I know that "deviations in the data are devil", and when the distribution is highly skewed, it is better to consider median as average rather than mean, but how to decide these hard-limits.

For example:

  • CASE 1:

    • Assume X = 10,20,30,40,50,60,70
    • In this case, I think that it is better to use mean and that it will give very accurate results.
  • CASE 2:

    • Assume X = 10,20,30,40,50,60,70,7000
    • In this case, I think that it is better to use median instead of using the mean.
  • CASE 3:

    • Assume X = 10,20,30,400,500,600,700
    • In this case, I think it is better to use IQR (Inter Quartile Range)

But I'm stuck with how to decide these hard-limits i.e. which to use in which condition, in general.

I've found a tool working on subjected principle, which takes context-less sample-distribution as input and determines whether mean is close/moderate or against the null-hypothesis.

Find References:-

What I'm really looking is a good answer which states how to derive these conclusions.

  • $\begingroup$ @Jeromy Anglim: Thanks for the editing and making me correct. $\endgroup$
    – user4331
    Apr 26, 2011 at 13:21
  • 4
    $\begingroup$ Asking for a "hard-limit" is misguided when you haven't specified the purpose of the statistics: the replies are going to push back on this. Even when the research objective is placed in evidence, there will always be mitigating factors modifying any rule of thumb. Using statistical procedures effectively and correctly always requires judgment. It's not blind application of algorithms. $\endgroup$
    – whuber
    Apr 26, 2011 at 14:52
  • $\begingroup$ @whuber Thank-you whuber for sharing the valuable thoughts. I totally agree with you that there is no rule of thumb, to decide whether to use mean/median for average. But it makes me to think when I used this tool: home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/MeanTest.htm Here, he is coming up with conclusions how far mean holds good with a given distribution. May be he is using Standard-Error, but what are these limits which changes the "conclusion" in his tool. $\endgroup$
    – user4331
    Apr 27, 2011 at 6:46
  • $\begingroup$ Just to note, in case 1 the mean and the median are equal. $\endgroup$ May 8, 2014 at 10:59

6 Answers 6


Framing the question

  • You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.

  • From a purely statistical perspective, the mean and median both provide different information about the central tendency of a sample of data. Thus, neither is correct or incorrect by definition.

  • From an applied perspective, we often want to say something meaningful about the central tendency of a sample, where central tendency maps onto some subjective notion of "typical".

General thoughts

  • When summarising what is typical in a sample, observations that are many standard deviations away from the mean (perhaps 3 or 4 SD) will have a large influence on the mean, but not the median. Such observations may lead the mean to deviate from what we think of as the "typical" value of the sample. This helps to explain the popularity of the median when it comes to reporting house prices and income, where a single island in the pacific or billionaire could dramatically influence the mean, but not the median. Such distributions can often include extreme outliers, and the distribution is positively skewed. In contrast, the median is robust.

  • The median can be problematic when the data takes on a limited number of values. For example, the median of a 5-point Likert item lacks the nuance possessed by the mean. For example, means of 2.8, 3.0, and 3.3 might all have a median of 3.

  • In general, the mean has the benefit of using more of the information from the data.

  • When skewed distributions exist, it is also possible to transform the distribution and report the mean of the transformed distribution.

  • When a distribution includes outliers, it is possible to use a trimmed mean, or remove the outliers, or adjust the value of the outlier to a less extreme value (e.g., 2 SD from the mean).

  • $\begingroup$ Thank you Jeromy for writing and understanding me more precise what I'm asking about. But as I said Jeromy, I want those hard-limits, and I came across with those hard-limits at: For mean: home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/MeanTest.htm For median:home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/… He has designed something around what I'm talking about. But I really couldn't find any pointers which states how to achieve this kind of solution, what he did in his tools. $\endgroup$
    – user4331
    Apr 27, 2011 at 6:55
  • $\begingroup$ @user4331 as far as I can see these links go to applets for doing significance testing related means and medians, and do not tell you when you should use one over the other. $\endgroup$ Apr 27, 2011 at 8:15
  • $\begingroup$ Yes, they are doing significance testing. But do you noticed that if you club these techniques and apply them in the illustrated fashion, than they can give you some admirable outputs:- for-ex: Start with a sample-distribution, find whether it passes the mean-test or not(using those applets); If 'yes' than go with mean; If 'no' than go for median test;Now find whether it passes the median test or not; If 'yes' than go with median; Else go with IQR. And I believe this approach can work out to find a better average, for in-general sample-distributions. $\endgroup$
    – user4331
    Apr 27, 2011 at 10:46
  • $\begingroup$ I didn't upvote because the idea of looking at coarse rating scales by mean is endorsed. Also, it's incorrect that medians are always influenced less by extreme values than means. I weigh these against the good advice whuber and Gael mentioned and give no vote. $\endgroup$
    – John
    Nov 13, 2017 at 14:18

You can read about measures of central tendency here: http://en.wikipedia.org/wiki/Central_tendency .

