Introduction to free statistics course online:

Statistics is done normally with the data collected for specific purposes.  The method used in Statistics for finding a representative value of the given data is called as the measure of central tendency. The three measures of central tendency of statistics were Mean, also median and mode. Statistics are all largely used in banking sectors, educationist, industrialist, economist, agriculturalists etc. Free statistics course online is nothing but free course material available on internet for the sake of helping students. Statistics is one of the important topics in mathematics. Let us see free statistics course online.

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Free statistics course online:


Example 1:

Find the mean deviation of the median for the following data: 6,12,7,5,14,12,20,6,8,21,23.

Solution :

Here the total number of observations is 11 which is odd. Arranging the data into ascending order,

5 , 6 , 6 , 7 , 8 , 12 , 12 , 14 , 20 , 21 , 23

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi − M| are

6 , 5 , 5 , 4 , 3 , 0 , 0 , 3 , 9 , 10 , 12
Therefore

∑ 11i-1 |xi - M| = 57

M.D.(M) =(1/11) ∑ 11i-1 |xi - M| = (1/11)*57  =5.181.

Example 2:

Find the mean deviation of  the median for the following data: 11,6, 6, 3, 13, 11, 18, 4, 7, 18, 22.
Solution:

Here the total number of observations is 11 which is odd. Arranging the data into

ascending order, we have 3,4,6,6,7,11,11,13,18,18,22

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi − M| are

8 , 7 , 5 , 5 ,4, 0 , 0, 2 , 7 , 7 , 11
Therefore
`sum` 11i-1 | - M| = `56`

and  M.D.(M) = `1/11` `sum` 11i-1 |xi - M| =`56/11` =5.6.

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Free statistics course online-examples:


Example 3:

Find the mean deviation of the mean for given data:

6,5,15,9,13,2,7,15

Solution:

Step 1 Mean of the given data are   `barx`


`barx` = `(6+5+15+9+13+2+7+15)/8`   =`72/8` =8

Step 2 The deviations of the respective observations from the mean x, i.e., xi– x are
6– 8,5–8,15–8,9–8,13–8,2–8,7–8,15–8      ( or )   –2,–3,7,1,5,–6,–1,7

Step 3 The absolute values of the deviations, i.e.,|xi − x |are  2,3,7,1,5,6,1,7


Step 4 The required mean deviation about the mean is

M.D. ( `barx` ) = `sum` 8 i-1 |xi-x| / 8

= `(2+3+7+1+5+6+1+7)/8` = `32/8` = 4

Example 4:

Find the mean deviation of the mean for given data :

12, 5, 19, 18, 4, 11, 18, 21, 20, 8, 15, 18, 2, 4, 15, 11, 3, 1, 10, 5

Solution:

find the mean ( `barx` ) of the given data

`barx`   =1/20`sum` 20i-1  xi = `220/20` = 11


The respective absolute values of the deviations from mean, i.e.,|x- `barx` | are

1,6,8,7,7,0,7,10,9,3,4,7,9,3,4,0,8,10,1,6

Therefore
20i-1 |xi - `barx` | = 118

and M.D. ( `barx` ) =`118/20` = 5.9