Respuesta :
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.
[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.