This is the problem written out by the teacher...The following are ages of people at a park. The youngest is 3. The adult is 18. The mean is 9. The MAD is 6. What are the four ages? 3,_,_,18...When I calculate ages to get a mean of 9, the MAD is not 6. If I calculate ages to have a MAD of 6, the mean is not 9. I keep ending up with numbers that include decimals. The teacher has instructed that there is not to be any decimals. For instance, if I input the ages as 3,4,11,18 the mean is indeed 9. When I calculate the MAD by using the absolute deviation of each value |3-9|+|4-9|+|11-9|+|18-9| divided by 4 which translates to 6+5+2+9 divided by 4 which is then 22 divided by 4 equaling 5.5. It seems that no matter what values I choose, either the mean or the MAD is different than what the teacher has written. Any help is greatly appreciated.
Mean and Mean Absolute Deviation Problem
- Thread starter mtilman
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Are you just doing some guess-n-check, or are you applying any of the theorems, formulas, etc, that you've learned in class? If the former, try the latter. If the latter, please reply showing an example of your efforts.
Note: I am assuming that "MAD" means "mean absolute deviation" as defined here:
. . . . .\(\displaystyle MAD\, =\, \dfrac{1}{n}\, \)\(\displaystyle \displaystyle \sum_{i=1}^n\, \left|\, x_i\, -\, \bar{x}\, \right|\)
...and that you're using the usual definition for the mean \(\displaystyle \bar{x}\):
. . . . .\(\displaystyle \bar{x}\, =\, \dfrac{1}{n}\,\)\(\displaystyle \displaystyle \sum_{i=1}^n\, x_i\)
In this case, of course, n = 4, so the second sum is simply (1/4)(3 + 18 + a + b), if we rename the missing values for simplicity's sake. And the value of a + b is found easily from the given mean value.
