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In the text book "New Comprehensive Mathematics for O Level" by Greer (1983), I see averaged deviation calculated like this:

Sum up absolute differences between single values and the mean. Then get its average. Througout the chapter the term mean deviation is used.

But I've recently seen several references that use the term standard deviation and this is what they do:

Calculate squares of differences between single values and the mean. Then get their average and finally the root of the answer.

I tried both methods on a common set of data and their answers differ. I'm not a statistician. I got confused while trying to teach deviation to my kids.

So in short, are the terms standard deviation and mean deviation the same or is my old text book wrong?

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    $\begingroup$ The two quantities differ. They weight the data differently. The standard deviation will be larger, and it is relatively more affected by larger values. The standard deviation (most particularly, the n-denominator version) can be thought of as a root-mean-square deviation. Standard deviations are more commonly used. $\endgroup$
    – Glen_b
    Jan 12, 2014 at 16:15
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    $\begingroup$ Very closely related: stats.stackexchange.com/questions/118/…. $\endgroup$
    – whuber ♦
    Jan 12, 2014 at 16:44
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    $\begingroup$ Incidentally, one reason that people tend to prefer standard deviation is because variances of sums of unrelated random variables add (and related ones also have a simple formula). That doesn't happen with mean deviation. $\endgroup$
    – Glen_b
    Sep 15, 2015 at 4:38
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    $\begingroup$ an important point is that the standard deviation derives from a model of squared errors (L2-norm, think about the normal distribution) while the mean of absolute differences corresponds to the L1-norm (think about the symmetrical exponential distribution): it is therefore more adapted (hear: sensitive) to outliers and sparse distirbutions $\endgroup$
    – meduz
    Sep 18, 2015 at 7:47
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    $\begingroup$ @Alexis the phrasing was poor. For independent random variables, Var(X+Y) = Var(X)+Var(Y). This fact is used all over the place (it leads to the familiar $\sqrt{n}\,$ terms when standardizing formulas involving means, like in one-sample t-statistics for example). There's no correspondingly general fact for mean deviation. $\endgroup$
    – Glen_b
    Jan 13, 2016 at 21:13

13 Answers 13

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Both answer how far your values are spread around the mean of the observations.

An observation that is 1 under the mean is equally "far" from the mean as a value that is 1 above the mean. Hence you should neglect the sign of the deviation. This can be done in two ways:

  • Calculate the absolute value of the deviations and sum these.

  • Square the deviations and sum these squares. Due to the square, you give more weight to high deviations, and hence the sum of these squares will be different from the sum of the means.

After calculating the "sum of absolute deviations" or the "square root of the sum of squared deviations", you average them to get the "mean deviation" and the "standard deviation" respectively.

The mean deviation is rarely used.

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  • $\begingroup$ So when one simply says 'deviation' do they mean 'standard deviation'? $\endgroup$
    – itsols
    Jan 12, 2014 at 10:46
  • $\begingroup$ I agree that 1 above or below would indicate a meaningful 'change' or 'dispersion' from a common-man's point-of-view. But squaring it would give larger values and that might not be my 'actual change'. Maybe I'm wrong but that's how I see it :/ $\endgroup$
    – itsols
    Jan 12, 2014 at 10:48
  • $\begingroup$ Most of the times the term standard deviation (square root of variance) is used. Calculating the squares is typically done, as it facilitates lots of other calculations. $\endgroup$
    – Kasper
    Jan 12, 2014 at 11:01
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    $\begingroup$ @itsols Technically, you should always specify which type of deviation statistic you are calculating for the data set -- the word deviation on its own should refer to the deviation of a single datapoint from the mean (in the way Kasper uses it in the answer). $\endgroup$
    – AmeliaBR
    Jan 12, 2014 at 16:51
  • $\begingroup$ @itsols, +1 to Amelia. Indeed, nobody says of a dataset statistic as just "deviation". A statistic is "mean absolute deviation" or "root of mean squared deviation" or such. $\endgroup$
    – ttnphns
    Jan 12, 2014 at 17:19
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Today, statistical values are predominantly calculated by computer programs (Excel, ...), not by hand-held calculators anymore . Hence, I would posit that calculating "mean deviation" is no more cumbersome than calculating "standard deviation". Although standard deviation may have "... mathematical properties that make it more useful in statistics", it is, in fact, a distortion of the concept of variance from a mean, since it gives extra weighting to data points far from the mean. It may take some time, but I, for one, hope statisticians evolve back to using "mean deviation" more often when discussing the distribution among data points -- it more accurately represents how we actually think of the distribution.

