Question

Lesson 4 Extra Practice Mean Absolute Deviation Determine the mean absolute deviation for e the nearest hundredth if necessary. Then des absolute deviation represents. 1. Number of Sibmas 2 5 8 9 7 6 3 5 1 &

Answer 1

Data set: 2 5 8 9 7 6 3 5 1

Number of data values: 9

Average: (2 + 5 + 8 + 9 + 7 + 6 + 3 + 5 + 1)/9 = 5.11

|2 - 5.11| = 3.11

|5 - 5.11| = 0.11

|8 - 5.11| = 2.89

|9 - 5.11| = 3.89

|7 - 5.11| = 1.89

|6 - 5.11| = 0.89

|3 - 5.11| = 2.11

|5 - 5.11| = 0.11

|1 - 5.11| = 4.11

Mean Absolute Deviation =

(3.11 + 0.11 + 2.89 + 3.89 + 1.89 + 0.89 + 2.11 + 0.11 + 4.11)/9 = 2.12