Mean Deviation Formula - Definition, Steps, Examples

Q: What is the mean deviation formula?

$\text{MD} = \frac{\sum \lvert x_i - \bar{x} \rvert}{n}$. Find each value's absolute distance from the mean, add them, and divide by the number of values.

Q: Should the mean deviation formula divide by n or n minus 1?

By $n$. The $n-1$ divisor belongs to the sample variance formula, where it corrects estimation bias, and it has no role here.

Key Points

The mean deviation formula is $\text{MD} = \frac{\sum \lvert x_i - \bar{x} \rvert}{n}$ , averaging absolute distances from the mean rather than squared ones.
The NIST e-Handbook of Statistical Methods calls this the average absolute deviation and divides by $N$ , not by $N-1$ .
Absolute values are essential, because signed deviations from the mean always sum to exactly zero and would give a mean deviation of zero for every data set.
Mean deviation can be measured about the median instead of the mean, and the median version is always smaller than or equal to the mean version.
The median absolute deviation is a different quantity, taking the median of the absolute deviations rather than their mean.

What Is the Mean Deviation Formula?

The mean deviation formula measures the average distance of values from their mean, ignoring direction, and is written

$\text{MD} = \frac{\sum_{i=1}^{n} \lvert x_i - \bar{x} \rvert}{n}$

The NIST/SEMATECH e-Handbook of Statistical Methods lists this quantity as the average absolute deviation, with the sum of absolute deviations divided by $N$ .

The Full Definition

Mean deviation answers the plainest possible question about spread: on average, how far is a value from the middle? It takes each distance, strips the sign, and averages the results.

That directness is its appeal. Unlike the standard deviation, it needs no squaring and no square root, and the answer is already in the data's own units.

The denominator is $n$ , the full count. There is no $n-1$ correction here, because the formula is being used as a description of the data in hand rather than as an unbiased estimate of a population parameter.

Why the Absolute Values Are Not Optional

Drop the absolute value bars and the formula collapses. Deviations from the mean are defined so that the positive and negative ones cancel exactly, which means $\sum (x_i - \bar{x}) = 0$ for every data set that has ever existed.

So the unsigned version would report zero spread for 1, 2, 3 and also for 1, 50, 99. The absolute values are what stop the cancellation and let the distances accumulate.

Squaring is the other way to stop the cancellation, and that is the route the variance takes. The two approaches differ in how harshly they treat large deviations: squaring punishes an outlier far more than taking its absolute value does, which is why mean deviation is the less outlier-sensitive of the two.

Variable Key

Symbol	Meaning	Unit or type	Notes
$\text{MD}$	mean deviation	same units as the data	the quantity being computed
$x_i$	the $i$ th observation	same units as the data	one value from the data set
$\bar{x}$	mean	same units as the data	or the median, for the median-based version
$\lvert \cdot \rvert$	absolute value	operator	discards the sign, keeps the size
$n$	number of observations	count	the divisor, with no correction

Examples of the Mean Deviation Formula

Example 1

Find the mean deviation of 2, 4, 6, 8, 10.

The mean is $\bar{x} = \frac{30}{5} = 6$ , and the absolute deviations are $4, 2, 0, 2, 4$ :

$\text{MD} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$

Final answer: $\text{MD} = 2.4$

Example 2

Find the mean deviation of 3, 5, 9 without using absolute values.

The deviations look like ordinary subtractions, so dropping the bars seems harmless. The mean is $\frac{17}{3} \approx 5.667$ , giving deviations of $-2.667$ , $-0.667$ and $3.333$ :

$\frac{-2.667 - 0.667 + 3.333}{3} = \frac{0}{3} = 0$

Zero. But the values are visibly spread between 3 and 9, so a spread of zero cannot be right. This is not an arithmetic slip, it is guaranteed: signed deviations from the mean always sum to zero.

Restoring the absolute values fixes it:

$\text{MD} = \frac{2.667 + 0.667 + 3.333}{3} = \frac{6.667}{3} \approx 2.222$

Final answer: $\text{MD} \approx 2.222$

Example 3

Find the mean deviation of 12, 15, 18, 21 about the mean.

The mean is $\frac{66}{4} = 16.5$ , and the absolute deviations are $4.5, 1.5, 1.5, 4.5$ :

$\text{MD} = \frac{12}{4} = 3$

Final answer: $\text{MD} = 3$

Example 4

Find the mean deviation of 4, 6, 7, 9, 24 about the mean, then about the median, and compare.

