Mean Deviation Questions in English

(B) The mean deviation is defined as the sum of absolute deviations from a central value divided by the number of observations.
It is a well-known mathematical property that the sum of absolute deviations $\sum |x_i - A|$ is minimized when $A$ is the median of the data set.
Therefore,the mean deviation from the median is the minimum possible value compared to the mean deviation measured from any other value.

(B) First,calculate the arithmetic mean $(\bar{x})$:
$\bar{x} = \frac{3 + 4 + 5 + 6 + 7}{5} = \frac{25}{5} = 5$
Next,calculate the mean deviation about the mean using the formula:
$\text{Mean Deviation} = \frac{\sum |x_i - \bar{x}|}{n}$
Substitute the values:
$= \frac{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}{5}$
$= \frac{|-2| + |-1| + |0| + |1| + |2|}{5}$
$= \frac{2 + 1 + 0 + 1 + 2}{5} = \frac{6}{5} = 1.2$

(C) The mean deviation about the mean for a frequency distribution is given by the formula $M.D.(\bar{x}) = \frac{\sum_{i=1}^{n} f_i |x_i - \bar{x}|}{\sum_{i=1}^{n} f_i}$.
Here,$d_i = x_i - \bar{x}$ represents the deviation of the $i$-th observation from the mean $\bar{x}$.
Thus,the formula can be written as $M.D. = \frac{\sum f|d|}{\sum f}$.
Therefore,option $C$ is correct.

(B) Step $1$: Calculate the mean of the observations.
Mean $(\bar{x}) = \frac{-1 + 0 + 4}{3} = \frac{3}{3} = 1$.
Step $2$: Calculate the mean deviation about the mean.
$M.D. = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|-1 - 1| + |0 - 1| + |4 - 1|}{3}$.
$M.D. = \frac{|-2| + |-1| + |3|}{3} = \frac{2 + 1 + 3}{3} = \frac{6}{3} = 2$.

(A) First,arrange the given data in ascending order:
$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$
Since the number of observations $n = 10$ (even),the median $M$ is the mean of the $5^{th}$ and $6^{th}$ terms:
$M = \frac{46 + 48}{2} = 47$
The mean deviation about the median is given by $\frac{1}{n} \sum |x_i - M|$.
Calculating $|x_i - 47|$ for each term:
$|34-47| = 13$
$|38-47| = 9$
$|42-47| = 5$
$|44-47| = 3$
$|46-47| = 1$
$|48-47| = 1$
$|54-47| = 7$
$|55-47| = 8$
$|63-47| = 16$
$|70-47| = 23$
Sum of absolute deviations = $13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86$
Mean deviation = $\frac{86}{10} = 8.6$.

(A) Given the set of numbers: $3, 4, 5, 6, 7$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{3 + 4 + 5 + 6 + 7}{5} = \frac{25}{5} = 5$.
Next,calculate the mean deviation about the mean using the formula: $\text{Mean Deviation} = \frac{\sum |x_i - \bar{x}|}{n}$.
$\text{Mean Deviation} = \frac{1}{5} [|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|]$.
$= \frac{1}{5} [|-2| + |-1| + |0| + |1| + |2|]$.
$= \frac{1}{5} [2 + 1 + 0 + 1 + 2] = \frac{6}{5} = 1.2$.

(C) First,arrange the observations in ascending order: $150, 210, 240, 300, 310, 320, 340$.
The number of observations $n = 7$,which is odd.
The median is the middle term,which is the $4^{th}$ term: $\text{Median} = 300$.
Now,calculate the mean deviation about the median using the formula $\frac{1}{n} \sum |x_i - \text{Median}|$:

$x_i$	$|x_i - 300|$
$150$	$|150 - 300| = 150$
$210$	$|210 - 300| = 90$
$240$	$|240 - 300| = 60$
$300$	$|300 - 300| = 0$
$310$	$|310 - 300| = 10$
$320$	$|320 - 300| = 20$
$340$	$|340 - 300| = 40$
Total	$\sum |x_i - 300| = 370$

Mean deviation about median $= \frac{370}{7} \approx 52.857 \approx 52.86$.

(B) The given set of numbers is $1, 2, 3, 4, 5$.
First,calculate the mean $(\overline{x})$:
$\overline{x} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$.
Next,calculate the mean deviation about the mean using the formula $\frac{\sum |x_i - \overline{x}|}{N}$:
$|1 - 3| = 2$
$|2 - 3| = 1$
$|3 - 3| = 0$
$|4 - 3| = 1$
$|5 - 3| = 2$
Sum of absolute deviations $= 2 + 1 + 0 + 1 + 2 = 6$.
Mean deviation $= \frac{6}{5} = 1.2$.

(D) The mean deviation of a frequency distribution is defined as the arithmetic mean of the absolute deviations of the observations from a central value (mean or median).
For a frequency distribution with frequencies $f_i$ and absolute deviations $|d_i|$,the formula is given by:
$\text{Mean Deviation} = \frac{\sum f_i |d_i|}{\sum f_i}$

(B) The coefficient of mean deviation about the mean is given by the formula: $\text{Coefficient of Mean Deviation} = \frac{\text{Mean Deviation}}{\text{Mean}}$.
Given,$\text{Mean Deviation} = 1.2$ and $\text{Mean} = 3$.
Therefore,$\text{Coefficient of Mean Deviation} = \frac{1.2}{3} = 0.4$.

