Mean Deviation
Mean Deviation about Mean of a Raw Data
Let \[{{x}_{1}},{{x}_{2}}{{x}_{3}},---,{{x}_{n}}\] be the n observation, then the mean of the data is given by:
\[\overline{X}=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+----+{{x}_{n}}}{n}\]
\[\Rightarrow \,\overline{X}\frac{1}{n}\sum\limits_{k=1}^{n}{{{X}_{k}}\Rightarrow \overline{X},=\frac{1}{n}}\sum\limits_{k=1}^{n}{{{X}_{k}}}\]
Then the deviation of the data from the mean is given by:
\[\left| \left. {{x}_{1}}-\overline{X} \right|,\left| {{x}_{2}},-\left. \overline{X} \right|,\left| \left. {{x}_{3}}-\overline{X} \right|,---,\left| \left. {{x}_{n}}-\overline{X} \right| \right. \right. \right. \right.\]
Now the mean deviation of the data is given by:
Mean Deviation \[=\,\Rightarrow M.D.=\frac{1}{n}\sum\limits_{k=1}^{n}{\left| \left. {{X}_{n}}-\overline{X} \right| \right.}\]
Mean Deviation about Mean of a Grouped Data Let
\[{{x}_{1}},{{x}_{2}},{{x}_{3}},---,{{x}_{n}}\] be the n - observation and \[{{f}_{1}},{{f}_{2}},{{f}_{3}},---{{f}_{n}}\] be the corresponding frequencies of the data. Then the mean of the data is given by:
\[\overline{X}=\frac{{{x}_{1}}{{f}_{1}}+{{x}_{2}}{{f}_{2}}+----+{{x}_{n}}{{f}_{n}}}{{{f}_{1}}+{{f}_{2}}++---+{{f}_{n}}}\,\] Or, \[\overline{X}=\frac{\sum\limits_{k=1}^{n}{{{X}_{k}}}{{f}_{k}}}{\sum\limits_{k=1}^{n}{{{f}_{k}}}}\]
Then the mean deviation is given by \[\Rightarrow \,M.D.=\frac{\sum\limits_{k=1}^{n}{{{f}_{k}}}\left| \left. {{X}_{k}}-\overline{X} \right| \right.}{\sum\limits_{k=1}^{n}{{{f}_{k}}}}\]
Mean Deviation about Median of a Ungrouped Data
The median of an ungrouped data is obtained by arranging the data in the ascending order. If the number of data is odd, then the median is obtained as \[\left( \frac{n+1}{2} \right)\] term of the data and if the number of data is even, then the median is obtained as:
\[\frac{{{\left( \frac{n}{2} \right)}^{th}}+{{\left( \frac{n}{2}+1 \right)}^{th}}}{2}\]
If M is the median of the data, then mean deviation is given by
\[\Rightarrow \,\,M.D.=\frac{1}{n}\sum\limits_{k=1}^{n}{\left| \left. {{X}_{n}}-M \right| \right.}\]
Mean Deviation about Median of a Grouped Data
Let \[{{x}_{1}},{{x}_{2}},{{x}_{3}},---,{{x}_{n}}\] be the n -observation and \[{{f}_{1}},{{f}_{2}},{{f}_{3}},---,{{f}_{n}}\] be the corresponding frequencies of the data. Then the mean deviation about the median of the data is given by \[\Rightarrow \,\,M.D.=\frac{\sum\limits_{k=1}^{n}{{{f}_{k}}\left| \left. {{X}_{k}}-M \right| \right.}}{\sum\limits_{k=1}^{n}{{{f}_{k}}}}\]
For the grouped data the median can be obtained by Median \[=\,I\,\,+\frac{\frac{N}{2}-C}{f}\times h\]
where,
I = lower limit of the median class
N = sum of all frequency y
C = cumulative frequency of preceding median class h = class width
Find the mean deviation of the data about the mean: 2, 4, 10, 12, 18, 16, 14, 20
(a) 2
(b) 4
(c) 5
(d) 8
(e) None of these
Answer: (c)
Explanation
The mean of the above data is given by:
\[\overline{X}=\frac{2+4+10+12+18+16+14+20}{8}\]
\[\overline{X}=12\]
Now the deviation about the mean is given by:
|2 - 12| = 10,|4 - 12| = 8,|10 ? 12| = 2,
|12 - 12| = 0,|18 - 12|= 6,|16 - 12|=4,
|14 - 12| = 2,|20 - 12| =8
Now mean deviation about the mean is given by:
\[M.D.=\frac{10+8+2+0+6+4+2+8}{8}\]
\[\Rightarrow \,\,M.D.=\frac{40}{8}=5\]
The score of 10 students of a class test is given as 44, 54, 46, 63, 55, 42, 34, 48, 70, 38. Calculate the mean deviation about the median.
(a) 8.6
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