Mean Deviation
Category : 10th Class
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
(b) 6.6
(c) 7.6
(d) 8.8
(e) None of these
Answer: (a)
Explanation
First we arrange the data in the increasing order as:
34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Now median of the data is given by: \[M=\frac{46+48}{2}=47\]
The deviation of the data is \[\left| {{x}_{i}}-47 \right|\]
|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
Therefore, mean deviation of the data is given by:
\[M.D.=\frac{13+9+5+3+1+1+7+8+16+23}{10}\]
\[\Rightarrow \,\,M.D.=\frac{86}{10}=8.6\]
The mean deviation about the mean of the following data is:
| \[{{x}_{i}}=\] | 3 | 9 | 17 | 23 | 27 |
| \[{{f}_{i}}=\] | 8 | 10 | 12 | 9 | 5 |
(a) 5.68
(b) 7.09
(c) 8.69
(d) 8.98
(e) None of these
Answer: (b)
Explanation
We have the data,
| \[{{x}_{i}}\] | \[{{f}_{i}}\] | \[{{f}_{i}}{{X}_{i}}\] | \[\left| \left. {{X}_{i}}-\overline{X} \right| \right.\] | \[{{f}_{i}}\left| \left. {{X}_{i}}-\overline{X} \right| \right.\] |
| 3 | 8 | 24 | 12 | 96 |
| 9 | 10 | 90 | 6 | 60 |
| 17 | 12 | 204 | 2 | 24 |
| 23 | 9 | 207 | 8 | 72 |
| 27 | 5 | 135 | 12 | 60 |
| N = 44 | 660 | 312 |
Now mean of the data is given by: \[\overline{X}=\frac{\sum{{{x}_{i}}{{f}_{i}}}}{N}=\frac{660}{44}=15\]
Thus the mean deviation of the data is given by:
\[M.D.=\frac{\sum{{{f}_{i}}\left| \left. {{x}_{i}}-15 \right| \right.}}{\sum{{{f}_{i}}}}=\frac{312}{44}=7.09\]
Calculate the mean deviation about the median of the following data:
| C.I. | 0-6 | 6-12 | 12-18 | 18-24 | 42-30 |
| Frequency | 8 | 10 | 12 | 9 | 5 |
(a) 5.689
(b) 7.297
(c) 6.318
(d) 8.982
(e) None of these
Answer: (c)
Calculate the mean deviation about the mean for the following data:
| Class: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Freq(f) | 2 | 3 | 8 | 14 | 8 |
(a) 8.98
(b) 9.96
(c) 10
(d) 9.39
(e) None of these
Answer: (d)
Variance and Standard Deviation
Since the mean deviations do not give accurate values and are not reliable, so we must have some other method of finding the measure of dispersion. Thus variance and standard deviation is a method to find the measure of dispersion. Let \[{{x}_{1}},{{x}_{2}},{{x}_{3}},----,{{x}_{n}}\] be n-observation and x be the mean of the data, then
\[\begin{align} & \Rightarrow \,\,\,{{\left( {{x}_{1}}-\overline{X} \right)}^{2}}+{{\left( {{x}_{2}}-\overline{X} \right)}^{2}}+{{\left( {{x}_{3}}-\overline{X} \right)}^{2}}+---- \\ & {{\left( {{x}_{n}}-\overline{X} \right)}^{2}}=\sum\limits_{k=1}^{n}{{{\left( {{x}_{k}}-\overline{X} \right)}^{2}}} \\ \end{align}\]
Thus the variance of the data is given by:
\[{{\sigma }^{2}}=\frac{1}{n}\sum\limits_{k=1}^{n}{{{\left( {{x}_{k}}-\overline{X} \right)}^{2}}}\]
The standard deviation of the data is given by:
\[\sigma =\sqrt{\frac{1}{n}\sum\limits_{k=1}^{n}{{{\left( {{x}_{k}}-\overline{X} \right)}^{2}}}}\]
Standard Deviation 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 variance and standard deviation of the data is given by \[{{\sigma }^{2}}=\frac{\sum\limits_{k=1}^{n}{{{f}_{k}}\left( {{x}_{k}}-\overline{X} \right)}}{\sum\limits_{k=1}^{n}{{{f}_{k}}}}\] and the standard deviation is given by:
\[{{\sigma }^{2}}\sqrt{\frac{\sum\limits_{k=1}^{n}{{{f}_{k}}}{{\left( {{x}_{k}}-\overline{X} \right)}^{2}}}{\sum\limits_{k=1}^{n}{{{f}_{k}}}}}\]
Standard Deviation of a Continuous Frequency Distribution
The another relation for continuous frequency distribution is given by
\[{{\sigma }^{2}}\frac{1}{n}\sqrt{N\sum\limits_{k=1}^{n}{{{f}_{k}}{{x}_{k}}^{2}-{{\left( \sum\limits_{k=1}^{n}{{{f}_{k}}{{x}_{k}}} \right)}^{2}}}}\]
Calculate the variance of the marks obtained by a group of 6 students in a class test in the college having maximum marks 25. The marks are 22 13, 12, 8, 15, 20.
