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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 numb

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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

#1
+80913
+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
Sort:

#1
+80913
+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

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