+1438 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.
+128474 Mmmm... let's see...
First....let's order the data by days missed
0,0,0,1,2,2,2,2,3,4,5,5,5,5,5
The median number of days missed is the middle value, 2 days
The mean (average) days missed is :
[ 3(0) + 1 + 4(2) + 3 + 4 + 5(5) ] / 15 =
41 / 15 days
So the mean number of days exceeds the median number of days by
41/15 - 2
41/15 - 30/15 =
11/15 days
