My question is: Is there a set of data which allow the median to be larger than the mode, the mode to be larger than the mean and the mean to be larger than the range? If so, is there a pattern, or a specific characteristic of a dataset to allow this situation (skewness of some kind maybe...)?

P.S I have corrected my typo mistake. Some of the answers already given relate to the opposite situation for which median

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    $\begingroup$ Your title has $>$ signs, but the text says "smaller" in each case. The answer will be the same either way, but best to make your question consistent. $\endgroup$
    – Nick Cox
    Commented Dec 12, 2016 at 17:10
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    $\begingroup$ There's not much meaning to such criteria. That's because (a) by concentrating a small probability within a very narrow band, you can create a mode at any value without appreciably changing the mean or median; (b) by putting a tiny probability on an extreme value you can put the mean anywhere within the range without changing the mode or median appreciably; and (c) by including extraordinarily large or small values with very tiny probabilities you can make the range as large as you please without appreciably changing any other properties. $\endgroup$
    – whuber
    Commented Dec 12, 2016 at 17:27
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    $\begingroup$ It also doesn't make sense to include range, a measure of the width of distribution, with three measures of central tendency. $\endgroup$ Commented Dec 12, 2016 at 17:28
  • $\begingroup$ @Giuseppe Biondi-Zoccai Your edit is intended to be helpful, but we don't usually edit questions whenever it is possible that the OP is confused on a technical point, even simple use of notation. $\endgroup$
    – Nick Cox
    Commented Dec 12, 2016 at 18:51
  • $\begingroup$ @prince_of_pears Dimensionally range has the same units as the other entities, so comparisons do make mathematical sense. I agree that off-hand I can't see statistical purpose to such comparisons, but that is a different matter and may be part of the question that the OP needs to clarify. Consider that there are plenty of contexts in which comparison of SD and mean makes sense and it's not customary then to object that one measures width and the other location. $\endgroup$
    – Nick Cox
    Commented Dec 12, 2016 at 18:55

3 Answers 3


The question has already been answered in the affirmative, but let's approach this from the point of view of construction -- how do we make a set of data that does this?

First, note that we can always make all three location-measures greater than the range. Simply construct a preliminary data set that has median > mode > mean and compute the range. Now add (range-mean) + $\epsilon$ (for some small positive $\epsilon$) to all of the data values to get the final data set, whereupon the three location-measures will all exceed the range.

So we have now reduced the problem to one of finding a data set where median > mode > mean .

Imagine we already had some data with a suitable median and mode. To make the mean smaller than the median and mode, you simply place a single value far enough below the bulk of the data that the mean is pulled down; we can place a second value just above the bulk of the data to keep the median where it was, without changing the mode. So now we can modify an existing data set that simply has median > mode and obtain one which has the mean where we want.

So let us create one with median > mode. We can do this by having one value repeated (if it's the only value that occurs twice, it's the sample mode) and then adding enough other values to make the median larger. This is an example:

 21, 21, 22, 23, 24

The median is 22 but the mode is 21.

Now let's add the two points as previously described, in such a way to make the mean 20 without changing the median or mode. The present points sum to 111, so we need two points that add to 140-111 = 29, and one of them should be just larger that 24. Let's make it 25. Then the smaller point is 29-25 = 4.

So now our data set is:

4, 21, 21, 22, 23, 24, 25

It has mean 20, mode 21 and median 22.

Now let's fix the relationship of those with the range. What's the range? It's 25-4=21, which is presently larger than the mean. We need simply add something to every data value to make the mean larger than 21, which leaves the range unaltered. Adding 2 will suffice. (Note that range-mean+1=2, so we can see that we took $\epsilon=1$)

So our final data set is

6, 23, 23, 24, 25, 26, 27

The range is still 21, the mean is now 22, the mode is 23, the median is 24

So this step by step approach is quite easy to use. In summary:

  1. Make a small data set with median > mode by repeating the smallest value and having all the larger values distinct (it's easiest to use sorted values). Having 5 points is convenient (since it lets you specify the median by moving the middle value) but 4 is feasible if needed.

  2. Obtain a mean below the median by adding two points that don't alter the median or mode (i.e. two distinct/singleton values will not disturb the mode, and placing them one either side the previous data will preserve the median; place the larger value just above all the present data and then compute the smallest so that the overall mean comes out just below the mode. This takes us to 7 data points.

  3. Compute the range. Add a constant (range - mean + $\epsilon$) to all the data values, which guarantees that the mean exceeds the range. This is the final data set.

