Mean Absolute Deviation Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
Calculate MAD for daily temperatures (°F): {68,72,65,75,70}\{68, 72, 65, 75, 70\}.

Solution

  1. 1
    Mean: (68+72+65+75+70)/5=350/5=70°F(68+72+65+75+70)/5 = 350/5 = 70°F
  2. 2
    Absolute deviations: ∣68−70∣=2|68-70|=2, ∣72−70∣=2|72-70|=2, ∣65−70∣=5|65-70|=5, ∣75−70∣=5|75-70|=5, ∣70−70∣=0|70-70|=0
  3. 3
    MAD=(2+2+5+5+0)/5=14/5=2.8°FMAD = (2+2+5+5+0)/5 = 14/5 = 2.8°F

Answer

MAD=2.8°FMAD = 2.8°F. Temperatures deviate from the mean by 2.8°F on average.
MAD of 2.8°F means the typical daily temperature is within 2.8°F of the weekly average. This is directly interpretable in the original unit (°F), making it more intuitive than variance (which would be in °F²).

About Mean Absolute Deviation

The average distance between each data value and the mean of the data set. Calculated by finding the mean, computing the absolute value of each deviation from the mean, and averaging those absolute deviations.

Learn more about Mean Absolute Deviation →

More Mean Absolute Deviation Examples

Example 1 easy

Calculate the Mean Absolute Deviation (MAD) for [formula] and explain what it measures.

Example 2 medium

Compare MAD and standard deviation for the data [formula]. Calculate both and explain when MAD is pr

Example 4 hard

Data set A: [formula] and Data set B: [formula]. Both have mean 5. Calculate MAD for each and explai

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