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Given a mean, median and total sum, how can you calculate how many elements in the collection will be more than a mean value? Here is an example; I have a stack of poles. The total hight of all the poles is 1472 in. The mean height is 64in and median 62 in. how many poles are 62in and shorter and how many are taller?
Based on the calculation, there are 23 poles with 64 in height. Since I know median is 62, the anything below 12th should be below 62 in. So is it safe to say, 12 elements must be smaller than 62 in and rest 11 must be greater than 62 inches?

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up vote 3 down vote accepted

If the median of 23 numbers is 62, then 11 of those numbers are less than or equal to 62 and 11 are greater than or equal to 62, and one of them, in the middle is exactly 62.

But if in addition to the above, one knows that the mean is 64, then we certainly know that at least one of the numbers is strictly more than 64. It could be only one. For example, if you had 22 of them equal to 62, then 23rd one would have to be 108 in order to make the mean 64. One the other hand, it could be that one of the numbers is 62 and 11 of them are more than 64. For example, if one of them is 62 and 11 of them are 67, and 11 of them are 61 + (2/11), then the mean is 64 and the median is 62. So the number of them that are more than the mean must be at least one and could be as high as 11.

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Check how "mean" and "median" are defined. The mean is just the sum divided by the number of values, "median" is the one such that half are larger and half less. From "mean" and "sum" you have the number of values, and you know half are larger than the median. But no way to relate to the mean.

Just play around with some values: What happens if they are all equal? Half of them is 0? Almost all are 0, one is large? If 10 are 0, one is 1, 10 are 50? If the middle one is 49 instead? If instead of 10 it is 100, or 1000?

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