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Mean Deviation Or Standard Mean Deviation

To understand Mean Deviation we need to first learn about absolute differences. Absolute differences are differences expressed without plus or minus. Example: mod (3-5) = 2 (absolute difference)

Mean Deviation or Average Deviation is the average of absolute differences between each value in a given set of values, and the average of all the values of the set.

Example: Find the mean deviation of the set 1,2,3,4,5

Step1: the average or the mean of the set of values 1, 2, 3, 4, 5 is 15/5 which is 3

Step2: the difference between this average (3) and each value in the set is 2,1,0,-1,-2

Step3: the absolute difference will be 2,1,0,1 and 2.

Step4: the average of these numbers is (2+1+0+1+2)/5, which is 1.2, this value is the required Mean deviation.

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Mean Deviation Definition

In statistical distribution, Mean Deviation is defined as the mean of the absolute values of the numerical differences between the values of the set and their mean.

Mean Deviation Formula

Mean Deviation is the mean of the absolute deviations of observations from some suitable average which may be the arithmetic mean[x(bar)]

We can deduce the formula of Mean Deviation as follows:

The difference (x- x (bar)) is called the deviation and when we take the absolute value, this deviation is mod[x-x(bar)].

The mean of these mods or absolute deviations is called the mean deviation or the mean absolute deviation

The Mean Deviation Formula is,

Dx(bar)= {mod[x1- x (bar)]+ mod[x2-x(bar)]+mod[xn- x(bar)]}/n

MD = Dx(bar) = sigma(modulus [x - x(bar)])/n

[x is each value in the given set, x(bar) =mean of the values, n= number of values]

Mean Deviation Examples

Example: Calculate mean deviation of the following distribution,

6,9,8,5,8,9,6,13

Solution: Mean Deviation= Dx(bar)= {mod[x1- x (bar)]+ mod[x2-x(bar)]+mod[xn- x(bar)]}/n

x(bar) = mean = (6+9+8+5+8+9+6+13)/8= 8

Dx(bar)=[mod(8-6)+mod(8-9)+mod(8-8)+mod(8-5)+mod(8-8)+mod(8-9)+mod(8-6)+mod(8-13)]/8

= [2+1+0+3+0+1+2+5]/8= 14/8 = 1.75

Example: A student took 5 class tests in Mathematics and scored 92, 78, 86, 90, 96. Find the mean deviation for the test scores

Solution: Given scores, 94, 78, 86, 90, 96

Let us find the mean of the scores = x(bar) = sigma (x)/n=(94+78+86+90+92)/5 =88

Now the following steps will help to find the mean deviation

Subtract mean score from each score and take the absolute value of each difference.

mod[x-x(bar)] = mod(94-88) = 6

mod(78-88) =10

mod(86-88) =2

mod(90-88) = 2

mod(96-88) =6

The next step would be to add up the absolute values

Sum of the absolute values = 6+10+2+2+6 = 26

and then divide the sum by the number of scores.

MD = sigma[mod(x-x(bar))]/n

= 26/5

= 5.2

by: nitinp

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