(B) First,calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

$x_i$	$f_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$3$	$8$	$24$	$|3 - 15| = 12$	$96$
$9$	$10$	$90$	$|9 - 15| = 6$	$60$
$17$	$12$	$204$	$|17 - 15| = 2$	$24$
$23$	$9$	$207$	$|23 - 15| = 8$	$72$
$27$	$5$	$135$	$|27 - 15| = 12$	$60$
Total	$N = 44$	$\sum f_i x_i = 660$	-	$\sum f_i |x_i - \bar{x}| = 312$

Mean $\bar{x} = \frac{660}{44} = 15$.
Mean deviation about mean $= \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{312}{44} \approx 7.09$.

(C) Given observations are $x_i = -1, 0, 4$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{-1 + 0 + 4}{3} = \frac{3}{3} = 1$.
Next,calculate the absolute deviations from the mean $|x_i - \bar{x}|$:
$|-1 - 1| = |-2| = 2$
$|0 - 1| = |-1| = 1$
$|4 - 1| = |3| = 3$
Sum of absolute deviations = $2 + 1 + 3 = 6$.
Mean deviation about the mean = $\frac{1}{n} \sum |x_i - \bar{x}| = \frac{6}{3} = 2$.

(B) The given series is an arithmetic progression: $5, 10, 15, \dots, 85$.
Here,the first term $a = 5$,common difference $d = 5$,and the last term $l = 85$.
Number of terms $n$ is given by $85 = 5 + (n - 1)5$,which implies $n = 17$.
The mean $\bar{x} = \frac{5 + 85}{2} = 45$.
The mean deviation about the mean is given by $M.D. = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
For an arithmetic progression with $n$ terms,the mean deviation about the mean is $M.D. = \frac{d(n^2 - 1)}{4n}$ if $n$ is odd.
Here $n = 17$ (odd) and $d = 5$.
$M.D. = \frac{5(17^2 - 1)}{4 \times 17} = \frac{5(289 - 1)}{68} = \frac{5 \times 288}{68} = \frac{1440}{68} \approx 21.176$.

(B) First,we arrange the observations in ascending order:
$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$
Here,$n = 10$ (even).
The median $(M) = \frac{(\frac{n}{2})^{th} \text{ term} + (\frac{n}{2} + 1)^{th} \text{ term}}{2} = \frac{46 + 48}{2} = 47$.
Now,calculate the sum of absolute deviations from the median:
$\Sigma |x_i - M| = |34-47| + |38-47| + |42-47| + |44-47| + |46-47| + |48-47| + |54-47| + |55-47| + |63-47| + |70-47|$
$= 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86$.
Mean deviation about the median $= \frac{\Sigma |x_i - M|}{n} = \frac{86}{10} = 8.6$.

(A) The given numbers are in an arithmetic progression: $1, 1+d, 1+2d, \dots, 1+100d$.
Here,the number of terms is $N = 101$.
The mean of these terms is $\bar{x} = \frac{1}{101} \sum_{i=0}^{100} (1 + id) = 1 + \frac{d}{101} \times \frac{100 \times 101}{2} = 1 + 50d$.
The mean deviation about the mean is given by $M.D. = \frac{1}{N} \sum_{i=0}^{100} |(1+id) - (1+50d)| = \frac{1}{101} \sum_{i=0}^{100} |(i-50)d| = \frac{|d|}{101} \sum_{i=0}^{100} |i-50|$.
Calculating the sum: $\sum_{i=0}^{100} |i-50| = (50+49+\dots+1) + 0 + (1+2+\dots+50) = 2 \times \frac{50 \times 51}{2} = 2550$.
Thus,$M.D. = \frac{|d|}{101} \times 2550 = 255$.
$|d| = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.

(A) The given numbers are $a, 2a, 3a, \dots, 50a$. The number of terms $n = 50$.
Since $n$ is even,the median $M$ is the average of the $25^{th}$ and $26^{th}$ terms:
$M = \frac{25a + 26a}{2} = 25.5a$.
The mean deviation about the median is given by $M.D. = \frac{1}{n} \sum_{i=1}^{n} |x_i - M|$.
$M.D. = \frac{1}{50} \sum_{i=1}^{50} |ia - 25.5a| = \frac{|a|}{50} \sum_{i=1}^{50} |i - 25.5|$.
Expanding the sum: $\sum_{i=1}^{50} |i - 25.5| = |1 - 25.5| + |2 - 25.5| + \dots + |50 - 25.5|$.
$= 24.5 + 23.5 + \dots + 0.5 + 0.5 + \dots + 23.5 + 24.5$.
$= 2 \times (0.5 + 1.5 + \dots + 24.5)$.
This is an arithmetic progression with $25$ terms where the first term $a_1 = 0.5$ and last term $l = 24.5$.
Sum $= 2 \times \left[ \frac{25}{2} (0.5 + 24.5) \right] = 25 \times 25 = 625$.
Given $M.D. = 50$,so $\frac{|a|}{50} \times 625 = 50$.
$|a| \times 12.5 = 50 \Rightarrow |a| = \frac{50}{12.5} = 4$.

