Quote:
Originally posted by Souk
Mode: In mathematics, the element that appears most frequently in a given set of data.
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)/2=8], but then we could not establish a mode of 8!
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You
could...
4, 8 , 8 , 7, 9, 8, 8, 12
Mean = 8 (4+8+8+7+9+8+8+12/8 = 64/8 = 8)
Median = 8 (7+9/2 = 16/2 = 8)
Mode = 8 (8 occurs 4 times)
Range = 8 (12-4=8)
Or you could use:
4, 8, 8, 5, 11, 8, 8, 12
4, 8, 8, 6, 10, 8, 8, 12
But there is a fundamental flaw with all these sets. You
must organize your data set from smallest to largest before making your median calculation.
Anyway, I may have given Don's answer a "thumbs up" too soon because it worked and it was simple. But why not this answer:
5, 6, 8, 8, 8, 8, 8, 13
(I know I could have thrown in a 7&9 combo, but we're really only looking for the highest possible number. Is 13 the highest we can go?)