Q: Find the mean deviation about the mean for the data:\$4,7,8,9,10,12,13,17\$

\$Mean(\overline{x})=\dfrac{4+7+8+9+10+12+13+17}{8}=\dfrac{80}{8}=10\$ \$x_i\$\$|x_i-\overline{x}|\$ \$4\$\$6\$ \$7\$\$3\$ \$8\$\$2\$ \$9\$\$1\$ \$10\$\$0\$ \$12\$\$2\$ \$13\$\$3\$ \$17\$\$7\$ \$\sum|x_i-\overline{x}|=24\$ \$\Rightarrow\$ Mean deviation about mean \$=\dfrac{24}{8}=3\$

Q: The scores of a batsman in ten innings are: \$38, 70, 48, 34, 42, 55, 63, 46, 54, 44\$. Find the mean deviation about median.

\$\begin{array}{l} 38,70,48,42,55,63,46,54,44 \\ Arranging\, :34,38,42,44,48,55,63,70 \\ Median=\frac { { \frac { N }{ 2 } ,\frac { n }{ 2 } +1 } }{ 2 } =\frac { { 5,6 } }{ 2 } = \\ =\frac { { 46+48 } }{ 2 } =47 \\ Mean\, \, deviation \\ =\left| { 34-47 } \right| +\left| { 38-47 } \right| +\left| { 42-47 } \right| +\left| { 44-47 } \right| \\ +\left| { 46-47 } \right| +\left| { 48-47 } \right| +\left| { 54-47 } \right| +\left| { 55-47 } \right| \\ \left| { 63-47 } \right| +\left| { 70-47 } \right| \\ \overline { 10 } \\ =13+9+5+3+1+1+7+8+16+23 \\ =\frac { { 86 } }{ { 10 } } =86 \end{array}\$

Q: Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

5, 8, 1, 15, 22\${\text{Moon = }}\dfrac{{{\text{sum of observation}}}}{{{\text{No}}{\text{. of obesuation}}}}\$\$ = \dfrac{{5 + 8 + 10 + 15 + 22}}{5}\$\$ = \dfrac{{60}}{5} = 12\$Sum of deviations\$ = |5 - 12| + |8 - 12| + |10 - 12| + |15 - 12| + |22 - 12|\$\$ = |7| + | - 4| + | - 2| + |3| + |10|\$\$ = 7 + 4 + 2 + 3 + 10\$\$ = 26\$

Q: Find the mean deviation about the median for the data \$36,72,46,42,60,45,53,46,51,49\$

Given =36,72,46,42,60,45,53,46,51,49 n = 10median \$ =\frac{(\frac{n}{2})^{th}obs+(\frac{n}{2}+1)^{th}}{2}obs\$\$\frac{5^{th}obs+6^{th}obs}{2}\$\$=\frac{46+49}{2}=\frac{95}{2}=47.5\$mean deviation above median =\$\frac{\epsilon \left | x:-m \right |}{10}\$= \$\frac{(136-47.51+142-47.51+145-47.51+146-47.51+146-47.51+149-47-51+151-47.51+153-47.51+160-47.51+172-47.51)}{10}\$= \$\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\$

Q: Find the mean deviation about median for the following data.\$18, 23, 9, 11, 26, 4, 14, 21\$

First arranging data in ascending order\$4, 9, 11, 14, 18, 21, 23, 26\$ No. of terms \$n= 8\$ ( even)Since \$n\$ is even, median \$\displaystyle=\left( \frac {n }{ 2 }\right )\$th term, \$=( \dfrac {8 }{ 2 } )\$th term, \$=4\$th term, \$=14\$Now, We calculate mean deviation about medianmedian \$(M)=14\$mean deviation about median \$=\displaystyle\frac{\sum |x_i - M|}{8}\$\$M.d.=\displaystyle\frac{(| 4-14 | + | 9-14 | + | 11-14 | + | 14-14 | + |18-14 | + | 21-14 | + | 23-14 | + |26-14|)}{8} \$ \$=\displaystyle\frac{(|-10| + |-5 | +|-3 |+|0|+|4| +|7|+|9|+|12|)}{8}\$ \$=\displaystyle\frac{10+5+3+4+7+9+12}{8}\$ \$=\displaystyle\frac{50}{8}\$ \$=6.25\$

Q: The mean deviation from the mean for the set of observations \$-1,0,4\$ is

We have \$\overset { \_ }{ x } =\dfrac { -1+0+4 }{ 3 } =1\$ \${ x }_{ i }\$ \$\left| { x }_{ i }-\overset { \_ }{ x } \right| \$ \$-1\$ \$|-1-1|=2\$ \$0\$ \$|0-1|=1\$ \$4\$ \$|4-1|=3\$Mean deviation from mean \$=\dfrac { \sum { \left| { x }_{ i }-\overset { \_ }{ x } \right| } }{ n } =\dfrac { 2+1+3 }{ 3 } =2\$So, option A is correct.

