Jensen_HW#6.docx

# 750000 problem 148 clearclc grant jensen aer e 161

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The mean without outliers of the vector vec is: 30.750000 Problem 14.8 clear,clc %Grant Jensen, AER E 161, Homework #6, Problem 14.8 %Mean vs median to replace a test score scores1 = [99,88,95] scores2 = [99,70,77]   mean1 = mean(scores1);  %Calculates the mean for the first set of scores median1 = median(scores1);  %Calculates the median for the first set of  scores   mean2 = mean(scores2);  %Calculates the mean for the second set of scores median2 = median(scores2);  %Calculates the median for the second set of  scores   fprintf( 'The replacement score from the first set of scores are %.2f  from the mean and %.2f from the median.\n' ,mean1,median1) fprintf( 'The replacement score from the second set of scores are %.2f  from the mean and %.2f from the median.\n' ,mean2,median2) Output: scores1 = 99 88 95 scores2 = 99 70 77 The replacement score from the first set of scores are 94.00 from the mean and 95.00 from the median. The replacement score from the second set of scores are 82.00 from the mean and 77.00 from the median. Therefore, for the first set of scores, an “average” from the median would be better as there isn’t the weight of the 88% dragging down the “average”.

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Name: Grant Jensen AERE 161, Spring 2018 HW #6 Due date: 03/07/2018 But for the second set of scores, the mean would be better for the “average” because the score of 99 offsets the damage done by the 70% and 77% scores.
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