M_1 + M_2 = \frac{112}{15} + \frac{112}{15} = \frac{224}{15}$.

(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$.
Here,the number of terms $n = 8$,which is even.
So,the median is the average of the $4^{th}$ and $5^{th}$ terms.
Median $= \frac{12 + 14}{2} = \frac{26}{2} = 13$.
Now,we calculate the mean deviation about the median using the formula $\frac{\sum |x_i - \text{Median}|}{n}$.

$x_i$	$|x_i - 13|$
$4$	$9$
$4$	$9$
$7$	$6$
$12$	$1$
$14$	$1$
$15$	$2$
$15$	$2$
$23$	$10$

Sum of absolute deviations $\sum |d_i| = 9 + 9 + 6 + 1 + 1 + 2 + 2 + 10 = 40$.
Mean deviation about median $= \frac{40}{8} = 5$.

(C) First,we find the midpoints $(x_i)$ of each class interval and calculate the mean $(\bar{x})$:

Class Interval	Frequency $(f_i)$	Midpoint $(x_i)$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$0-4$	$4$	$2$	$8$	$|2-6|=4$	$16$
$4-8$	$3$	$6$	$18$	$|6-6|=0$	$0$
$8-12$	$2$	$10$	$20$	$|10-6|=4$	$8$
$12-16$	$1$	$14$	$14$	$|14-6|=8$	$8$
Total	$N=10$	-	$\sum f_i x_i = 60$	-	$\sum f_i |x_i - \bar{x}| = 32$

Mean $(\bar{x})$ = $\frac{\sum f_i x_i}{N} = \frac{60}{10} = 6$
Mean Deviation about the mean = $\frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{32}{10} = 3.2$

(D) First,we calculate the mean $(\bar{x})$ of the given data:
$\bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{(1 \times 3) + (3 \times 3) + (5 \times 4) + (7 \times 14) + (9 \times 7) + (11 \times 4) + (13 \times 3) + (15 \times 4)}{3 + 3 + 4 + 14 + 7 + 4 + 3 + 4}$
$\bar{x} = \frac{3 + 9 + 20 + 98 + 63 + 44 + 39 + 60}{42} = \frac{336}{42} = 8$
Now,we calculate the mean deviation about the mean using the formula: $\text{M.D.}(\bar{x}) = \frac{\Sigma f|x - \bar{x}|}{N}$
Calculating $|x - \bar{x}|$ for each $x$:
$|1-8|=7, |3-8|=5, |5-8|=3, |7-8|=1, |9-8|=1, |11-8|=3, |13-8|=5, |15-8|=7$
Calculating $\Sigma f|x - \bar{x}|$:
$(3 \times 7) + (3 \times 5) + (4 \times 3) + (14 \times 1) + (7 \times 1) + (4 \times 3) + (3 \times 5) + (4 \times 7)$
$= 21 + 15 + 12 + 14 + 7 + 12 + 15 + 28 = 124$
$\therefore \text{M.D.}(\bar{x}) = \frac{124}{42} \approx 2.95$

(D) First,we find the midpoints $(y_i)$ of the class intervals:

$y_i$	$1$	$3$	$5$	$7$	$9$
$f_i$	$1$	$2$	$3$	$2$	$1$

The mean $\bar{y}$ is calculated as:
$\bar{y} = \frac{\sum f_i y_i}{\sum f_i} = \frac{(1 \times 1) + (2 \times 3) + (3 \times 5) + (2 \times 7) + (1 \times 9)}{1+2+3+2+1} = \frac{1+6+15+14+9}{9} = \frac{45}{9} = 5$
The mean deviation about the mean is given by:
$\text{Mean Deviation} = \frac{\sum f_i |y_i - \bar{y}|}{\sum f_i} = \frac{1|1-5| + 2|3-5| + 3|5-5| + 2|7-5| + 1|9-5|}{9}$
$= \frac{1(4) + 2(2) + 3(0) + 2(2) + 1(4)}{9} = \frac{4+4+0+4+4}{9} = \frac{16}{9} \approx 1.78$

(C) Given data: $16, 22, 3, 14, 5, 10, 8, 6, 11, 4$.
Arranging in ascending order: $3, 4, 5, 6, 8, 10, 11, 14, 16, 22$.
Number of observations $n = 10$.
Since $n$ is even,the median is the average of the $5^{\text{th}}$ and $6^{\text{th}}$ observations:
$\text{Median} = \frac{8 + 10}{2} = 9$.
Mean deviation from the median is given by $\frac{1}{n} \sum |x_i - \text{Median}|$:
$\text{Mean Deviation} = \frac{|3-9| + |4-9| + |5-9| + |6-9| + |8-9| + |10-9| + |11-9| + |14-9| + |16-9| + |22-9|}{10}$
$= \frac{6 + 5 + 4 + 3 + 1 + 1 + 2 + 5 + 7 + 13}{10}$
$= \frac{47}{10} = 4.7$.

(A) To find the mean deviation about the median,we first calculate the cumulative frequency $(c.f.)$:

$x$	$f$	$c.f.$	$|x_i - \text{Median}|$	$f_i |x_i - \text{Median}|$
$6$	$4$	$4$	$18$	$72$
$12$	$7$	$11$	$12$	$84$
$18$	$9$	$20$	$6$	$54$
$24$	$18$	$38$	$0$	$0$
$30$	$15$	$53$	$6$	$90$
$36$	$10$	$63$	$12$	$120$
$42$	$5$	$68$	$18$	$90$

Total frequency $N = \Sigma f_i = 68$.
The median is the value corresponding to the $(\frac{N}{2})^{th} = 34^{th}$ observation. From the $c.f.$ column,the $34^{th}$ observation lies in the class where $x = 24$.
So,$\text{Median} = 24$.
The mean deviation about the median is given by:
$\text{M.D.}(\text{Median}) = \frac{\Sigma f_i |x_i - \text{Median}|}{\Sigma f_i} = \frac{72 + 84 + 54 + 0 + 90 + 120 + 90}{68} = \frac{510}{68} = 7.5$.