Identify the outlier in the following list. then find the mean and median of the list with and without the outlier. 10, 4, 20, 5, 11, 150, 7 a. outlier = 10 mean with outlier = 29.6 median with outlier = 10 mean without outlier = 44.1 median without outlier = 10.5 c. outlier = 150 mean with outlier = 9.5 median with outlier = 8.5 mean without outlier = 29.6 median without outlier = 10 b. outlier = 10 mean with outlier = 10 median with outlier = 29.6 mean without outlier = 8.5 median without outlier = 9.5 d. outlier = 150 mean with outlier = 29.6 median with outlier = 10 mean without outlier = 9.5 median without outlier = 8.5

Outlier: 10
Mean with outlier: 29.6
Median with outlier: 10
Mean without outlier: 9.5
Median without outlier: 8.5

 

d. Outlier = 150 Mean with outlier = 29.6 Median with outlier = 10
Mean without outlier = 9.5 Median without outlier = 8.5

Outlier: 150

The value in the data set that is far greater or less than most of the other values in the data set.

Mean w/ Outlier: 29.6

4 + 5 + 7 + 10 + 11 + 20 + 150 = 207

207 / 7 = 29.6

Median w/ Outlier: 10

The 4th value out of the 7 total in the data set, 10; it is the median because it is the middle-most value in the data set.

Mean w/o Outlier: 9.5

4 + 5 + 7 + 10 + 11 + 20 = 57

57 / 6 = 9.5

Median w/o Outlier: 8.5

When there are two middle-most values in a data set, add them together and divide by to ( average them ) to get the median.

7 + 10 = 17

17 / 2 = 8.5

I'm fairly sure it's C - even as a prediction, if you add (4x2) and (2x2) it's 12 - and the closest is thirteen, adding on the fractions.