The mean is $\frac{50}{5} = 10$ and the median is 7. About the mean, the absolute deviations are $6, 4, 3, 1, 14$ :

$\text{MD}_{\text{mean}} = \frac{28}{5} = 5.6$

About the median, they are $3, 1, 0, 2, 17$ :

$\text{MD}_{\text{median}} = \frac{23}{5} = 4.6$

Final answer: $5.6$ about the mean and $4.6$ about the median.

The median version is smaller, and that is not a coincidence. The median is the point that minimises the sum of absolute deviations, so the mean deviation about the median is the smallest such average any centre can give.

Example 5

A forecast predicts 20, 22, 25 units and actual demand is 18, 26, 24. Find the mean absolute error.

Here the deviations are between paired values rather than from a mean, but the averaging is identical. The absolute errors are $2, 4, 1$ :

$\text{MAE} = \frac{2+4+1}{3} \approx 2.333$

Final answer: about $2.333$ units

This is the mean absolute error used in forecasting, and it is the mean deviation formula applied to forecast-minus-actual pairs.

Don't Confuse This With the Median Absolute Deviation

Quantity	Formula	What it averages
Mean deviation	$\frac{\sum \lvert x_i - \bar{x} \rvert}{n}$	the mean of absolute deviations from the mean
Median absolute deviation	$\text{median}(\lvert x_i - \tilde{x} \rvert)$	the median of absolute deviations from the median
Sample standard deviation	$\sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$	squared deviations, then rooted
Range	largest minus smallest	only the two extreme values

Both of the first two are abbreviated MAD in different textbooks, which is the single biggest source of confusion on this topic. The NIST e-Handbook reserves MAD for the median absolute deviation and calls the first one the average absolute deviation.

Mean Deviation Pitfalls and How to Avoid Each One

Mistake 1: Letting the signed deviations cancel

Where it slips in: The subtraction is done first and the absolute values are meant to be applied afterwards, then forgotten.

Don't do this: $\frac{\sum (x_i - \bar{x})}{n}$ , which is zero for every data set.

The correct way: Take the absolute value of each deviation as you compute it, not as a later pass. An answer of exactly zero on visibly spread data is the signature of this slip.

Students who write the deviations in a column with the bars already drawn almost never lose them, because the notation does the remembering. Doing it as a second sweep is what invites the omission.

Mistake 2: Dividing by n minus 1

Where it slips in: The $n-1$ habit carries over from variance and standard deviation, where it is correct.

Don't do this: $\frac{\sum \lvert x_i - \bar{x} \rvert}{n-1}$

The correct way: Divide by $n$ . The $n-1$ correction exists to remove bias in estimating a population variance, and the mean deviation is not doing that job.

Mistake 3: Confusing mean deviation with median absolute deviation

Where it slips in: Both are called MAD, and software functions named mad usually return the median version.

Don't do this: Treat a reported MAD as the mean deviation without checking which was computed.

The correct way: Check whether the outer operation is a mean or a median. On 4, 6, 7, 9, 24 the mean deviation about the median is 4.6, while the median absolute deviation is just 2, so the two are not interchangeable.

Bottom Line

The mean deviation formula is $\text{MD} = \frac{\sum \lvert x_i - \bar{x} \rvert}{n}$ , the average absolute distance from the mean.
The absolute values are structural, not cosmetic, because signed deviations always cancel to zero.
The divisor is $n$ , with no $n-1$ correction, since the formula describes data rather than estimating a population parameter.
Mean deviation and median absolute deviation are both abbreviated MAD and are different quantities, so always check which is meant.
To work through measures of spread with a teacher, explore an 8th grade math tutor or math classes online.

Where to Go From Here

Find the mean deviation of 5, 8, 11, 16. Then compute it about the median as well, and check that the median version is the smaller of the two.

Want a live Bhanzu trainer to walk through more statistics problems? Book a free demo class.

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Frequently Asked Questions

What is the mean deviation formula?

\text{MD} = \frac{\sum \lvert x_i - \bar{x} \rvert}{n}

. Find each value's absolute distance from the mean, add them, and divide by the number of values.

Why does the mean deviation formula use absolute values?

Because signed deviations from the mean always sum to zero, so without the absolute values the formula would report no spread for any data set at all.

Should the mean deviation formula divide by n or n minus 1?

n

. The

n-1

divisor belongs to the sample variance formula, where it corrects estimation bias, and it has no role here.

Is mean deviation the same as standard deviation?

No. Mean deviation averages absolute distances while standard deviation averages squared distances and then takes a root. Squaring makes the standard deviation react much more strongly to extreme values, so mean deviation is the steadier description of a data set with outliers.

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Bhanzu Team

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Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.

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