Question 1

The sum of the deviations of the variates  from the arithmetic mean is always.

Accepted Answer

The sum of the deviations of a given set of observations from their arithmetic mean is always zero. It is due to the property that the arithmetic mean is characterised as the centre of gravity. i.e. sum of positive deviation from the mean is equal to the sum of negative deviations.For example:$3,4,6,8,14$$\overline{x}=\dfrac{3+4+6+8+14}{5}=7$$x_{i}$       $x_{i}-\overline{x}$$3$          $-4$$4$          $-3$$6$          $-1$$8$             $1$$14$           $7$ $\sum (x_{i} - \overline{x})=-8+8=0$Hence, the sum of the deviations about mean is $0$.

Question 2

Find the mean deviation about the mean for the data:$4,7,8,9,10,12,13,17$

Accepted Answer

$Mean(\overline{x})=\dfrac{4+7+8+9+10+12+13+17}{8}=\dfrac{80}{8}=10$ $x_i$$|x_i-\overline{x}|$  $4$$6$  $7$$3$  $8$$2$  $9$$1$  $10$$0$  $12$$2$  $13$$3$  $17$$7$  $\sum|x_i-\overline{x}|=24$ $\Rightarrow$  Mean deviation about mean $=\dfrac{24}{8}=3$

Question 3

Find the mean deviation about the median for the data $13,17,16,14,11,13,10,16,11,18,12,17$

Accepted Answer

Given: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 19n = 12median = $\frac{6^{th}obs+7^{th}obs}{2}$$=\frac{13+14}{2}=\frac{27}{2}=13.5$mean deviation about median =$\frac{(110-13.51+111.13.51+111-13.51+112-13.51+113-13.51+113-13.51+114-13.51+116-13.5+116-13.51+117-13.51+117-13.51+118-13.51)}{12}$=$\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}$$=\frac{28}{12}=2.23$

Question 4

The scores of a batsman in ten innings are: $38, 70, 48, 34, 42, 55, 63, 46, 54, 44$. Find the mean deviation about median.

Accepted Answer

$\begin{array}{l} 38,70,48,42,55,63,46,54,44 \ Arranging\, :34,38,42,44,48,55,63,70 \ Median=\frac { { \frac { N }{ 2 } ,\frac { n }{ 2 } +1 } }{ 2 } =\frac { { 5,6 } }{ 2 } = \ =\frac { { 46+48 } }{ 2 } =47 \ Mean\, \, deviation \ =\left| { 34-47 } \right| +\left| { 38-47 } \right| +\left| { 42-47 } \right| +\left| { 44-47 } \right|  \ +\left| { 46-47 } \right| +\left| { 48-47 } \right| +\left| { 54-47 } \right| +\left| { 55-47 } \right|  \ \left| { 63-47 } \right| +\left| { 70-47 } \right|  \ \overline { 10 }  \ =13+9+5+3+1+1+7+8+16+23 \ =\frac { { 86 } }{ { 10 } } =86 \end{array}$

Question 5

Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

Accepted Answer

5, 8, 1, 15, 22${\text{Moon  =  }}\dfrac{{{\text{sum of observation}}}}{{{\text{No}}{\text{. of obesuation}}}}$$ = \dfrac{{5 + 8 + 10 + 15 + 22}}{5}$$ = \dfrac{{60}}{5} = 12$Sum&#10;of deviations$ = |5 - 12| + |8 - 12| + |10 - 12| + |15 - 12| + |22 - 12|$$ = |7| + | - 4| + | - 2| + |3| + |10|$$ = 7 + 4 + 2 + 3 + 10$$ = 26$

Question 6

Find the mean deviation about the median for the data $36,72,46,42,60,45,53,46,51,49$

Accepted Answer

Given =36,72,46,42,60,45,53,46,51,49 n = 10median $ =\frac{(\frac{n}{2})^{th}obs+(\frac{n}{2}+1)^{th}}{2}obs$$\frac{5^{th}obs+6^{th}obs}{2}$$=\frac{46+49}{2}=\frac{95}{2}=47.5$mean deviation above median =$\frac{\epsilon \left | x:-m \right |}{10}$= $\frac{(136-47.51+142-47.51+145-47.51+146-47.51+146-47.51+149-47-51+151-47.51+153-47.51+160-47.51+172-47.51)}{10}$= $\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$= $\frac{70}{10}=7$