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  • $\begingroup$ You make an extraordinary claim about how (statistically qualified) people "actually think." What is the source of your information about that? $\endgroup$
    – whuber ♦
    Dec 16, 2014 at 18:26
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    $\begingroup$ The source is simply people whom I have questioned on this topic, as well as myself. When asked: How do you picture the variation in this data set? the response was always expressed in terms of the linear distance from the mean -- the response never included squares or square roots. Granted, I am an engineer, not a "statistician", but I would ask anyone else to challenge themselves on this topic. Yes, we love the math of standard deviation -- it's fun, but is this really how you picture the deviation from the mean? $\endgroup$
    – andyl
    Dec 16, 2014 at 22:03
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    $\begingroup$ It depends on the purpose. For data exploration, I tend to use robust, rank-based estimates of dispersion, such as median deviations from medians, which in spirit are close to your proposal. But for much other work, especially when assessing (even mentally) the potential for statistical significance, estimating appropriate sample sizes, figuring out the value of information, and deciding among competing statistical procedures, thinking in terms of variances (and therefore standard deviations) is essential. Mean deviations are no substitute, as the mathematics clearly shows. $\endgroup$
    – whuber ♦
    Dec 16, 2014 at 23:07
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    $\begingroup$ Check out this paper. $\endgroup$
    – Pete
    Dec 8, 2015 at 16:28
  • $\begingroup$ @Pete How you reached there? $\endgroup$
    – Vicrobot
    Jan 7, 2019 at 19:08
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Both measure the dispersion of your data by computing the distance of the data to its mean.

  1. the mean absolute deviation is using norm L1 (it is also called Manhattan distance or rectilinear distance)
  2. the standard deviation is using norm L2 (also called Euclidean distance)

The difference between the two norms is that the standard deviation is calculating the square of the difference whereas the mean absolute deviation is only looking at the absolute difference. Hence large outliers will create a higher dispersion when using the standard deviation instead of the other method. The Euclidean distance is indeed also more often used. The main reason is that the standard deviation have nice properties when the data is normally distributed. So under this assumption, it is recommended to use it. However people often do this assumption for data which is actually not normally distributed which creates issues. If your data is not normally distributed, you can still use the standard deviation, but you should be careful with the interpretation of the results.

Finally you should know that both measures of dispersion are particular cases of the Minkowski distance, for p=1 and p=2. You can increase p to get other measures of the dispersion of your data.

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They both measure the same concept, but are not equal.

You are comparing $\frac{1}{n} \sum |x_i-\bar{x}|$ with $\sqrt{\frac{1}{n} \sum (x_i-\bar{x})^2}$. They aren't equal for two reasons:

Firstly the square-root operator is not linear, or $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$. Therefore the sum of absolute deviations is not equal to the square root of the sum of squared deviations, even though the absolute function can be represented as the square function followed by a square root:
$\sum|x_i-\bar{x}| = \sum \sqrt{(x_i-\bar{x})^2} \neq \sqrt{\sum(x_i-\bar{x})^2}$
as the square root is taken after the sum has been calculated.

Secondly, $n$ is now also under the square root in the standard deviation calculation.

Try calculating $\frac{1}{n}\sum \sqrt{(x_i-\bar{x})^2}$ - it should yield the same answer as the mean deviation and help you to understand.

The reason why the standard deviation is preferred is because it is mathematically easier to work with later on, when calculations become more complicated.

