The table lists the test scores William and Andre received on five math assessments William's scores 89 97 78 81 91 Andre's score 90 74 73 87 82 Which statement best describes the difference of the mean of the two sets?A)It is equal to about 0.5 times the mean absolute deviation of either data set. B)It is equal to about 1 times the mean absolute deviation of either data set. C)It is equal to about 1.5 times the mean absolute deviation of either data set. D)It is equal to about 2 times the mean absolute deviation of either data set.

Start out by finding the mean of William's set.  To find the mean, add all of the data points together and divide by the number of data points:
[tex]\frac{89+97+78+81+91}{5}=\frac{436}{5}=87.2[/tex]
To find his mean absolute deviation, find the absolute values of the differences between each data point and the mean; then find the average of these:
[tex]\frac{|89-87.2|+|97-87.2|+|78-87.2|+|81-87.2|+|91-87.2|}{5} \\ \\=\frac{1.8+9.8+9.2+6.2+3.8}{5}=\frac{30.8}{5}=6.16[/tex]
Now find the mean of Andre's data (again, add all of the data points together and divide by the number of data points):
[tex]\frac{90+74+73+87+82}{5}=\frac{406}{5}=81.2[/tex]
Find his mean absolute distance (find the absolute value of the difference between each data point and the mean, then average these):
[tex]\frac{|90-81.2|+|74-81.2|+|73-81.2|+|87-81.2|+|82-81.2|}{5} \\ \\=\frac{8.8+7.2+8.2+5.8+0.8}{5}=\frac{30.8}{5}=6.16[/tex]
Both mean absolute deviations are the same.  The difference between the means of the two sets is
87.2-81.2=6.
6 is about 1 times the mean absolute deviation of either set.