(a) 15.68
(b) 21.2
(c) 22.6
(d) 18.93
(e) None of these
Answer: (c)
Explanation
We have, mean of the data:
\[\overline{X}=\frac{22+13+12+8+15+20}{6}=\frac{90}{6}=15\]
Here n = 6, therefore, variance of the data is given by
| Data | \[{{X}_{i}}\overline{X}\] | \[{{\left( {{X}_{i}}\overline{X} \right)}^{2}}\] |
| 22 | 7 | 49 |
| 13 | -2 | 4 |
| 12 | -3 | 9 |
| 8 | -7 | 49 |
| 15 | 0 | 0 |
| 20 | 5 | 25 |
Therefore, \[\sum {{\left( {{X}_{i}}-\overline{X} \right)}^{2}}=136\]
Hence variance of the data is\[\frac{136}{6}=22.\overline{6}\]
Calculate the standard deviation of the data obtained for the selling of different articles in a shop in a certain week of a month. The data is given as follows: 50, 60, 62, 60, 45, 48, 65, 68, 58, 44.
(a) 7.68
(b) 9.25
(c) 15.6
(d) 14.58
(e) None of these
Answer: (a)
Explanation
Here we have the mean of the data:
\[\begin{align} & \overline{X}=\frac{50+60+62+60+45+65+68+58+48}{9} \\ & =\frac{516}{9}=57.\overline{3} \\ \end{align}\]
Now we calculate the variance of the data as:
| Data | \[\left( {{X}_{i}}-\overline{X} \right)\] | \[{{\left( {{X}_{i}}-\overline{X} \right)}^{2}}\] |
| 50 | -7.3 | 53.29 |
| 60 | 2.7 | 7.29 |
| 62 | 4.7 | 22.09 |
| 60 | 2.7 | 7.29 |
| 45 | -12.3 | 151.29 |
| 65 | 7.7 | 59.29 |
| 58 | 0.7 | 0.49 |
| 68 | 10.7 | 144.49 |
| 48 | -9.3 | 86.49 |
Therefore, \[\sum {{\left( {{X}_{i}}-\overline{X} \right)}^{2}}=532.01\]
Hence variance of the data is given by:
\[{{\sigma }^{2}}=\frac{\Sigma {{\left( {{X}_{i}}-\overline{X} \right)}^{2}}}{n}\]
\[\Rightarrow {{\sigma }^{2}}=\frac{532.01}{9}=59.11\]
\[\Rightarrow \,\,\sigma =7.68\]
If the five out of the seven data is 14, 12, 10, 4, 2 has mean and variance as 8 and 16 respectively. Out of seven data, five data is given above. Hence find the remaining two data.
(a) (7 & 6)
(b) (9 & 10)
(c) (6 & 9)
(d) (8 & 6)
(e) None of these
Answer: (d)
Explanation
Let the two observation be 'm' and 'n'.
Then
\[\begin{align} & \overline{X}=\frac{14+12+10+4+2+m+n}{7} \\ & =\frac{42+m+n}{7}=8 \\ \end{align}\]
\[\Rightarrow \,\,m\,+n=14----(1)\]
Now variance is given by:
\[\begin{align} & {{\sigma }^{2}}=\frac{{{14}^{2}}+{{12}^{2}}+{{10}^{2}}+{{4}^{2}}+{{2}^{2}}+{{m}^{2}}+n}{7} \\ & -{{\left( \overline{X} \right)}^{2}}=16 \\ \end{align}\]
\[\Rightarrow \,\,{{m}^{2}}+{{n}^{2}}=100\] \[\Rightarrow \,\,m-n=-2-----(2)\]
Solving equation (1) and (2) we get,
m = 6 and n = 8
The variance of 15 observation is found to be 4. If each observation is increased by 9, then find the new variance of the observation.
(a) 4
(b) 9
(c) 13
(d) 6.5
(e) None of these
Answer: (a)
The incorrect mean and standard deviation of 20 data was found to be 10 and 2. If the incorrect observation was found to be 8, then find the correct standard deviation when this data is committed.
(a) 7.9
(b) 8
(c) 2
(d) 5
(e) None of these
Answer: (d)
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