(B) Arrange the observations in ascending order: $40, 54, 62, 68, 76, 90$.
Number of terms $(n) = 6$ (even).
Median $(M) = \frac{(\frac{n}{2})^{th} \text{ term} + (\frac{n}{2} + 1)^{th} \text{ term}}{2} = \frac{62 + 68}{2} = 65$.
$\Sigma |x_i - M| = |40-65| + |54-65| + |62-65| + |68-65| + |76-65| + |90-65| = 25 + 11 + 3 + 3 + 11 + 25 = 78$.
Mean deviation about median $= \frac{\Sigma |x_i - M|}{n} = \frac{78}{6} = 13$.
Coefficient of mean deviation $= \frac{\text{Mean Deviation}}{\text{Median}} = \frac{13}{65} = 0.2$.

(D) Given observations are $4, 7, 2, 8, 6, a$. The number of observations $n = 6$.
The mean is given as $7$.
$\frac{4 + 7 + 2 + 8 + 6 + a}{6} = 7$
$\frac{27 + a}{6} = 7$
$27 + a = 42$
$a = 15$.
The observations are $4, 7, 2, 8, 6, 15$.
Arranging in ascending order: $2, 4, 6, 7, 8, 15$.
Since $n = 6$ (even),the median $M = \frac{\text{value of } (n/2)^{th} \text{ term} + \text{value of } (n/2 + 1)^{th} \text{ term}}{2} = \frac{6 + 7}{2} = 6.5$.
Mean deviation about median $= \frac{\sum |x_i - M|}{n}$.

$x_i$	$|x_i - 6.5|$
$2$	$4.5$
$4$	$2.5$
$6$	$0.5$
$7$	$0.5$
$8$	$1.5$
$15$	$8.5$

Sum of deviations $= 4.5 + 2.5 + 0.5 + 0.5 + 1.5 + 8.5 = 18$.
Mean deviation $= \frac{18}{6} = 3$.

(B) The number of terms in the series is $N = 2n + 1$.
The mean $\bar{x}$ is given by $\bar{x} = \frac{a + (a + d) + (a + 2d) + \dots + (a + 2nd)}{2n + 1}$.
Using the sum of an arithmetic progression,$\bar{x} = \frac{1}{2n + 1} \left[ \frac{2n + 1}{2} (a + a + 2nd) \right] = a + nd$.
The mean deviation about the mean is $MD = \frac{1}{N} \sum_{i=1}^{N} |x_i - \bar{x}|$.
$\sum |x_i - \bar{x}| = |a - (a + nd)| + |(a + d) - (a + nd)| + \dots + |(a + 2nd) - (a + nd)|$
$= |-nd| + |(1 - n)d| + \dots + |0| + \dots + |nd| = 2|d| [n + (n - 1) + \dots + 1] = 2|d| \frac{n(n + 1)}{2} = n(n + 1)|d|$.
Therefore,$MD = \frac{n(n + 1)|d|}{2n + 1}$.

(B) The mean deviation of a data set is defined as the average of the absolute deviations of the observations from a central value $M$.
Mathematically,the mean deviation about $M$ is given by $MD(M) = \frac{1}{n} \sum_{i=1}^{n} |x_i - M|$.
It is a well-established statistical property that the sum of absolute deviations $\sum |x_i - M|$ is minimized when $M$ is the median of the data set.
Therefore,the mean deviation about the median is the minimum possible mean deviation compared to any other central value.

(B) The given series $1, 1+d, 1+2d, \dots, 1+100d$ is an $A.P.$ with $n = 101$ terms.
The mean of this series is $\bar{x} = \frac{\sum_{r=0}^{100} (1+rd)}{101} = \frac{101 + d \frac{100 \times 101}{2}}{101} = 1 + 50d$.
The mean deviation from the mean is given by $\frac{1}{101} \sum_{r=0}^{100} |(1+rd) - (1+50d)| = \frac{1}{101} \sum_{r=0}^{100} |(r-50)d|$.
Assuming $d > 0$,this becomes $\frac{d}{101} [\sum_{r=0}^{50} (50-r) + \sum_{r=51}^{100} (r-50)]$.
$= \frac{d}{101} [ (50+49+\dots+0) + (1+2+\dots+50) ] = \frac{d}{101} [ 2 \times \frac{50 \times 51}{2} ] = \frac{50 \times 51 \times d}{101}$.
Given the mean deviation is $255$,we have $\frac{50 \times 51 \times d}{101} = 255$.
$d = \frac{255 \times 101}{50 \times 51} = \frac{5 \times 101}{50} = \frac{101}{10} = 10.1$.

(B) The given numbers are $a, 2a, 3a, \dots, 50a$. The total number of terms is $n = 50$.
Since $n$ is even,the median is the average of the $25^{th}$ and $26^{th}$ terms:
$\text{Median} = \frac{25a + 26a}{2} = 25.5a$.
Mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \text{Median}| = 50$.
$\frac{1}{50} \sum_{i=1}^{50} |ia - 25.5a| = 50$.
$\sum_{i=1}^{50} |i - 25.5| |a| = 2500$.
$|a| (|25.5 - 1| + |25.5 - 2| + \dots + |25.5 - 50|) = 2500$.
$|a| (24.5 + 23.5 + \dots + 0.5 + 0.5 + \dots + 24.5) = 2500$.
$|a| \times 2 \times (0.5 + 1.5 + \dots + 24.5) = 2500$.
$|a| \times 2 \times \frac{25}{2} (0.5 + 24.5) = 2500$.
$|a| \times 25 \times 25 = 2500$.
$|a| \times 625 = 2500$.
$|a| = 4$.

