Mean Deviation Questions in English

(D) We have the following data.
First,calculate the mean $\bar{x}$:
$\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{(2 \times 2) + (4 \times 4) + (5 \times 10) + (7 \times 8) + (9 \times 6)}{2 + 4 + 10 + 8 + 6}$
$\bar{x} = \frac{4 + 16 + 50 + 56 + 54}{30} = \frac{180}{30} = 6$
Now,calculate the mean deviation about the mean using the formula $\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$:
$\text{M.D.}(\bar{x}) = \frac{2|2-6| + 4|4-6| + 10|5-6| + 8|7-6| + 6|9-6|}{30}$
$\text{M.D.}(\bar{x}) = \frac{2(4) + 4(2) + 10(1) + 8(1) + 6(3)}{30}$
$\text{M.D.}(\bar{x}) = \frac{8 + 8 + 10 + 8 + 18}{30} = \frac{52}{30} = 1.733$

(D) First,we calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(2 \times 1) + (4 \times 2) + (6 \times 3) + (8 \times 2) + (10 \times 1)}{1 + 2 + 3 + 2 + 1} = \frac{2 + 8 + 18 + 16 + 10}{9} = \frac{54}{9} = 6$.
Now,calculate the mean deviation about the mean $(MD_{\bar{x}})$:
$MD_{\bar{x}} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{1|2-6| + 2|4-6| + 3|6-6| + 2|8-6| + 1|10-6|}{9} = \frac{4 + 4 + 0 + 4 + 4}{9} = \frac{16}{9}$.
Next,calculate the median $(M)$:
Total frequency $N = 9$. The median is the value corresponding to the $\frac{N+1}{2}$-th observation,which is the $5$-th observation. Looking at the cumulative frequencies $(1, 3, 6, 8, 9)$,the $5$-th observation falls in the group where $x_i = 6$. Thus,$M = 6$.
Calculate the mean deviation about the median $(MD_M)$:
$MD_M = \frac{\sum f_i |x_i - M|}{\sum f_i} = \frac{1|2-6| + 2|4-6| + 3|6-6| + 2|8-6| + 1|10-6|}{9} = \frac{16}{9}$.
The sum of the mean deviation from the mean and the mean deviation from the median is:
$\frac{16}{9} + \frac{16}{9} = \frac{32}{9}$.

(A) Given,the mean $(\bar{x}) = 10$ and the number of observations $n = 8$.
The sum of the observations is $6 + 7 + 11 + 12 + 13 + \alpha + 12 + 16 = 77 + \alpha$.
Since $\bar{x} = \frac{\sum x_i}{n}$,we have $10 = \frac{77 + \alpha}{8}$.
$80 = 77 + \alpha \Rightarrow \alpha = 3$.
The data set is $6, 7, 11, 12, 13, 3, 12, 16$.
The mean deviation about the mean is $\frac{1}{n} \sum |x_i - \bar{x}|$.
$\text{MD}(\bar{x}) = \frac{|6-10| + |7-10| + |11-10| + |12-10| + |13-10| + |3-10| + |12-10| + |16-10|}{8}$.
$\text{MD}(\bar{x}) = \frac{|-4| + |-3| + |1| + |2| + |3| + |-7| + |2| + |6|}{8}$.
$\text{MD}(\bar{x}) = \frac{4 + 3 + 1 + 2 + 3 + 7 + 2 + 6}{8} = \frac{28}{8} = 3.5$.

(D) First,we calculate the class marks $(x_i)$ and the mean $(\bar{x})$:

Class interval	$x_i$	$f_i$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$0-2$	$1$	$1$	$1$	$4$	$4$
$2-4$	$3$	$3$	$9$	$2$	$6$
$4-6$	$5$	$5$	$25$	$0$	$0$
$6-8$	$7$	$3$	$21$	$2$	$6$
$8-10$	$9$	$1$	$9$	$4$	$4$
Total		$13$	$65$		$20$

The mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{65}{13} = 5$.
The mean deviation about the mean is $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{20}{13}$.

(D) The given data is $44, 5, 27, 20, 8, 54, 9, 14, 35$. The number of observations $N = 9$.
First,calculate the mean $\bar{x} = \frac{44+5+27+20+8+54+9+14+35}{9} = \frac{216}{9} = 24$.
The mean deviation about the mean $M_1 = \frac{1}{N} \sum |x_i - \bar{x}|$.
$|44-24| + |5-24| + |27-24| + |20-24| + |8-24| + |54-24| + |9-24| + |14-24| + |35-24| = 20 + 19 + 3 + 4 + 16 + 30 + 15 + 10 + 11 = 128$.
So,$M_1 = \frac{128}{9}$.
Now,arrange the data in ascending order: $5, 8, 9, 14, 20, 27, 35, 44, 54$.
The median $M$ is the $\left(\frac{9+1}{2}\right)^{\text{th}} = 5^{\text{th}}$ term,which is $20$.
The mean deviation about the median $M_2 = \frac{1}{N} \sum |x_i - M|$.
$|5-20| + |8-20| + |9-20| + |14-20| + |20-20| + |27-20| + |35-20| + |44-20| + |54-20| = 15 + 12 + 11 + 6 + 0 + 7 + 15 + 24 + 34 = 124$.
So,$M_2 = \frac{124}{9}$.
Therefore,$M_1 - M_2 = \frac{128}{9} - \frac{124}{9} = \frac{4}{9}$.