Generally, you analyse a sample in order to tell something about a (much larger) population. Often you know more about the population than merely the data in your sample, usually something that motivated you to take a sample in the first place. If you know that the population has normal distribution then the sample mean will be the best estimator of the expected value even if the sample does not look normal (using small sample sizes like the above you can't really characterise the distribution anyway). You can reliably estimate the mean if you have a lot of data even if the distribution is not normal (see T-test for non normal when N>50?).

In cases of distributions that can not be described parametrically median and the IQR may tell much more. IQR is a dispersion measure as opposed to the mean and median that are location measures. You can read about dispersion parameters here: http://en.wikipedia.org/wiki/Statistical_dispersion .

A further aspect to consider is that some of your data may be outliers (see Rigorous definition of an outlier?).

  • $\begingroup$ Thanks a lot for your quick response, but still it doesn't answer the basic question. How to define those hard-limits?? $\endgroup$
    – user4331
    Apr 26, 2011 at 13:21
  • $\begingroup$ There are no "hard-limits". If you have no more information than a sample of 7 data points you probably don't know enough about the problem to achieve meaningful results. $\endgroup$
    – GaBorgulya
    Apr 26, 2011 at 13:32
  • $\begingroup$ I agree with you, that these 7 data-points are almost meaning-less if there is no context around them. But GaBorgulya, I'm not very concerned about the information around these sample data-points, but what I'm concerned about is- how these data-points should be treated in-general. And as I quoted in few of my comments, I found these hard-limits at: bit.ly/ibQhWo, bit.ly/fWhRTD I'm not saying what he is doing is very optimal or good solution, but what I'm saying is he came up with interesting conclusions, which are really making some sense for me. $\endgroup$
    – user4331
    Apr 27, 2011 at 7:03

There are no hard and fast rules. They convey different information and have different properties. You select the statistic that best conveys what you want to convey. Or better yet, select statistics that best describe the data. Keep this same thing in mind when you're selecting a measure of central tendency to analyze.

(snipped a bunch of stuff repeating Mike Lawrence's answer)

Note that Mike Lawrence is referring to something that's surprising for a lot of people. In the behavioural sciences there's a lot of folk wisdom that you use medians with small sample sizes. But in actual fact that's exactly the wrong thing to do because the median quickly becomes more biased than the mean with small samples.

  • $\begingroup$ If that's your desire then you need to find out what the limits are for your data. There are no general rules. $\endgroup$
    – John
    Apr 26, 2011 at 13:28
  • $\begingroup$ @user4331 It sounds to me like you really should not add an auto-switch to your routine. It's an admirable goal, but as has been pointed out, it's a tricky question that depends on a knowledge of the data, your goals, and what is "normal/usual" for the data, and an auto-switch can't possibly know those things. So your auto-switch routine could be terribly misleading to its users, who would assume that the routine could somehow know better than they when to switch. $\endgroup$
    – Wayne
    Apr 26, 2011 at 14:53
  • $\begingroup$ I might also add that if you want help for a specific kind of data for which you have some estimate of skewness expected and help for a specific plan for handling that data then you should post that question. That might get you the answers you want. $\endgroup$
    – John
    Apr 26, 2011 at 14:57
  • $\begingroup$ I'm talking about context-less data. I'm talking about any data in-general. $\endgroup$
    – user4331
    Apr 27, 2011 at 7:12
  • $\begingroup$ Then your question has been answered. There are no rules and having such a switch is very unlikely to be a good idea. $\endgroup$
    – John
    Apr 28, 2011 at 19:09

Be careful with medians: they are biased estimators and the degree of bias can change depending on the skew of the distribution and the sample size (see Miller, 1988). This means that if you are comparing two conditions that have either different skew or different sample sizes, you may find a difference that is in fact attributable to bias rather than a real difference, or you may fail to find a difference when there is a real one when the direction of the difference in bias is opposite to the direction of the real difference between the conditions.

  • 3
    $\begingroup$ I think this response needs some qualification. So far the OP has not stated the target of the estimation--the estimand--so we cannot possibly know whether a statistic is biased or not. For instance, the median is an unbiased estimator of the geometric mean of a (usually highly skewed) lognormal distribution. $\endgroup$
    – whuber
    Apr 26, 2011 at 14:46
  • 1
    $\begingroup$ That's right for a large enough N (depends on degree of skew what large enough is). But if the N is very small (like the examples posted) then medians are pulled by the tail more than means. Bias in the mean is constant with N but varies with N for median. That's what the poster is talking about. $\endgroup$
    – John
    Apr 26, 2011 at 14:54
  • 2
    $\begingroup$ @John Nothing of what you say comports with my understanding of these estimators, but perhaps that's because nobody has yet stipulated what is being estimated, nor have they clearly stated any distributional assumptions except that the distribution is "highly skewed." $\endgroup$
    – whuber
    Apr 26, 2011 at 18:25
  • 1
    $\begingroup$ Doing a quick simulation of looking at medians across N with skewed distributions will show it. $\endgroup$
    – John
    Apr 28, 2011 at 19:08

"deviations in the data are the devil" is just not true I think - well I don't agree with it at least. I'd say its more like "chilli" than the "devil" - as much as you can reasonably handle is good, but it can get nasty if there is too much.