(C) The total number of observations is $3n$.
There are $n$ observations equal to $a$ and $2n$ observations equal to $-2a$.
The mean $(\bar{X})$ is calculated as:
$\bar{X} = \frac{n(a) + 2n(-2a)}{3n} = \frac{na - 4na}{3n} = \frac{-3na}{3n} = -a$.
The mean deviation about the mean is given by $\frac{1}{3n} \sum_{i=1}^{3n} |x_i - \bar{X}|$.
Substituting the values:
Mean deviation $= \frac{n|a - (-a)| + 2n|-2a - (-a)|}{3n}$
$= \frac{n|2a| + 2n|-a|}{3n}$
$= \frac{2na + 2na}{3n} = \frac{4na}{3n} = \frac{4a}{3}$.

(A) The given observations are $x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x$. Since there are $n = 10$ observations,the median is the average of the $5^{th}$ and $6^{th}$ terms: $\text{Median} = \frac{5x + 6x}{2} = 5.5|x|$.
The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{10} |x_i - \text{Median}| = 30$.
Substituting the values: $\frac{1}{10} (|x - 5.5x| + |2x - 5.5x| + |3x - 5.5x| + |4x - 5.5x| + |5x - 5.5x| + |6x - 5.5x| + |7x - 5.5x| + |8x - 5.5x| + |9x - 5.5x| + |10x - 5.5x|) = 30$.
This simplifies to: $\frac{1}{10} (| -4.5x | + | -3.5x | + | -2.5x | + | -1.5x | + | -0.5x | + | 0.5x | + | 1.5x | + | 2.5x | + | 3.5x | + | 4.5x |) = 30$.
Since $|-a| = |a|$,we have: $\frac{2}{10} (4.5 + 3.5 + 2.5 + 1.5 + 0.5) |x| = 30$.
$\frac{2}{10} (12.5) |x| = 30$.
$2.5 |x| = 30$.
$|x| = \frac{30}{2.5} = 12$.

(B) The mean deviation of a set of observations $x_1, x_2, ..., x_n$ about a value $a$ is defined as $\frac{1}{n} \sum_{i=1}^{n} |x_i - a|$.
It is a well-known mathematical property that the sum of absolute deviations $\sum |x_i - a|$ is minimized when $a$ is the median of the data set.
Therefore,the mean deviation is least when it is calculated about the median.

(A) The given numbers are $1, 1+d, 1+2d, \dots, 1+100d$. This is an arithmetic progression with $n = 101$ terms.
The mean $\bar{x}$ is given by:
$\bar{x} = \frac{1}{101} \sum_{k=0}^{100} (1 + kd) = \frac{1}{101} [101 + d \times \frac{100 \times 101}{2}] = 1 + 50d$.
The mean deviation from the mean is:
$MD = \frac{1}{101} \sum_{k=0}^{100} |(1 + kd) - (1 + 50d)| = \frac{1}{101} \sum_{k=0}^{100} |(k - 50)d| = \frac{|d|}{101} \sum_{k=0}^{100} |k - 50|$.
Calculating the sum: $\sum_{k=0}^{100} |k - 50| = |0-50| + |1-50| + \dots + |50-50| + \dots + |100-50| = 50 + 49 + \dots + 0 + 1 + \dots + 50 = 2 \times \frac{50 \times 51}{2} = 2550$.
Thus,$MD = \frac{|d|}{101} \times 2550 = 255$.
$|d| = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.

(C) Given $n = 5$,mean $\bar{x} = 5$,and variance $\sigma^2 = 124$.
Let the observations be $x_1=1, x_2=2, x_3=6, x_4, x_5$.
Since $\bar{x} = \frac{\sum x_i}{5} = 5$,we have $1+2+6+x_4+x_5 = 25$,so $x_4+x_5 = 16$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 124$.
$\frac{1^2+2^2+6^2+x_4^2+x_5^2}{5} - 5^2 = 124$ $\Rightarrow \frac{1+4+36+x_4^2+x_5^2}{5} = 149$ $\Rightarrow x_4^2+x_5^2 = 745 - 41 = 704$.
We have $(x_4+x_5)^2 = x_4^2+x_5^2 + 2x_4x_5$ $\Rightarrow 16^2 = 704 + 2x_4x_5$ $\Rightarrow 256 = 704 + 2x_4x_5$ $\Rightarrow 2x_4x_5 = -448$ $\Rightarrow x_4x_5 = -224$.
Solving $x_4+x_5=16$ and $x_4x_5=-224$,the quadratic equation $t^2 - 16t - 224 = 0$ gives $t = \frac{16 \pm \sqrt{256 + 896}}{2} = \frac{16 \pm \sqrt{1152}}{2} = 8 \pm 12\sqrt{2}$.
Mean deviation $M$.$D$. $= \frac{1}{5} \sum |x_i - 5| = \frac{1}{5} (|1-5| + |2-5| + |6-5| + |x_4-5| + |x_5-5|)$.
$|x_4-5| + |x_5-5| = |x_4+x_5-10| = |16-10| = 6$ (since $x_4, x_5$ are on opposite sides of $5$).
$M$.$D$. $= \frac{4+3+1+6}{5} = \frac{14}{5} = 2.8$.

