Question 784117
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The total of the 5 scores, given a mean of 83, is 415.  Since the median is 85, of the 5 scores, two must be greater than 85 and two must be less than 85.  But since the mode is 90, there must be more 90s than any other score.  Hence there are 2 90s and one of every other score.  Since there are only two scores larger than 85, they must represent the two 90s.  So now we have the 1st, 2nd, and 3rd highest scores as 90, 90, and 85.  The sum of these three scores is 265, leaving 415 minus 265 = 150 to be accounted for with the two remaining scores, both of which must be less than or equal to 85.  So of these two scores, the most one of them could be would be 85 (if it was 0.00001 points higher, then 85 would not have been the median, 85.00001 would have been).  So if the higher of the two lowest scores was 85, the lowest must be 150 minus 85 = 65.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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