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## Main Question or Discussion Point

What are the advantages of using the absolute mean deviation over the standard deviation. Is it possible to show a simple example where the former is more (or less) appropriate?

Also, related to the mean deviation is my own variation. Does it have a name? Instead of calculating the absolute differences from the mean for each number, my technique would instead find the average of all the absolute differences for each number against each other number.

So for example: given the numbers 3,7,7,19

Average is: 9

Absolute Mean deviation is: 5

My 'special' deviation is: 6

This is found thusly:

(|3-3| + |7-3| + |7-3| + |19-3| +

|3-7| + |7-7| + |7-7| + |19-7| +

|3-7| + |7-7| + |7-7| + |19-7| +

|3-19| + |7-19| + |7-19| + |19-19| ) / 16

= 6

As you can see, everything is compared against everything else. What do people here think? One could also remove the 3-3, 7-7, 7-7, and 19-19 bits, and then divide by 12 for a similar variation (results in 8 by the way).

Could this method be usefully applied in stats?

Also, related to the mean deviation is my own variation. Does it have a name? Instead of calculating the absolute differences from the mean for each number, my technique would instead find the average of all the absolute differences for each number against each other number.

So for example: given the numbers 3,7,7,19

Average is: 9

Absolute Mean deviation is: 5

My 'special' deviation is: 6

This is found thusly:

(|3-3| + |7-3| + |7-3| + |19-3| +

|3-7| + |7-7| + |7-7| + |19-7| +

|3-7| + |7-7| + |7-7| + |19-7| +

|3-19| + |7-19| + |7-19| + |19-19| ) / 16

= 6

As you can see, everything is compared against everything else. What do people here think? One could also remove the 3-3, 7-7, 7-7, and 19-19 bits, and then divide by 12 for a similar variation (results in 8 by the way).

Could this method be usefully applied in stats?