(D) Given observations are $4, 7, 2, 8, 6, a$ and the mean is $7$.
We know that Mean $= \frac{4 + 7 + 2 + 8 + 6 + a}{6}$.
$7 = \frac{27 + a}{6}$ $\Rightarrow 42 = 27 + a$ $\Rightarrow a = 15$.
Now,the observations in ascending order are $2, 4, 6, 7, 8, 15$.
Since the number of observations $n = 6$ (even),the median is the average of the $3^{rd}$ and $4^{th}$ observations.
Median $= \frac{6 + 7}{2} = 6.5$.
Mean deviation from median $= \frac{\sum |x_i - 6.5|}{6}$.
$= \frac{|2 - 6.5| + |4 - 6.5| + |6 - 6.5| + |7 - 6.5| + |8 - 6.5| + |15 - 6.5|}{6}$.
$= \frac{4.5 + 2.5 + 0.5 + 0.5 + 1.5 + 8.5}{6} = \frac{18}{6} = 3$.

(A) Step $1$: Calculate the mean of the given data:
$\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$
Step $2$: Find the absolute deviations of the observations from the mean $|x_i - \bar{x}|$:
$|6-9| = 3$
$|7-9| = 2$
$|10-9| = 1$
$|12-9| = 3$
$|13-9| = 4$
$|4-9| = 5$
$|8-9| = 1$
$|12-9| = 3$
Step $3$: Calculate the mean deviation about the mean:
$M.D.(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{3+2+1+3+4+5+1+3}{8} = \frac{22}{8} = 2.75$

(A) First,we calculate the mean $(\bar{x})$ of the given data:
$\bar{x} = \frac{1}{20} \sum_{i=1}^{20} x_i = \frac{12+3+18+17+4+9+17+19+20+15+8+17+2+3+16+11+3+1+0+5}{20} = \frac{201}{20} = 10.05$.
Next,we find the absolute deviations $|x_i - \bar{x}|$:
$|12-10.05| = 1.95, |3-10.05| = 7.05, |18-10.05| = 7.95, |17-10.05| = 6.95, |4-10.05| = 6.05, |9-10.05| = 1.05, |17-10.05| = 6.95, |19-10.05| = 8.95, |20-10.05| = 9.95, |15-10.05| = 4.95, |8-10.05| = 2.05, |17-10.05| = 6.95, |2-10.05| = 8.05, |3-10.05| = 7.05, |16-10.05| = 5.95, |11-10.05| = 0.95, |3-10.05| = 7.05, |1-10.05| = 9.05, |0-10.05| = 10.05, |5-10.05| = 5.05$.
Sum of absolute deviations:
$\sum |x_i - \bar{x}| = 1.95 + 7.05 + 7.95 + 6.95 + 6.05 + 1.05 + 6.95 + 8.95 + 9.95 + 4.95 + 2.05 + 6.95 + 8.05 + 7.05 + 5.95 + 0.95 + 7.05 + 9.05 + 10.05 + 5.05 = 124$.
Mean Deviation about the mean:
$M.D.(\bar{x}) = \frac{1}{20} \sum |x_i - \bar{x}| = \frac{124}{20} = 6.2$.

(A) The number of observations $n = 11$,which is odd.
Arranging the data in ascending order: $3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$.
Median $(M) = \left(\frac{n+1}{2}\right)^{\text{th}}$ observation $= 6^{\text{th}}$ observation $= 9$.
The absolute deviations $|x_i - M|$ are:
$|3-9|=6, |3-9|=6, |4-9|=5, |5-9|=4, |7-9|=2, |9-9|=0, |10-9|=1, |12-9|=3, |18-9|=9, |19-9|=10, |21-9|=12$.
Sum of absolute deviations $\sum |x_i - M| = 6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12 = 58$.
Mean Deviation about median $= \frac{1}{n} \sum |x_i - M| = \frac{58}{11} \approx 5.27$.

(A) First,we calculate the mean $\bar{x}$ of the given data:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(2 \times 2) + (5 \times 8) + (6 \times 10) + (8 \times 7) + (10 \times 8) + (12 \times 5)}{2 + 8 + 10 + 7 + 8 + 5} = \frac{4 + 40 + 60 + 56 + 80 + 60}{40} = \frac{300}{40} = 7.5$
Next,we calculate the mean deviation about the mean using the formula $M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N}$:

$x_i$	$f_i$	$|x_i - 7.5|$	$f_i |x_i - 7.5|$
$2$	$2$	$5.5$	$11$
$5$	$8$	$2.5$	$20$
$6$	$10$	$1.5$	$15$
$8$	$7$	$0.5$	$3.5$
$10$	$8$	$2.5$	$20$
$12$	$5$	$4.5$	$22.5$
Total	$40$	-	$92$

$M.D.(\bar{x}) = \frac{92}{40} = 2.3$

The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies $(c.f.)$ to the given data:

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$
$c.f.$	$3$	$7$	$12$	$14$	$18$	$23$	$27$	$30$

Now,$N = 30$,which is even.
Median is the mean of the $15^{\text{th}}$ and $16^{\text{th}}$ observations. Both of these observations lie in the cumulative frequency $18$,for which the corresponding observation is $13$.
Therefore,Median $M = \frac{15^{\text{th}} \text{ observation} + 16^{\text{th}} \text{ observation}}{2} = \frac{13+13}{2} = 13$.
Now,absolute values of the deviations from median,$|x_i - M|$,are calculated:

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

We have $\sum f_i = 30$ and $\sum f_i|x_i - M| = 30+28+20+2+0+10+32+27 = 149$.
Therefore,$M.D.(M) = \frac{1}{N} \sum f_i|x_i - M| = \frac{149}{30} = 4.966... \approx 4.97$.

