# In left skewed data, what is the relationship between mean and median?

I found this question on the open MOOC from Stanford, however the answer is not present.

I think the median $\leq$ mean.

Is this the case?

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Which open MOOC course is that? What do the course materials suggest the answer should be? –  Glen_b Mar 9 at 23:35
class.stanford.edu/courses/Medicine/HRP258/… . The course is over. –  Kunjan Kshetri Mar 10 at 3:40
Thanks, that's some context at least, though all that's left there are weekly readings which don't throw much light on this issue. I do wonder what the course had to say on the topic. –  Glen_b Mar 10 at 4:18

It's a nontrivial question (certainly not as trivial as I would guess the people asking the question think).

The difficulty is ultimately caused by the fact that we don't really know what we mean by 'skewness' - a lot of the time it's kind of obvious, but sometimes it really isn't. Given the difficulty in pinning down what we mean by 'location' and 'spread' in nontrivial cases (for example, the mean isn't always what we mean when we talk about location), it should be no great surprise that a more subtle concept like skewness is at least as slippery. So this leads us to try various algebraic definitions of what we mean, and they don't always agree with each other.

1) If you measure skewness by the second Pearson skewness coefficient, then the mean will be less than the median (i.e. in this case you have it backwards).

The (population) second Pearson skewness is $$\frac{3(\mu-\stackrel{\sim}{\mu})}{\sigma}\,,$$ and will be negative ("left skew") when $\mu<\stackrel{\sim}{\mu}$.

The sample versions of these statistics work similarly.

The reason for the necessary relationship between mean and median in this case is because that's how the skewness measure is defined.

Here's a left-skewed density (by both the second Pearson measure and the more common measure in (2) below):

The median is marked in the lower margin in green, the mean in red.

So I expect the answer they want you to give is that the mean is less than the median. It's usually the case with the sorts of distributions we tend to give names to.

(But read on, and see why that's not actually correct as a general statement.)

2) If you measure it by the more usual standardized third moment, then it is often, but by no means always, the case that the mean will be less than the median.

That is, it's possible to construct examples where the opposite is true, or where one skewness measure is zero while the other is non-zero.

Which is to say, there's no necessary relationship between the locations of the mean, median and the moment-skewness.

Consider, for example, the following sample (the same example can be constructed as a discrete probability distribution):

  2.7 15.0 15.0 15.0 30.0 30.0

mean: 17.95
median: 15


Yet the (Fisher, third-moment) skewness coefficient is negative (i.e. by its lights, we have left-skew data) since the sum of the cubes of the deviations from the mean is negative.

So in that case, left-skew, but mean>median.

(On the other hand, if you change 2.7 in the above example to 3, then you have an example where the moment-skewness is zero, yet the mean exceeds the median. If you make it 3.3, then the moment-skewness is positive, and the mean exceeds the median - i.e. is finally in the 'anticipated' direction.)

If you use the first Pearson skewness instead of either of the above definitions, you have a similar issue to this case - the direction of the skewness does not pin down the relation between mean and median in general.

Edit: in answer to a question in comments -- an example where the mean and median are equal, but the moment-skewness is negative. Consider the following data (as before, it also counts as an example for a discrete population; consider writing the numbers on the faces of a die).

 1  5  6  6  8 10


the mean and the median are both 6, but the sum of cubes of deviations from the mean are negative, so the third moment skewness is negative.

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@Peter Sorry for the slow reply, I was busy constructing just such examples and didn't see your question. –  Glen_b Mar 9 at 22:40
I've seen a lot of textbook definitions and none mentioned this. Cool. –  Peter Flom Mar 9 at 22:48
@Peter Unfortunately, a lot of elementary textbooks simply repeat incorrect information from other textbooks without actually doing any real investigation themselves, and so a basic misconception gets propagated. Counterexamples are, as you see, relatively easy to construct (I just make them by hand as needed). Kendall and Stuart (Advanced theory of Statistics, Vol I - don't let the title put you off, it's quite readable), at least the third and fourth editions, have good information. More recent editions are by Stuart and Ord. I've actually posted about this issue on CV a number of times. –  Glen_b Mar 9 at 22:53
Binomials ${5 \choose k} 0.8^k 0.2^{5 - k}$ and ${5 \choose k} 0.2^k 0.8^{5 - k}$ show that mean $=$ median is perfectly consistent with asymmetry. A point about this example is that no-one can convincingly dismiss it as obscure or pathological. –  Nick Cox Mar 9 at 23:45
@Nick Yes, binomials with integer mean are great examples. –  Glen_b Mar 10 at 2:41

No. Left skewed data has a long tail on the left (low end) so the mean will usually be less than the median. (But see @Glen_b 's answer for an exception). Casually, I think data that "looks" left skewed will have mean less than median.

Right skewed data is more common; for instance, income. There the mean is greater than the median.

R code

set.seed(123)  #set random seed
normdata <- rnorm(1000) #Normal data, skew = 0
extleft <- c(rep(-10, 5), rep(-20, 5)) #Some data to make skew left
alldata <- c(normdata,extleft)

library(moments)
skewness(alldata) #-6.77
mean(alldata) #-0.13
median(alldata) #-0.001

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Can the mean ever be equal to the median? –  Kunjan Kshetri Mar 9 at 22:13
unj2 I added an example to my answer where the third-moment skewness is negative but mean=median. –  Glen_b Mar 9 at 23:18