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Mean Deviation — Definition, Formula & Examples

Mean deviation is the average of the absolute differences between each data value and the mean of the data set. It tells you, on average, how far the data points are spread from the center.

For a data set of nn observations x1,x2,…,xnx_1, x_2, \ldots, x_n with arithmetic mean xˉ\bar{x}, the mean deviation (also called mean absolute deviation or MAD) is defined as 1n∑i=1n∣xi−xˉ∣\frac{1}{n}\sum_{i=1}^{n}|x_i - \bar{x}|. The absolute value ensures that deviations above and below the mean do not cancel each other out.

Key Formula

Mean Deviation=1n∑i=1n∣xi−xˉ∣\text{Mean Deviation} = \frac{1}{n}\sum_{i=1}^{n}|x_i - \bar{x}|
Where:
  • xix_i = Each individual data value
  • xˉ\bar{x} = The arithmetic mean of the data set
  • nn = The total number of data values
  • ∣  ∣|\;| = Absolute value — makes every difference positive

How It Works

To calculate mean deviation, first find the mean of your data set. Then subtract the mean from each data value and take the absolute value of each difference — this removes negative signs. Finally, average those absolute differences. The result is a single number that summarizes how spread out the data is. A small mean deviation means the values cluster tightly around the mean, while a large one means they are widely scattered.

Worked Example

Problem: Find the mean deviation of the data set: 2, 6, 8, 10, 14.
Step 1: Find the mean of the data set.
xˉ=2+6+8+10+145=405=8\bar{x} = \frac{2 + 6 + 8 + 10 + 14}{5} = \frac{40}{5} = 8
Step 2: Subtract the mean from each value and take the absolute value.
∣2−8∣=6,∣6−8∣=2,∣8−8∣=0,∣10−8∣=2,∣14−8∣=6|2-8|=6,\quad |6-8|=2,\quad |8-8|=0,\quad |10-8|=2,\quad |14-8|=6
Step 3: Sum all the absolute deviations.
6+2+0+2+6=166 + 2 + 0 + 2 + 6 = 16
Step 4: Divide by the number of data values to get the mean deviation.
Mean Deviation=165=3.2\text{Mean Deviation} = \frac{16}{5} = 3.2
Answer: The mean deviation is 3.2. On average, each data value is 3.2 units away from the mean of 8.

Another Example

Problem: A student's test scores are 70, 75, 80, 85, 90. What is the mean deviation?
Step 1: Calculate the mean.
xˉ=70+75+80+85+905=4005=80\bar{x} = \frac{70+75+80+85+90}{5} = \frac{400}{5} = 80
Step 2: Find each absolute deviation from the mean.
∣70−80∣=10,∣75−80∣=5,∣80−80∣=0,∣85−80∣=5,∣90−80∣=10|70-80|=10,\quad |75-80|=5,\quad |80-80|=0,\quad |85-80|=5,\quad |90-80|=10
Step 3: Average the absolute deviations.
Mean Deviation=10+5+0+5+105=305=6\text{Mean Deviation} = \frac{10+5+0+5+10}{5} = \frac{30}{5} = 6
Answer: The mean deviation of the test scores is 6 points.

Visualization

Why It Matters

Mean deviation appears in high school statistics and AP Statistics courses as an accessible way to quantify data spread before students encounter standard deviation. Quality control teams in manufacturing use it to monitor how consistently a product meets specifications. Because it is based on absolute values rather than squares, it is more intuitive and less sensitive to extreme outliers than standard deviation.

Common Mistakes

Mistake: Forgetting to take the absolute value, so positive and negative deviations cancel out to zero.
Correction: Always apply absolute value to each difference before summing. Without it, the deviations will always sum to zero regardless of the data's spread.
Mistake: Dividing by n−1n - 1 instead of nn.
Correction: Mean deviation uses nn in the denominator, not n−1n - 1. The n−1n - 1 adjustment (Bessel's correction) applies to sample standard deviation and sample variance, not to mean deviation.

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