Mean and Mean Absolute Deviation

[nsmith96]

i need help constructing a data set of 8 items that has a mean of 5 and a mean absolute deviation of 4

 

[mmm4444bot]

You posted nothing about what you already understand or what you've already tried, so I don't know how to help you.

I can make some observations; they might spur your thought process to forge a way to either a solution or a specific question.

S1 = {5, 5, 5, 5, 5, 5, 5, 5} yields mean: 5 and average absolute deviation: 0

If you don't understand this, stop now.

S2 = {1, 5, 5, 5, 5, 5, 5, 9} yields mean: 5 and average absolute deviation: 1

What I did here is this:

5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 40

To maintain the mean of 5, the data set must always sum to 40.

Change 5 + 5 into 1 + 9. That won't alter the sum.

Replacing two elements of 5 with elements 1 and 9 gives set S2: {1, 5, 5, 5, 5, 5, 5, 9}.

The mean is still 5, and now two elements of S2 deviate from the mean by 4 each.

(4 + 0 + 0 + 0 + 0 + 0 + 0 + 4)/8 = 8/8 = 1

In other words, there is 2*4 in absolute deviation, and 2*4/8 is the average absolute deviation: 1.

Where might this train-of-thought go next ?

I mean, what might happen, if we replaced more pairs of element 5 with 1s and 9s ?