(D) We make the following table from the given data:

Marks obtained	Number of students $(f_i)$	Mid-points $(x_i)$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$10-20$	$2$	$15$	$30$	$30$	$60$
$20-30$	$3$	$25$	$75$	$20$	$60$
$30-40$	$8$	$35$	$280$	$10$	$80$
$40-50$	$14$	$45$	$630$	$0$	$0$
$50-60$	$8$	$55$	$440$	$10$	$80$
$60-70$	$3$	$65$	$195$	$20$	$60$
$70-80$	$2$	$75$	$150$	$30$	$60$
Total	$N=40$	-	$1800$	-	$400$

Here,$N = \sum f_i = 40$ and $\sum f_i x_i = 1800$.
The mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1800}{40} = 45$.
The mean deviation about the mean is given by $M.D.(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.
$M.D.(\bar{x}) = \frac{400}{40} = 10$.

(10.16) To calculate the mean deviation about the median,we first construct the frequency distribution table:

Class	Freq $(f_i)$	$c.f.$	Mid-point $(x_i)$	$|x_i - M|$	$f_i |x_i - M|$
$0-10$	$6$	$6$	$5$	$23$	$138$
$10-20$	$7$	$13$	$15$	$13$	$91$
$20-30$	$15$	$28$	$25$	$3$	$45$
$30-40$	$16$	$44$	$35$	$7$	$112$
$40-50$	$4$	$48$	$45$	$17$	$68$
$50-60$	$2$	$50$	$55$	$27$	$54$
Total	$N=50$	-	-	-	$508$

Here,$N = 50$,so $\frac{N}{2} = 25$. The cumulative frequency just greater than $25$ is $28$,which corresponds to the class $20-30$.
Thus,the median class is $20-30$.
Using the formula for median: $M = l + \frac{\frac{N}{2} - C}{f} \times h$
Where $l = 20, C = 13, f = 15, h = 10$.
$M = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{120}{15} = 20 + 8 = 28$.
Mean deviation about median $= \frac{1}{N} \sum f_i |x_i - M| = \frac{508}{50} = 10.16$.

(B) The given data is $4, 7, 8, 9, 10, 12, 13, 17$.
The mean of the data,$\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$.
The deviations of the respective observations from the mean $\bar{x}$,i.e.,$|x_{i} - \bar{x}|$,are:
$|4-10| = 6$
$|7-10| = 3$
$|8-10| = 2$
$|9-10| = 1$
$|10-10| = 0$
$|12-10| = 2$
$|13-10| = 3$
$|17-10| = 7$
The mean deviation about the mean is given by:
$M.D.(\bar{x}) = \frac{\sum_{i=1}^{8} |x_{i} - \bar{x}|}{8}$
$M.D.(\bar{x}) = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8} = \frac{24}{8} = 3$.

(A) The given data is $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} = 50$.
Next,find the absolute deviations $|x_i - \bar{x}|$ for each observation:
$|38-50| = 12$
$|70-50| = 20$
$|48-50| = 2$
$|40-50| = 10$
$|42-50| = 8$
$|55-50| = 5$
$|63-50| = 13$
$|46-50| = 4$
$|54-50| = 4$
$|44-50| = 6$
The sum of absolute deviations is $12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84$.
The mean deviation about the mean is $\frac{\sum |x_i - \bar{x}|}{n} = \frac{84}{10} = 8.4$.

(A) The given data is $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$.
Here,the number of observations is $n = 12$,which is even.
Arranging the data in ascending order,we obtain:
$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$.
Median $M = \frac{(\frac{n}{2})^{th} \text{ observation} + (\frac{n}{2} + 1)^{th} \text{ observation}}{2} = \frac{6^{th} \text{ observation} + 7^{th} \text{ observation}}{2} = \frac{13 + 14}{2} = 13.5$.
The absolute deviations $|x_i - M|$ are:
$|10 - 13.5| = 3.5$
$|11 - 13.5| = 2.5$
$|11 - 13.5| = 2.5$
$|12 - 13.5| = 1.5$
$|13 - 13.5| = 0.5$
$|13 - 13.5| = 0.5$
$|14 - 13.5| = 0.5$
$|16 - 13.5| = 2.5$
$|16 - 13.5| = 2.5$
$|17 - 13.5| = 3.5$
$|17 - 13.5| = 3.5$
$|18 - 13.5| = 4.5$
Sum of absolute deviations = $3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28$.
Mean deviation about median = $\frac{\sum |x_i - M|}{n} = \frac{28}{12} \approx 2.33$.

(A) The given data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.
Here,the number of observations is $n = 10$,which is even.
Arranging the data in ascending order,we obtain:
$36, 42, 45, 46, 46, 49, 51, 53, 60, 72$.
Median $M = \frac{(\frac{n}{2})^{\text{th}} \text{ observation} + (\frac{n}{2} + 1)^{\text{th}} \text{ observation}}{2}$.
$M = \frac{5^{\text{th}} \text{ observation} + 6^{\text{th}} \text{ observation}}{2} = \frac{46 + 49}{2} = \frac{95}{2} = 47.5$.
The absolute deviations $|x_i - M|$ are:
$|36 - 47.5| = 11.5$
$|42 - 47.5| = 5.5$
$|45 - 47.5| = 2.5$
$|46 - 47.5| = 1.5$
$|46 - 47.5| = 1.5$
$|49 - 47.5| = 1.5$
$|51 - 47.5| = 3.5$
$|53 - 47.5| = 5.5$
$|60 - 47.5| = 12.5$
$|72 - 47.5| = 24.5$
The mean deviation about the median is $\frac{\sum |x_i - M|}{n} = \frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10} = \frac{70}{10} = 7.0$.