The most general procedure I know of to "choose a statistic" to report your data is a combination of two things

  1. Bayesian inference (describing what is known)
  2. Decision theory (taking actions under uncertainty)

However, both of these methods are only partially "algorithmic" so to speak. You have to supply the inputs though. Perhaps the most important part of this stage is that you have to ask a question that your procedure is going to answer. Naturally, different questions get different answers. As the saying goes "I have just derived a very elegant and beautiful answer. All I have to do now is figure out the question." This is a common problem that I have seen with many statistical procedures, is that there is not always a clear statement of the class of problems that it is the best procedure to use.

Bayesian inference requires you to specify your prior information in a mathematical framework. This involves

  1. Specifying the hypothesis space - what possibilities am I going to consider?
  2. Assigning probabilities to each part of the space
  3. Using the rules of probability theory to manipulate the assigned probabilities

this is basically an open-ended problem (you can always analyse a given English statement more deeply, to extract more or different information from it). Decision theory also requires you to specify a loss function - and there are basically no rules or principles by which to do this, at least as far as I know (computational simplicity is a key driver).

One useful question to ask yourself though is "what information about the sample do I convey by presenting this statistic?" or "how much of the complete data set can I recover from using just this set of statistics?"

One way you could use Bayesian statistics to help you here is to propose a hypothesis:

$$\begin{array}{l l} H_{mean}:\text{The mean is the best statistic} \\H_{med}:\text{The median is the best statistic} \\H_{IQR}:\text{The IQR is the best statistic} \end{array} $$

Now these are not "mathematically well posed" hypothesis, but if we use them anyway, and see what parts of the maths are required to make it well posed. The first part is the prior probabilities, without any data, how likely is each hypothesis? The usual answer is equal probabilities (but not always - may have some theoretical reason to support one hypothesis being more likely - the CLT is perhaps one for $H_{mean}$ being higher than the others).

So we use Bayes theorem to update each probability ($I$=prior information, $D$= data set):

$$P(H_{i}|D,I)=P(H_{i}|I)\frac{P(D|H_{i},I)}{P(D|I)}\implies \frac{P(H_{i}|D,I)}{P(H_{j}|D,I)}=\frac{P(H_{i}|I)}{P(H_{j}|I)}\frac{P(D|H_{i},I)}{P(D|H_{j},I)}$$

So if the prior probabilities are equal, then the relative probabilities are given by the likelihood ratio. So you also need to specify a probability distribution for what type of data sets you would be likely to see if the mean was the best statistic, etc. Note that each hypothesis doesn't actually state what the specific value of the mean, median, or IQR actually is. Therefore, the probability cannot depend on the exact value of the mean. Hence in the likelihoods these must have been "integrated out" using the sum and product rules

$$P(D|H_{i},I)=\int P(\theta_{i}|H_{i},I)P(D|\theta_{i},H_{i},I)d\theta_{i}$$

So you have the prior $P(\theta_{i}|H_{i},I)$ which can be interpreted for i=mean as "given that the mean is the best statistic, and prior to seeing the data, what values of the mean are we likely to see?" and the likelihood $P(D|\theta_{i},H_{i},I)$ can be similarly interpreted as "given the mean is best, and equal to $\theta_{mean}$ how likely is the data that was observed?". This may help you come up with some kinds of features that your distribution should have.

This describes the inference - now it is time to apply decision theory. This is particularly simple because your decision doesn't influence the state of nature - the statistic won't change if you do or don't use it. So we can describe the decisions ($A$ for "action" because $D$ is already taken):

$$\begin{array}{l l} A_{mean}:\text{The mean is the reported statistic} \\A_{med}:\text{The median is the reported statistic} \\A_{IQR}:\text{The IQR is the reported statistic} \end{array} $$

And now you need to specify a loss matrix $L_{ij}$ which relates the action/decision $A_{i}$ to the state of nature $H_{j}$ - what is the loss if I report the mean, but the median is actually the best statistic? In most cases the diagonal elements will be zero - taking the correct action means no loss. You may also have that all non-diagonal elements are equal - how you are wrong doesn't matter, only whether or not you are wrong.

You then proceed by calculating the average loss for each action, weighted by their probabilities:


And you then choose the action with the smallest average loss.


In your examples you seem to make the underlying assumption that you are "blind" to how the data you are going to analyze were generated. In other words, whether mean or median is the right tool for say something robust about central tendency of a sample of data largely depends on your (prior) knowledge of the data generating process. That's probably the main reason why most of us dealing with statistics are steadily focused on the same research fields (or their nearest outskirsts). As a sidelight, in real practice the 7000 value of your in your CASE 2 example would sound more like a mistaken data entry than a genuine outlier.


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