Question 7

Find the mean deviation about median for the following data.$18, 23, 9, 11, 26, 4, 14, 21$

Accepted Answer

First arranging data in ascending order$4, 9, 11, 14, 18, 21, 23, 26$ No. of terms $n= 8$ ( even)Since $n$ is even, median $\displaystyle=\left( \frac {n  }{ 2 }\right )$th term,            $=( \dfrac {8  }{ 2 } )$th term,            $=4$th term,            $=14$Now, We  calculate mean deviation about medianmedian $(M)=14$mean deviation about median $=\displaystyle\frac{\sum |x_i - M|}{8}$$M.d.=\displaystyle\frac{(| 4-14 | + | 9-14 | + | 11-14 | + | 14-14 | + |18-14 | + | 21-14 | + | 23-14 | + |26-14|)}{8} $           $=\displaystyle\frac{(|-10| + |-5 | +|-3 |+|0|+|4| +|7|+|9|+|12|)}{8}$           $=\displaystyle\frac{10+5+3+4+7+9+12}{8}$           $=\displaystyle\frac{50}{8}$            $=6.25$

Question 8

The mean deviation about median from the data$340,150,210,240,300,310,320$ is

Accepted Answer

Arrange in increasing order150,210,240,300,310,320,340           N=number of data = 7median (M) =300The mean deviation about median $=  \frac{150+90+60+0+10+20+40}{7 }= 52.857$

Question 9

Mean deviation can be calculated from _______.

Accepted Answer

The mean deviation is calculated either from mean or median, but only median is&#10;preferred because when the signs are ignored, the sum of deviation of the sets&#10;taken from median is minimum.

Question 10

The mean deviation from the mean for the set of observations $-1,0,4$ is

Accepted Answer

We have $\overset { \_  }{ x } =\dfrac { -1+0+4 }{ 3 } =1$ ${ x }_{ i }$ $\left| { x }_{ i }-\overset { \_  }{ x }  \right| $ $-1$ $|-1-1|=2$ $0$ $|0-1|=1$ $4$ $|4-1|=3$Mean deviation from mean $=\dfrac { \sum { \left| { x }_{ i }-\overset { \_  }{ x }  \right|  }  }{ n } =\dfrac { 2+1+3 }{ 3 } =2$So, option A is correct.

Revise with Concepts

Mean Deviation

The sum of the deviations of the variates from the arithmetic mean is always.

Find the mean deviation about the mean for the data:
$4, 7, 8, 9, 10, 12, 13, 17$

Find the mean deviation about the median for the data
$13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

The scores of a batsman in ten innings are: $38, 70, 48, 34, 42, 55, 63, 46, 54, 44$ . Find the mean deviation about median.

Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

Find the mean deviation about the median for the data
$36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

Find the mean deviation about median for the following data.
$18, 23, 9, 11, 26, 4, 14, 21$

The mean deviation about median from the data
$340, 150, 210, 240, 300, 310, 320$ is

Mean deviation can be calculated from _______.

The mean deviation from the mean for the set of observations $- 1, 0, 4$ is

Revise with Concepts

Mean Deviation

The sum of the deviations of the variates from the arithmetic mean is always.

Find the mean deviation about the mean for the data:4,7,8,9,10,12,13,17

Find the mean deviation about the median for the data 13,17,16,14,11,13,10,16,11,18,12,17

The scores of a batsman in ten innings are: 38,70,48,34,42,55,63,46,54,44. Find the mean deviation about median.

Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

Find the mean deviation about the median for the data 36,72,46,42,60,45,53,46,51,49

Find the mean deviation about median for the following data. 18,23,9,11,26,4,14,21

The mean deviation about median from the data 340,150,210,240,300,310,320 is

Mean deviation can be calculated from _______.

The mean deviation from the mean for the set of observations −1,0,4 is

Find the mean deviation about the mean for the data:
$4, 7, 8, 9, 10, 12, 13, 17$

Find the mean deviation about the median for the data
$13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

The scores of a batsman in ten innings are: $38, 70, 48, 34, 42, 55, 63, 46, 54, 44$ . Find the mean deviation about median.

Find the mean deviation about the median for the data
$36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

Find the mean deviation about median for the following data.
$18, 23, 9, 11, 26, 4, 14, 21$

The mean deviation about median from the data
$340, 150, 210, 240, 300, 310, 320$ is

The mean deviation from the mean for the set of observations $- 1, 0, 4$ is