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    $\begingroup$ The absolute value of a sum is not, in general, the same as the sum of absolute values! Neither square, square root, nor absolute functions are linear, which is why the sum after applying the function is different from applying the function after taking the sum. $\endgroup$
    – AmeliaBR
    Jan 12, 2014 at 16:28
  • $\begingroup$ @AmeliaBR you are of course perfectly correct! $\endgroup$
    – ltronneberg
    Jan 12, 2014 at 16:47
  • $\begingroup$ The rest of the argument was good, though, which is why I decided to edit out the problematic statement. $\endgroup$
    – AmeliaBR
    Jan 12, 2014 at 16:52
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@itsols, I'll add to Kasper's important notion that The mean deviation is rarely used. Why is standard deviation considered generally a better measure of variability than mean absolute deviation? Because arithmetic mean is the locus of minimal sum of squared (and not sum of absolute) deviations from it.

Suppose you want to assess the degree of altruism. Then you probably won't ask a person about how much he is ready to give money in "general situation" of life. Rather, you'll choose to ask how much he is ready to do it in the constained situation, where he has minimal possible resourses for his own living. I.e. what is the amount of individual altruism in the situation when that amount is individual's minimal?

Likewise, what is the degree of variability of these data? Intuitively, the best measuring index for it is the one which is minimized (or maximized) down to the limit in this context. The context is "around the arithmetic mean". Then st. deviation is the best choice in this sense. If the context were "around the median" then mean |deviation| would be the best choice, because median is the locus of minimal sum of absolute deviations from it.

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    $\begingroup$ Your justification for SD based on Locus is circular. You are justifying SD by putting special importance on arithmetic mean - all this shows is that they have a relationship, not that SD is special. One can similarly place importance on median, which is the locus of minimal sum of absalute loss. The real reasons why SD is used more often is because the maths is easier to work with ... furthermore, it's easier computationally (both because medians require "sorting", and because squares are faster to compute than branch statements). Philosophically absalute deviation has greater value. $\endgroup$
    – samthebest
    Sep 24, 2015 at 10:49
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One thing worth adding is that the most likely reason your 30-year-old textbook used the absolute mean deviation as opposed to standard deviation is that it is easier to calculate by hand (no squaring / square roots). Now that calculators are readily accessible to high school students, there is no reason not to ask them to calculate standard deviation.

There are still some situations where absolute deviations are used instead of standard deviations in complex model fitting. Absolute deviations are less sensitive to extreme outliers (values far from the mean/trendline) compared to standard deviations because they don't square that distance before adding it to the values from other data points. Since model fitting methods aim to reduce the total deviation from the trendline (according to whichever method deviation is calculation), methods that use standard deviation can end up creating a trendline that diverges away from the majority of points in order to be closer to an outlier. Using absolute deviations reduces this distortion, but at the cost of making calculation of the trendline more complicated.

That's because, as others have noted, the standard deviation has mathematical properties and relationships which generally make it more useful in statistics. But "useful" should never be confused with perfect.

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    $\begingroup$ just curious, what are the "math properties" that makes SD more useful than mean absolute deviation? excellent answer by the way. $\endgroup$
    – Weipeng
    Jul 22, 2017 at 0:21
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    $\begingroup$ @pongba Standard deviation is intrinsic to many statistical models that assume random variation from multiple effects that can cancel each other out (aka, normally distributed data). This includes the sampling accuracy (margin of error) when using a survey from a large population. If your data meets this model, you can estimate the probability of getting a value from the number of SD from the mean. You can calculate the SD of multiple independent effects from the SD of individual components. See also: en.m.wikipedia.org/wiki/Standard_deviation $\endgroup$
    – AmeliaBR
    Jul 23, 2017 at 4:51
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    $\begingroup$ "absolute mean deviation" would be better as "mean absolute deviation" $\endgroup$
    – Nick Cox
    Sep 6, 2020 at 10:01
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They are similar measures that try to quantify the same notion. Typically you use st. deviation since it has nice properties, if you make some assumption about the underlying distribution.

