[MotoSook]

OK. Don has it, but it is not the only solution.

Start with the definitions:

Mean: Average of a set of n numbers, obtained by adding the numbers and dividing by n. For example, the arithmetic mean of the set of 5 numbers 1, 3, 6, 8, and 12 is (1 + 3 + 6 + 8 + 12) ÷ 5 = 30 ÷ 5 = 6.

Median: In mathematics and statistics, the middle number of an ordered group of numbers. If there is no middle number (because there is an even number of terms), the median is the mean (average) of the two middle numbers. For example, the median of the group 2, 3, 7, 11, 12 is 7; that of 3, 4, 7, 9, 11, 13 is 8 (the mean of 7 and 9). The median together with the mode and arithmetic mean make up the average of a set of data. In addition it is useful to know the range or spread of the data.

Mode: In mathematics, the element that appears most frequently in a given set of data. For example, the mode for the data 0, 0, 9, 9, 9, 12, 87, 87 is 9.

Range: In statistics, a measure of dispersion in a frequency distribution, equalling the difference between the largest and smallest values of the variable. The range is sensitive to extreme values in the sense that it will give a distorted picture of the dispersion if one measurement is unusually large or small. The interquartile range is often preferred.

Here's how it works:

There are 8 numbers (terms), that's the easy part. We know the mean has to be 8, so the sum of the terms must be 64 (8X8=64 or 64/8=8).

We know the range has to also be 8, so the difference between the max and min has to equal 8, 12-4=8. That works.

Now the median, has to also equal 8, but because there are even number of terms, the average of the two middle numbers most be 8. We could us 7 and 9 [(7+9)/8=8], but then we could not establish a mode of 8! So the two middle numbers must be 8 and 8. Eight twice and most often of all the other terms allows the mode to equal 8, and also satisfies the median.

We're now at 4, _ , _, 8, 8, _, _, 12

Fill in the blanks with anything you want as long as the sequence works and the sum of the terms equals 64.