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Mean Absolute Deviation — How to Find and Use MAD

By MathPal Editorial Team·Published

Grade: 6-7 | Topic: Statistics

What You Will Learn​

The mean absolute deviation (MAD) is a measure of how spread out a data set is. While the mean tells you the typical value, the MAD tells you how far the typical data point is from that mean. This guide walks through the formula step by step, explains what the result means, and shows how to compare variability between two groups.

Theory​

Why variability matters​

Consider two exam classes:

  • Class A scores: 70, 70, 70, 70, 70 — mean 70
  • Class B scores: 50, 60, 70, 80, 90 — mean 70

Both classes have the same mean, but Class A has zero spread while Class B has wide spread. A measure of variability captures this difference.

The MAD formula​

MAD=∑∣xi−xˉ∣n\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n}

where:

  • xˉ\bar{x} = mean of the data
  • xix_i = each individual value
  • ∣xi−xˉ∣|x_i - \bar{x}| = absolute deviation (how far each value is from the mean, always positive)
  • nn = number of values

Step-by-step process:

  1. Find the mean xˉ\bar{x}.
  2. Subtract the mean from each value: xi−xˉx_i - \bar{x}.
  3. Take the absolute value of each difference: ∣xi−xˉ∣|x_i - \bar{x}|.
  4. Add all the absolute deviations.
  5. Divide by nn (the number of values).

Interpreting MAD​

  • MAD = 0: all values are identical (no spread).
  • Small MAD: values cluster close to the mean.
  • Large MAD: values are spread far from the mean.
  • MAD is expressed in the same units as the original data.

Worked Examples​

Example 1 — Basic MAD calculation​

Data set: 4, 7, 13, 2, 9

Step 1: Mean = 4+7+13+2+95=355=7\dfrac{4+7+13+2+9}{5} = \dfrac{35}{5} = 7.

Step 2 & 3: Absolute deviations:

| Value xix_i | xi−7x_i - 7 | ∣xi−7∣|x_i - 7| | |-------------|-----------|-------------| | 4 | -3 | 3 | | 7 | 0 | 0 | | 13 | 6 | 6 | | 2 | -5 | 5 | | 9 | 2 | 2 |

Step 4: Sum = 3+0+6+5+2=163+0+6+5+2 = 16.

Step 5: MAD = 165=3.2\dfrac{16}{5} = 3.2.

Interpretation: On average, each value is 3.2 units away from the mean of 7.

Example 2 — Class A vs Class B comparison​

Class A: 70, 70, 70, 70, 70 — mean = 70

All deviations = 0. MAD = 0. No spread.

Class B: 50, 60, 70, 80, 90 — mean = 70

Deviations: 20, 10, 0, 10, 20. Sum = 60. MAD = 60/5 = 12.

Interpretation: Class B scores are, on average, 12 marks away from the mean — much more varied than Class A.

Example 3 — Which team is more consistent?​

Team X sprint times (seconds): 11.2, 11.4, 11.3, 11.5, 11.1

Mean = 56.55=11.3\dfrac{56.5}{5} = 11.3 s.

Absolute deviations: 0.1, 0.1, 0.0, 0.2, 0.2. Sum = 0.6. MAD = 0.12 s.

Team Y sprint times: 10.8, 11.5, 11.0, 12.0, 10.7

Mean = 56.05=11.2\dfrac{56.0}{5} = 11.2 s.

Absolute deviations: 0.4, 0.3, 0.2, 0.8, 0.5. Sum = 2.2. MAD = 0.44 s.

Team X (MAD = 0.12 s) is more consistent than Team Y (MAD = 0.44 s), even though both have similar means.

Common Mistakes​

Mistake 1 — Forgetting the absolute value​

❌ Finding deviations as xi−xˉx_i - \bar{x} (signed), then averaging them. Positive and negative deviations cancel to give 0 or a misleading result.

✅ Always take the absolute value: ∣xi−xˉ∣|x_i - \bar{x}|. Deviations are distances — distances are always positive.

Mistake 2 — Using median instead of mean​

❌ Calculating deviations from the median.

✅ MAD specifically measures deviations from the mean (xˉ\bar{x}), not the median.

Mistake 3 — Dividing by the wrong number​

❌ Dividing the sum of deviations by n−1n-1 (common in standard deviation, not MAD).

✅ For MAD, divide by nn — the total number of data values.

Practice Problems​

Problem 1: Find the MAD of: 3, 8, 5, 12, 7.

Show Answer

Mean = 355=7\dfrac{35}{5} = 7.

Absolute deviations: 4, 1, 2, 5, 0. Sum = 12.

MAD = 125=2.4\dfrac{12}{5} = \mathbf{2.4}

Problem 2: A set of temperatures (°C) for one week: 18, 22, 19, 25, 21, 20, 23. Find the MAD.

Show Answer

Mean = 1487≈21.14\dfrac{148}{7} \approx 21.14.

Deviations: 3.14, 0.86, 2.14, 3.86, 0.14, 1.14, 1.86. Sum ≈ 13.14.

MAD ≈ 13.147≈1.88\dfrac{13.14}{7} \approx \mathbf{1.88} °C.

Problem 3: Data set A has MAD = 1.5 and Data set B has MAD = 4.2. Both have the same mean. Which data set has values more tightly clustered around the mean?

Show Answer

Data set A — a smaller MAD means values are closer to the mean on average.

Problem 4: Can the MAD ever be greater than the range of the data? Explain.

Show Answer

No. The range is the maximum possible distance between any value and the mean (roughly). The MAD is the average of all deviations, so it can never exceed the range.

Summary​

  • MAD measures the average distance from each data value to the mean.
  • Formula: MAD=∑∣xi−xˉ∣n\text{MAD} = \dfrac{\sum |x_i - \bar{x}|}{n} — find mean, find absolute deviations, average them.
  • A small MAD = tightly clustered data. A large MAD = widely spread data.
  • MAD lets you compare variability between two data sets that may have similar means.
  • Always use absolute values — deviations are distances, so they are never negative.

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