(D) The mean of the absolute deviations of observations $x_1, x_2, \ldots, x_n$ from their mean $\bar{x}$ is defined as the mean deviation about the mean.
Mathematically,it is given by:
$\text{Mean Deviation} = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$
where $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$.

(D) The given observations are $1, 3, 5, 7, 11, 13, 17, 19, 23$.
First,we calculate the mean $(\bar{x})$:
$\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \frac{99}{9} = 11$.
The mean deviation about the mean is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
Mean deviation $= \frac{1}{9} [|1-11| + |3-11| + |5-11| + |7-11| + |11-11| + |13-11| + |17-11| + |19-11| + |23-11|]$.
Mean deviation $= \frac{1}{9} [10 + 8 + 6 + 4 + 0 + 2 + 6 + 8 + 12]$.
Mean deviation $= \frac{56}{9} = 6 \frac{2}{9}$.

(B) Given observations are $2, 7, 5, 6, 4, 3, 11, 17, 8$.
First,calculate the arithmetic mean $(\bar{x})$:
$\bar{x} = \frac{2 + 7 + 5 + 6 + 4 + 3 + 11 + 17 + 8}{9} = \frac{63}{9} = 7$.
Now,calculate the absolute deviations $d_i = |x_i - \bar{x}|$ for each observation:
$|2 - 7| = 5$
$|7 - 7| = 0$
$|5 - 7| = 2$
$|6 - 7| = 1$
$|4 - 7| = 3$
$|3 - 7| = 4$
$|11 - 7| = 4$
$|17 - 7| = 10$
$|8 - 7| = 1$
Sum of deviations $\Sigma d_i = 5 + 0 + 2 + 1 + 3 + 4 + 4 + 10 + 1 = 30$.
Mean deviation ($M$.$D$.) = $\frac{\Sigma d_i}{N} = \frac{30}{9} = \frac{10}{3}$.

(B) First,we calculate the cumulative frequency $(cf)$ and the total frequency $(N)$:

$x_i$	$f_i$	$cf$
$6$	$4$	$4$
$12$	$7$	$11$
$18$	$9$	$20$
$24$	$18$	$38$
$30$	$15$	$53$
$36$	$10$	$63$
$42$	$5$	$68$

Here,$N = 68$,which is even. The median is the average of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observations,i.e.,the $34^{th}$ and $35^{th}$ observations.
Looking at the cumulative frequency,both the $34^{th}$ and $35^{th}$ observations fall in the class where $x_i = 24$.
Therefore,$\text{Median} = 24$.
Now,we calculate the mean deviation about the median using the formula $\text{MD}(\text{Median}) = \frac{\sum f_i |x_i - \text{Median}|}{N}$:
$\sum f_i |x_i - 24| = 4|6-24| + 7|12-24| + 9|18-24| + 18|24-24| + 15|30-24| + 10|36-24| + 5|42-24|$
$= 4(18) + 7(12) + 9(6) + 18(0) + 15(6) + 10(12) + 5(18)$
$= 72 + 84 + 54 + 0 + 90 + 120 + 90 = 510$
$\text{MD}(\text{Median}) = \frac{510}{68} = 7.5$.

(C) The given numbers are $3x, 6x, 9x, \ldots, 81x$. This is an arithmetic progression with $n = 27$ terms.
Since $n = 27$ is odd,the median is the $\frac{n+1}{2}$-th term,which is the $14$-th term.
The $14$-th term is $3x \times 14 = 42x$.
The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \text{Median}|$.
Here,$\text{MD} = \frac{1}{27} \sum_{k=1}^{27} |3kx - 42x| = \frac{3|x|}{27} \sum_{k=1}^{27} |k - 14| = \frac{|x|}{9} [\sum_{k=1}^{13} (14-k) + \sum_{k=15}^{27} (k-14)]$.
Calculating the sums: $\sum_{k=1}^{13} (14-k) = 13+12+\ldots+1 = \frac{13 \times 14}{2} = 91$.
Similarly,$\sum_{k=15}^{27} (k-14) = 1+2+\ldots+13 = 91$.
So,$\text{MD} = \frac{|x|}{9} (91 + 91) = \frac{|x|}{9} \times 182 = 91$.
Thus,$\frac{|x|}{9} \times 2 = 1 \implies |x| = \frac{9}{2}$.

