If the data is in order and is represented by a1, a2, ..., an for dataset size n, where n is an odd number then a[(n+1)/2]=5. Also an=a1+10. It's also clear that a1<5 and an<15 since 5 is the median and the range is 10. If n is even then the median is the average of a[n/2] and a[(n+2)/2].
The mode is 10 and that implies at least two tens in the dataset.
Mean, median and mode are different versions of the average, but mean=40. This is not consistent with the requirements, particularly because the range is 10 and the lowest datum has to be less than 5, the median. All the data values "left" of the median must be less than the median and those to the "right" of the median must be greater, by definition of the median.
If we assume that the range at best is approximately 5 to 15, then the mean=40 lies outside the range which suggests an error in the question. The mean, or average, has to be within the range of the data.
- Let us suppose that 40 is the sum of the data rather than the mean, which is the sum of the data divided by the size of the dataset. If the least of the data is a1 then the greatest is a1+10. We know that a1<5 so a1+10<15 and ≥10 so a1≥0. There have to be at least 2 tens in the data because the mode is 10. The minimum size of the data is 7 consisting of a1, a2, a3, 5, 10, 10, a1+10. We assumed the sum was 40 so 2a1+a2+a3+35=40 making 2a1+a2+a3=5. If a1=0, then a2+a3=5 and we know a3>a2>a1, so a2 and a3 could be 1 and 4, 2 and 3, 1.5 and 3.5, etc. This gives us the data: 0, 1, 4, 5, 10, 10, 10 where the median is 5, the mode is 10, the range is 10 and the mean is 40/7. We could also have: 0, 2, 3, 5, 10, 10, 10, etc. If we put a1=0.5 then a2 and a3 could be 1 and 3, 1.5 and 2.5, etc., giving us, for example: 0.5, 1.5, 2.5, 5, 10, 10, 10.5. This also meets all requirements with a mean of 40/7.
- If we keep the mean at 40 then we need to adjust the range. For the sake of illustration let the range be 100, so the greatest data value is 100+a1. More to follow...