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Mean Absolute Deviation

Understanding Mean Absolute Deviation (MAD)

In statistics, Mean Absolute Deviation (MAD) is a way to measure how spread out or scattered a set of data is. It tells us the average distance between each data point and the mean of the data set.

A larger MAD means the data points are far away from the mean (more spread out). A smaller MAD means the data points are clustered closely around the mean (less spread out).

How to Calculate MAD

Finding the MAD takes three simple steps:

  1. Find the mean (average) of the data set.
  2. Find the distance of each data point from the mean. This is the absolute value of the difference: data pointmean| \text{data point} - \text{mean} |.
  3. Find the mean of those distances by adding them all up and dividing by the number of data points.

Example Problems

Example 1: Find the MAD of 2, 4, 6, 8, 10

Step 1: Find the mean. Mean=2+4+6+8+105=305=6\text{Mean} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6

Step 2: Find the distance of each point from the mean (6).

  • 26=4|2 - 6| = 4
  • 46=2|4 - 6| = 2
  • 66=0|6 - 6| = 0
  • 86=2|8 - 6| = 2
  • 106=4|10 - 6| = 4

Step 3: Find the mean of these distances. MAD=4+2+0+2+45=125=2.4\text{MAD} = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4

The Mean Absolute Deviation is 2.42.4.

Example 2: Calculate the MAD for 10, 12, 8, 15, 5

Step 1: Find the mean. Mean=10+12+8+15+55=505=10\text{Mean} = \frac{10 + 12 + 8 + 15 + 5}{5} = \frac{50}{5} = 10

Step 2: Find the distances from the mean (10).

  • 1010=0|10 - 10| = 0
  • 1210=2|12 - 10| = 2
  • 810=2|8 - 10| = 2
  • 1510=5|15 - 10| = 5
  • 510=5|5 - 10| = 5

Step 3: Average the distances. MAD=0+2+2+5+55=145=2.8\text{MAD} = \frac{0 + 2 + 2 + 5 + 5}{5} = \frac{14}{5} = 2.8

The Mean Absolute Deviation is 2.82.8.

Interpreting MAD

Question: Dataset A has a MAD of 1.51.5 and Dataset B has a MAD of 4.24.2. Which is more spread out?

Answer: Dataset B is more spread out. Because the MAD represents the average distance from the mean, a higher number (4.2>1.54.2 > 1.5) indicates that the values in Dataset B are scattered further away from their average compared to Dataset A.