Mean Deviation Practice Problems

Mean Deviation is a measure of dispersion that indicates the average distance of the data values from a central value, usually the mean or the median.

Example 1: Find the mean deviation of the following data: 7, 5, 1, 3, 6, 4, 10.

Solution:

For Finding the mean deviation,

First, calculate the mean value of the given data

Mean = Sum of all the observations / Number of observations

= \frac{7 \ + \ 5 \ + \ 1 \ + \ 3 \ + \ 6 \ + \ 4 \ + \ 9}{7}

Mean = 5.14 ≅ 5

Now finding the absolute deviation.

Given Value	Absolute Deviation of Mean
7	|7 - 5| = |2| = 2
5	|5 - 5| = |0| = 0
1	|1 - 5| = |-4| = 4
3	|3 - 5| = |-2| = 2
6	|6 - 5| = |1| = 1
4	|4 - 5| = |-1| = 1
10	|10 - 5| = |5| = 5

Further the mean of Absolute Values

= 2, 0, 4, 2, 1, 1, 5

Mean = \frac{2 \ + \ 0 \ + \ 4 \ + \ 2 \ + \ 1 \ + \ 1 \ + \ 5}{7}

Mean Deviation = 2.14

Therefore,
Mean deviation for 7, 5, 1, 3, 6, 4, 10 is 2.14

Example 2: Find the mean deviation of the following data: 11, 9, 7, 3, 2, 8, 10, 12, 15, 13.

Solution:

For Finding the mean deviation,

First, calculate the mean value of the given data

Mean =  = \frac{11 \ + \ 9\ + \ 7 \ + \ 3 \ + \ 2 \ + \ 8 \ + \ 10 \ + \ 12 \ + \ 15 \ + \ 13}{10}

Mean = 9

Now finding the absolute deviation.

Given Value	Absolute Deviation of Mean
11	|11 - 9| = |2| = 2
9	|9 - 9| = |0| = 0
7	|7 - 9| = |-2| = 2
3	|3 - 9| = |-6| = 6
2	|2 - 9| = |-7| = 7
8	|8 - 9| = |-1| = 1
10	|10 - 9| = |1| = 1
12	|12 - 9| = |3| = 3
15	|15 - 9| = |6| = 6
13	|13 - 9| = |4| = 4

Further find the mean of these values obtained are 2, 0, 2, 6, 7, 1, 1, 3, 6, 4.  

Mean Deviation ⇒\frac{2 \ + \ 0 \ + \ 2 \ + \ 6 \ + \ 7 \ + \ 1 \ + \ 1 \ + \ 3 \ + \ 6 \ + \ 4 }{10}\\=\frac{32}{10}

⇒ Mean = 3.2

Therefore, Mean deviation for 11, 9, 7, 3, 2, 8, 10, 12, 15, 13 is 3.2.

Example 3: Find the mean deviation of the following data table:

Solution:

The mean of the following data is,

Class Interval	Frequency(fi)	Mid Point(xi)	fixi
5-15	8	10	80
15-25	12	20	240
25-35	6	30	180
35-45	4	40	160
	∑fi = 30		∑fixi = 660

x̄ = ∑f_ix_i / ∑f_i = 660/30 = 22

Class Interval	Frequency(fi)	Mid Point(xi)	|xi - x̄|
5-15	8	10	12
15-25	12	20	2
25-35	6	30	8
35-45	4	40	18
	∑fi = 30		∑|xi - x̄| = 40

Mean Deviation = 40/30 = 1.33