(A) The frequency distribution is given by $f_i = 2i$ for $x_i = i$ where $i \in \{1, 2, 3, 4, 5, 6\}$.

$x_i$	$1$	$2$	$3$	$4$	$5$	$6$
$f_i$	$2$	$4$	$6$	$8$	$10$	$12$

Total frequency $N = \sum f_i = 2(1+2+3+4+5+6) = 2(21) = 42$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2(1^2+2^2+3^2+4^2+5^2+6^2)}{42} = \frac{2(91)}{42} = \frac{182}{42} = \frac{13}{3}$.
Mean deviation from mean $MD = \frac{\sum f_i |x_i - \bar{x}|}{N}$.
$MD = \frac{2|1-\frac{13}{3}| + 4|2-\frac{13}{3}| + 6|3-\frac{13}{3}| + 8|4-\frac{13}{3}| + 10|5-\frac{13}{3}| + 12|6-\frac{13}{3}|}{42}$.
$MD = \frac{2(\frac{10}{3}) + 4(\frac{7}{3}) + 6(\frac{4}{3}) + 8(\frac{1}{3}) + 10(\frac{2}{3}) + 12(\frac{5}{3})}{42}$.
$MD = \frac{20 + 28 + 24 + 8 + 20 + 60}{3 \times 42} = \frac{160}{126} = \frac{80}{63}$.

(A) Let the two missing observations be $a$ and $b$. The sum of the $10$ observations is $10 \times 9 = 90$.
Sum of given $8$ observations $= 2+3+5+10+11+13+15+21 = 80$.
So,$a+b = 90 - 80 = 10$.
Given variance $\sigma^2 = 34.2$. The formula for variance is $\frac{\Sigma x_i^2}{n} - (\bar{x})^2 = 34.2$.
$\frac{2^2+3^2+5^2+10^2+11^2+13^2+15^2+21^2+a^2+b^2}{10} - 9^2 = 34.2$.
$\frac{4+9+25+100+121+169+225+441+a^2+b^2}{10} - 81 = 34.2$.
$1094 + a^2 + b^2 = 1152 \Rightarrow a^2 + b^2 = 58$.
Since $a+b=10$ and $a^2+b^2=58$,we solve to get $a=3$ and $b=7$.
The $10$ observations are $2, 3, 3, 5, 7, 10, 11, 13, 15, 21$.
The median is the average of the $5^{th}$ and $6^{th}$ terms: $\frac{7+10}{2} = 8.5$.
Mean deviation about median $= \frac{\Sigma |x_i - 8.5|}{10} = \frac{|2-8.5| + |3-8.5| + |3-8.5| + |5-8.5| + |7-8.5| + |10-8.5| + |11-8.5| + |13-8.5| + |15-8.5| + |21-8.5|}{10}$.
$= \frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10} = \frac{50}{10} = 5$.

(B) The given numbers are $k, 2k, 3k, \dots, 1000k$. Here $n = 1000$ (even).
Median $X_M = \frac{(\frac{n}{2})k + (\frac{n}{2} + 1)k}{2} = \frac{500k + 501k}{2} = 500.5k$.
Mean deviation about median $= \frac{1}{n} \sum_{i=1}^{n} |x_i - X_M| = \frac{1}{1000} \sum_{i=1}^{1000} |ik - 500.5k| = \frac{k}{1000} \sum_{i=1}^{1000} |i - 500.5|$.
This sum is $2 \times (0.5 + 1.5 + 2.5 + \dots + 499.5) = 2 \times \frac{500}{2} (0.5 + 499.5) = 500 \times 500 = 250000$.
Mean deviation $= \frac{k \times 250000}{1000} = 250k$.
Given $250k = 500$,so $k = 2$.
Therefore,$k^{2} = 2^{2} = 4$.

(C) First,calculate the total frequency $N = \sum f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26$.
Next,calculate the sum of the products $\sum f_i x_i = (5 \times 8) + (7 \times 6) + (9 \times 2) + (10 \times 2) + (12 \times 2) + (15 \times 6) = 40 + 42 + 18 + 20 + 24 + 90 = 234$.
The mean $\bar{x}$ is given by $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{234}{26} = 9$.
The mean deviation about the mean is calculated as $MD = \frac{\sum f_i |x_i - \bar{x}|}{N}$.
$MD = \frac{8|5-9| + 6|7-9| + 2|9-9| + 2|10-9| + 2|12-9| + 6|15-9|}{26}$.
$MD = \frac{8(4) + 6(2) + 2(0) + 2(1) + 2(3) + 6(6)}{26} = \frac{32 + 12 + 0 + 2 + 6 + 36}{26} = \frac{88}{26} = \frac{44}{13}$.