On the other hand the absolute value in mean deviation causes some issues from a mathematical perspective since you can't differentiate it and you can't analyse it easily. Some discussion here.

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No. You are wrong. Just kidding. There are, however, many viable reasons why one would want to compute mean deviation rather than formal std, and in this way I am in agreement with the viewpoint of my engineering Brethren. Certainly if I am computing statistics to compare with a body of existing work which is expressing qualitative as well as quantitative conclusions, I woud stick with std. But, for example, assume I am trying to run some fast anomaly-detection algorithms on binary, machine-generated data. I'm not after academic comparisons as my final goal. But I am interested in the fundamental inference about the "spread" of a particular flow of data about its mean. I'm also interested in computing this iteratively, and as efficiently as possible. In digital electronic hardware, we play dirty tricks all the time -- we distill multiplications and divisions into left and right shifts, respectively, and for "computing" absolute values, we simply drop the sign bit (and compute one's or two's complement if necessary, both easy transforms). So, my choice is to compute it in the most knuckle-dragging way I can, and apply linear thresholds to my computations for fast anomaly detection over desired time windows.

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    $\begingroup$ The standard deviation can be computed efficiently and simply with an online algorithm, just as any moment can be (including the mean absolute deviation). Thus, the requirement for fast or simple calculation would not rule this out (nor would it rule out any moment-based estimators of spread). $\endgroup$
    – whuber ♦
    Sep 14, 2015 at 17:52
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Amar Sagoo has a very good article explaining this.

To add my own attempt at an intuitive understanding:

Mean deviation is a decent way of asking how far a hypothetical "average" point is from the mean, but it doesn't really work for asking how far all the points are from each other, or how "spread out" the data are.

Standard deviation is asking how far apart all the points are, so in incorporates more useful information than just the mean deviation (which is why mean deviation is usually only used as a stepping stone toward understanding standard deviation).

A good analogy is the Pythagorean Theorem. The Pythagorean Theorem tells us the distance between points in two dimensions by taking the horizontal distance and the vertical distance, squaring them, adding the squares, and taking the square root of the total.

If you look at it closely, the formula for (population) Standard Deviation is basically the same as the Pythagorean Theorem, but with a lot more than two dimensions (and using distance from each point to the mean as the distance in each dimension). As such it gives the most accurate picture of the "distance" between all the points in your data set.

To push that analogy a little further, the mean absolute deviation would be like taking the average of the horizontal and vertical distances, which is shorter than the total distance, while the sum absolute deviation would be adding the horizontal and vertical distances, which is longer than the actual distance.

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    $\begingroup$ I assume when you say mean deviation you mean mean absolutely deviation which is what the OP was talking about. Terminology is important because mean deviation is always 0. Regarding the difference between mean absolute deviation & standard deviation the both involve the deviation of ALL the points from the mean. One involves the sum of the absolute deviations from the mean while the is the square root if the sum of the squared deviation.. $\endgroup$
    – Michael R. Chernick
    Sep 18, 2019 at 21:14
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The standard deviation represents dispersion due to random processes. Specifically, many physical measurements which are expected to be due to the sum of many independent processes have normal (bell curve) distributions.

The normal probability distribution is given by: $ \Large Y = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}} $

Where $Y$ is the probability of getting a value $x$ given a mean $\mu$ and $\sigma$…the standard deviation!

In other words, the standard deviation is a term that arises out of independent random variables being summed together. So, I disagree with some of the answers given here - standard deviation isn't just an alternative to mean deviation which "happens to be more convenient for later calculations". Standard deviation is the right way to model dispersion for normally distributed phenomena.

If you look at the equation, you can see the standard deviation more heavily weights larger deviations from the mean. Intuitively, you can think of the mean deviation as measuring the actual average deviation from the mean, whereas the standard deviation accounts for a bell shaped aka "normal" distribution around the mean. So if your data is normally distributed, the standard deviation tells you that if you sample more values, ~68% of them will be found within one standard deviation around the mean.

