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For the data whose frequency histogram is shown, by how many days is the mean number of days missed per student greater than the median number of days missed per student for the 15 students? Express your answer as a common fraction.

AWESOMEEE  May 30, 2015

Best Answer 

 #1
avatar+88839 
+5

0,0,0,1,2,2,2,2,3,4,5,5,5,5,5   The median is 2 days

 

The average is

(3(0) + 1(1) + 4(2) + 1(3) + 1(4) + 5(5)) / 15  = 41/15   days

 

So ....... 41/15 - 2 =  [41 - 30]/15 = 11/15 days

 

 

CPhill  May 31, 2015
 #1
avatar+88839 
+5
Best Answer

0,0,0,1,2,2,2,2,3,4,5,5,5,5,5   The median is 2 days

 

The average is

(3(0) + 1(1) + 4(2) + 1(3) + 1(4) + 5(5)) / 15  = 41/15   days

 

So ....... 41/15 - 2 =  [41 - 30]/15 = 11/15 days

 

 

CPhill  May 31, 2015

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