First,calculate the mean $\bar{x}$:

$x_i$	$f_i$	$f_i x_i$
$5$	$7$	$35$
$10$	$4$	$40$
$15$	$6$	$90$
$20$	$3$	$60$
$25$	$5$	$125$
Total	$N = 25$	$\sum f_i x_i = 350$

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$
Now,calculate the mean deviation about the mean:

$x_i$	$f_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$5$	$7$	$|5-14| = 9$	$63$
$10$	$4$	$|10-14| = 4$	$16$
$15$	$6$	$|15-14| = 1$	$6$
$20$	$3$	$|20-14| = 6$	$18$
$25$	$5$	$|25-14| = 11$	$55$
Total	$25$	-	$158$

$M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$

(A) First,calculate the mean $\bar{x}$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(10 \times 4) + (30 \times 24) + (50 \times 28) + (70 \times 16) + (90 \times 8)}{4 + 24 + 28 + 16 + 8} = \frac{40 + 720 + 1400 + 1120 + 720}{80} = \frac{4000}{80} = 50$.
Now,calculate the mean deviation about the mean $M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$:

$x_i$	$f_i$	$|x_i - 50|$	$f_i |x_i - 50|$
$10$	$4$	$40$	$160$
$30$	$24$	$20$	$480$
$50$	$28$	$0$	$0$
$70$	$16$	$20$	$320$
$90$	$8$	$40$	$320$
Total	$80$	-	$1280$

$M.D.(\bar{x}) = \frac{1280}{80} = 16$.

(A) The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies $(c.f.)$ of the given data,we obtain the following table:

${x_i}$	${f_i}$	$c.f.$
$5$	$8$	$8$
$7$	$6$	$14$
$9$	$2$	$16$
$10$	$2$	$18$
$12$	$2$	$20$
$15$	$6$	$26$

Here,$N = \sum f_i = 26,$ which is even.
Median is the mean of the $13^{th}$ and $14^{th}$ observations. Both of these observations lie in the cumulative frequency $14,$ for which the corresponding observation is $7.$
$\therefore \text{Median} (M) = \frac{7 + 7}{2} = 7.$
The absolute values of the deviations from median,i.e.,$|x_i - M|,$ are calculated as follows:

$x_i$	$|x_i - M|$	$f_i$	$f_i|x_i - M|$
$5$	$2$	$8$	$16$
$7$	$0$	$6$	$0$
$9$	$2$	$2$	$4$
$10$	$3$	$2$	$6$
$12$	$5$	$2$	$10$
$15$	$8$	$6$	$48$

$\sum f_i |x_i - M| = 16 + 0 + 4 + 6 + 10 + 48 = 84.$
$\text{Mean Deviation} (M) = \frac{1}{N} \sum f_i |x_i - M| = \frac{84}{26} \approx 3.23.$

(A) The given observations are already in ascending order.
Adding a column for cumulative frequencies $(c.f.)$:

$x_i$	$f_i$	$c.f.$
$15$	$3$	$3$
$21$	$5$	$8$
$27$	$6$	$14$
$30$	$7$	$21$
$35$	$8$	$29$

Here,$N = \sum f_i = 29$,which is odd.
Median $= \left(\frac{N+1}{2}\right)^{th}$ observation $= \left(\frac{29+1}{2}\right)^{th} = 15^{th}$ observation.
This observation lies in the cumulative frequency $21$,for which the corresponding value is $x_i = 30$.
So,Median $(M) = 30$.
Calculating $|x_i - M|$ and $f_i|x_i - M|$:

$x_i$	$f_i$	$|x_i - 30|$	$f_i|x_i - 30|$
$15$	$3$	$15$	$45$
$21$	$5$	$9$	$45$
$27$	$6$	$3$	$18$
$30$	$7$	$0$	$0$
$35$	$8$	$5$	$40$

Sum $\sum f_i|x_i - M| = 45 + 45 + 18 + 0 + 40 = 148$.
Mean Deviation about Median $= \frac{1}{N} \sum f_i|x_i - M| = \frac{148}{29} \approx 5.10$.

(A) The mean deviation about the mean is calculated as follows:

Income	$f_i$	$x_i$	$f_ix_i$	$|x_i - \bar{x}|$	$f_i|x_i - \bar{x}|$
$0-100$	$4$	$50$	$200$	$308$	$1232$
$100-200$	$8$	$150$	$1200$	$208$	$1664$
$200-300$	$9$	$250$	$2250$	$108$	$972$
$300-400$	$10$	$350$	$3500$	$8$	$80$
$400-500$	$7$	$450$	$3150$	$92$	$644$
$500-600$	$5$	$550$	$2750$	$192$	$960$
$600-700$	$4$	$650$	$2600$	$292$	$1168$
$700-800$	$3$	$750$	$2250$	$392$	$1176$
Total	$50$	-	$17900$	-	$7896$

$N = \sum f_i = 50$
Mean $\bar{x} = \frac{\sum f_ix_i}{N} = \frac{17900}{50} = 358$
Mean Deviation about mean $= \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{7896}{50} = 157.92$

(A) The mean deviation about the mean is calculated as follows:

Height $(cm)$	$f_i$	$x_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$95-105$	$9$	$100$	$900$	$25.3$	$227.7$
$105-115$	$13$	$110$	$1430$	$15.3$	$198.9$
$115-125$	$26$	$120$	$3120$	$5.3$	$137.8$
$125-135$	$30$	$130$	$3900$	$4.7$	$141.0$
$135-145$	$12$	$140$	$1680$	$14.7$	$176.4$
$145-155$	$10$	$150$	$1500$	$24.7$	$247.0$
Total	$N=100$	-	$12530$	-	$1128.8$

$\text{Mean } \bar{x} = \frac{\sum f_i x_i}{N} = \frac{12530}{100} = 125.3$
$\text{Mean Deviation } (M.D.) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{1128.8}{100} = 11.28$

(N/A) To find the mean deviation about the median,we first calculate the median.

Marks	$f_{i}$	$C_{f}$	$x_{i}$	$|x_{i}-M|$	$f_{i}|x_{i}-M|$
$0-10$	$6$	$6$	$5$	$22.86$	$137.16$
$10-20$	$8$	$14$	$15$	$12.86$	$102.88$
$20-30$	$14$	$28$	$25$	$2.86$	$40.04$
$30-40$	$16$	$44$	$35$	$7.14$	$114.24$
$40-50$	$4$	$48$	$45$	$17.14$	$68.56$
$50-60$	$2$	$50$	$55$	$27.14$	$54.28$
Total	$N=50$	-	-	-	$517.16$

Here,$N=50$,so $\frac{N}{2} = 25$.
The cumulative frequency just greater than $25$ is $28$,so the median class is $20-30$.
Median $M = l + \frac{\frac{N}{2} - C_{f-1}}{f_{m}} \times h = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + 7.86 = 27.86$.
Mean deviation about median $= \frac{\sum f_{i}|x_{i}-M|}{N} = \frac{517.16}{50} = 10.3432 \approx 10.34$.

(A) The given data is not continuous. We convert it into a continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval.

Age	Frequency $(f_i)$	Cumulative Frequency $(c.f.)$	Mid-point $(x_i)$	$|x_i - \text{Median}|$	$f_i |x_i - \text{Median}|$
$15.5-20.5$	$5$	$5$	$18$	$20$	$100$
$20.5-25.5$	$6$	$11$	$23$	$15$	$90$
$25.5-30.5$	$12$	$23$	$28$	$10$	$120$
$30.5-35.5$	$14$	$37$	$33$	$5$	$70$
$35.5-40.5$	$26$	$63$	$38$	$0$	$0$
$40.5-45.5$	$12$	$75$	$43$	$5$	$60$
$45.5-50.5$	$16$	$91$	$48$	$10$	$160$
$50.5-55.5$	$9$	$100$	$53$	$15$	$135$
Total	$N=100$	-	-	-	$735$

Here,$N = 100$,so $\frac{N}{2} = 50$. The cumulative frequency just greater than $50$ is $63$,which corresponds to the class interval $35.5-40.5$.
Median $= l + \frac{\frac{N}{2} - C}{f} \times h = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{13 \times 5}{26} = 35.5 + 2.5 = 38$.
Mean Deviation about Median $= \frac{1}{N} \sum f_i |x_i - \text{Median}| = \frac{735}{100} = 7.35$.

First,calculate the mean $\bar{x}$:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{(20 \times 6) + (21 \times 4) + (22 \times 5) + (23 \times 1) + (24 \times 4)}{6 + 4 + 5 + 1 + 4} = \frac{120 + 84 + 110 + 23 + 96}{20} = \frac{433}{20} = 21.65$
Next,calculate the mean deviation about the mean:

$\text{Size } (x_i)$	$\text{Freq } (f_i)$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$20$	$6$	$1.65$	$9.90$
$21$	$4$	$0.65$	$2.60$
$22$	$5$	$0.35$	$1.75$
$23$	$1$	$1.35$	$1.35$
$24$	$4$	$2.35$	$9.40$
$\text{Total}$	$20$	-	$25.00$

$\text{Mean Deviation} = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i} = \frac{25}{20} = 1.25$

(C) First,we find the cumulative frequency $(cf)$ and the median $(M_e)$.
$ \begin{array}{|c|c|c|c|c|} \hline \text{Marks} (x_i) & f_i & cf & |x_i - M_e| & f_i |x_i - M_e| \\ \hline 10 & 2 & 2 & |10-12|=2 & 4 \\ \hline 11 & 3 & 5 & |11-12|=1 & 3 \\ \hline 12 & 8 & 13 & |12-12|=0 & 0 \\ \hline 14 & 3 & 16 & |14-12|=2 & 6 \\ \hline 15 & 4 & 20 & |15-12|=3 & 12 \\ \hline \text{Total} & \Sigma f_i = 20 & & & \Sigma f_i |x_i - M_e| = 25 \\ \hline \end{array} $
Total number of observations $N = \Sigma f_i = 20$.
Since $N$ is even,the median $M_e$ is the average of the $10^{th}$ and $11^{th}$ observations. Looking at the $cf$ column,both the $10^{th}$ and $11^{th}$ observations fall in the value $12$.
Therefore,$M_e = 12$.
The mean deviation about the median is given by:
$MD(M_e) = \frac{\Sigma f_i |x_i - M_e|}{\Sigma f_i} = \frac{25}{20} = 1.25$.