On the other hand, if you have a single random variable, the distribution might look like a rectangle, with an equal probability of values appearing anywhere within a range. In this case, the mean deviation might be more appropriate.

TL;DR if you have data that are due to many underlying random processes or which you simply know to be distributed normally, use standard deviation function.

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Consider three set of data having same mean and MD but their ranges are changing. It is interesting to see how SD changes with change in the range of the data.

SET 1: 1, 3,5,7,9,11,13,15,17,19 Range:1-19 Mean=10, MD=5 SD= 6.05

SET 2: 2,3,5,7,7,9,13,15,14,23 Range: 1-23 Mean=10 MD=5 SD=6.28

SET 3: 3,5,5,7,7,8,10,12,13,30 Range: 1-30 Mean =10 MD=5 SD=7.70

It can be observed that all the three sets have same mean and MD. It is to be highlighted that while MD do not change with change in range, SD show changes with every change in ranges. This clearly establishes the supremacy of SD as compared to MD in dealing with variation in the data.

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    $\begingroup$ NOPE! (1) In your SET 2, you should change 14 to 17 in order to have Mean=10; currently, Mean(SET 2)=9.8! (2) Let's assume we made the correction I specified in my "(1)". The equality of (Mean; MD)=(10, 5) in all of the three sets shows that the three sets have essentially the same distribution from MATHEMATICAL point of view. That is correct. However, SD gives different results! That is due to giving more weight to the extreme values (from the mean) by squaring the absolute deviations. You lost the idea of why Gauss employed squaring in his 1821 study. Gauss, C. F. (1821). $\endgroup$
    – Erdogan CEVHER
    Jun 7, 2021 at 8:26
  • $\begingroup$ (cont'd)... Theory of the combination of observations least subject to error. Classics in Applied Mathematics. (Translator: G.W.Stewart). epubs.siam.org/doi/book/10.1137/1.9781611971248?mobileUi=0 Squaring is JUST for computational easiness during comparisons! HOWEVER, it violates scientific correctness ("PERFECTNESS" acc. to @AmeliaBR above). Your example just shows the SD's WEAKNESS, NOT SD's supremacy! $\endgroup$
    – Erdogan CEVHER
    Jun 7, 2021 at 8:29
  • $\begingroup$ (cont'd2) SD=7.70 in the 3rd set violates the idea that "Extreme values are ALSO part of the life", and the artificial introduction of "Normality" idea. See TALEB's (2014) edge.org/response-detail/25401 and TALEB's (2021) Incerto for more detailed discussion on this topic. $\endgroup$
    – Erdogan CEVHER
    Jun 7, 2021 at 8:53
  • $\begingroup$ I don't think this establishes supremacy nor inferiority. Rather, if you wish to account for range or relativity, use percentage calculations instead of absolute ones; or, provide a range calculation off to the side. $\endgroup$
    – Andrew
    Oct 26, 2021 at 13:08
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The two measures differ indeed. The first is often referred to as Mean Absolute Deviation (MAD) and the second is Standard Deviation (STD). In embedded applications with severely limited computing power and limited program memory, avoiding the square root calculations can be very desirable.

From a quick rough test it seems that MAD = f * STD with f somewhere between 0.78 and 0.80 for a set of gaussian distributed random samples.

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Each of the three parameters - Mean (M), Mean Absolute Deviation (MAD) and Standard Deviation (σ), calculated for a set, provide some unique information about the set which the other two parameters don't. σ loosely includes the information provided by MAD, but it isn't vice versa. Hence, σ is conveniently used everywhere.

M => around which number the observations are centered. But a set can have its observations quite far from the mean, on an average, as compared to another set having the same mean. In order to get that information (i.e. the average distance of observations from its mean), we move to MAD.

MAD => how far each observation individually is from the mean of all observations, but it doesn't tell how the observations are arranged in relation to one another. To get that information (i.e. the average distance of the set itself from its mean, which depends upon how the observations are arranged in relation to one another), we move to σ.

σ => how far the complete set is from its mean (or, how far the observations are from each other).

If you want to go deeper, have a look at my article here.

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