Checking those calculations in R:

x <- c(6, 23, 23, 24, 25, 26, 27)

  range mean mode median
1    21   22   23     24

(note that if we somehow happened to generate more than one mode, this calculation tries to find the largest of them)

  • $\begingroup$ Thank you, this explanation is absolutely fantastic. Is there a theoretical characteristic of such a series of numbers? What I mean is, it is known that if mean>median>mode, the distribution is positive skewed. On the other hand, if mean<median<mode it is negatively skewed. Is there a distributional characteristic in this constraint? $\endgroup$
    – BlueSigma
    Commented Dec 13, 2016 at 6:59
  • $\begingroup$ 1. If you define skewness in terms of a relationship between mean and median (like the second Pearson skewness coefficient, median-skewness) or in terms of a relationship between mode and mean (like the first Pearson skewness coefficient; mode-skewness) then a distribution with mean>median>mode is positively skewed. Otherwise, it's not necessarily the case -- for example if I define skewness in terms of the third central moment of a standardized variable (moment-skewness) ... ctd $\endgroup$
    – Glen_b
    Commented Dec 13, 2016 at 23:00
  • $\begingroup$ ctd ... or in terms of quartile-skewness then it is not necessarily the case that mean>median>mode implies positive skewness. 2. I'm not 100% sure what you're asking ... the relevant characteristic of the set of numbers is that the statistics you mentioned are in the desired order, by construction. Statistics derived from differences in those statistics (like the median skewness or mode skewness) will be of the implied sign, but it's a consequence of meeting the conditions. Outside that I am not sure what you seek here. $\endgroup$
    – Glen_b
    Commented Dec 13, 2016 at 23:04

Yes, it's not hard to come up with such a set.

S = {0, 1, 2, 3, 4, 4, 1000}

Median = 3, Mode = 4, Mean = 144.85, Range = 1000

Data of this kind will skew to the right, since your mean is higher than the median, implying that on average, values above the median are further away than values below.

  • 2
    $\begingroup$ Be advised that the question has been clarified to have the inequalities go the other way - but that's easily fixed with a slight adjustment of the values: {1000, 1996, 1996, 1997, 1998, 1999, 2000 } : median 1997, mode = 1996, mean = 1855.14, range = 1000 $\endgroup$
    – R.M.
    Commented Dec 12, 2016 at 22:04
  • $\begingroup$ Conceptually I think it is better to think of this in terms of probability distributions. The mean the median and the mode are numerical values for the random variable. The range is the length of the set of possible values. It is not comparable to the other parameter. For a normal distribution the mean, median and mode are all equal. There is an issue of definition about the mode. If there is just one peak in the density there is no ambiguity. But if you have more than one peak some define the mode to be the highest peak while others say all peaks are modes; $\endgroup$ Commented Dec 12, 2016 at 22:11
  • $\begingroup$ In the case of a uniform distribution there are no modes because.there are no peaks. For symmetric unimodal distributions with finite mean the mean equals the mode. For skewed distributions satisfying the conditions for mean and median and mode to all exist and be unique any order is possible but they cannot all be equal. $\endgroup$ Commented Dec 12, 2016 at 22:17
  • $\begingroup$ @Michael Chernick Not so. 0, 0, 1, 1, 1, 1, 3 has mean, median and mode identical at 1 but it is not symmetric. Lest this be thought contrived, the binomial ${10 \choose k} 0.1^k 0.9^{10 - k}, k = 0, \dots, 10$ is manifestly skewed and has mean, median and mode identical at 1, and there are other such cases. $\endgroup$
    – Nick Cox
    Commented Dec 13, 2016 at 9:26
  • $\begingroup$ On uniform distributions, I agree that there is no useful mode but it would also be possible to argue that every possible value is a mode. $\endgroup$
    – Nick Cox
    Commented Dec 13, 2016 at 9:27

Irrespective of what the order is, the answer is yes. Data sets that are subsets of distributions, whose left tails are heavier than their right tails will frequently have the mode smaller than the median and the median smaller than the mean and the mean smaller than the range. A beta distribution with the mode greater 1/2 would have that property. If one wants to have the mode in any particular position, one can make a mixture distribution by adding in a small percentage of a narrow (small) standard deviation but tall distribution, e.g., Dirac $\delta$, wherever one wants to put that mode.

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    $\begingroup$ Though for unimodal distributions in a particular sense, having the mean between between the mode and median might be thought more common with discrete random variables than with continuous random variables: see Paul T. von Hippel, Mean, Median, and Skew: Correcting a Textbook Rule, Journal of Statistics Education Volume 13, Number 2 (2005) or my thoughts. The textbook rule, being too confident about a statement by Karl Pearson, is to have the median between mode and mean $\endgroup$
    – Henry
    Commented Dec 12, 2016 at 21:42
  • $\begingroup$ Good point (+1), and the mode can be anywhere. $\endgroup$
    – Carl
    Commented Dec 12, 2016 at 22:03

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