I don't think median $\pm$ mad is appropriate in general.
You can easily build distributions where 50% of the data are fractionally lower than the median, and 50% of the data are spread out much greater than the median - e.g. (4.9,4.9,4.9,4.9,5,1000000,1000000,100000,1000000). The 5 $\pm$ 0.10 notation seems to suggest that there's some mass around (median + mad ~= 5.10), and that's just not always the case, and you've got no idea that there's a big mass over near 1000000.
Quartiles/quantiles give a much better idea of the distribution at the cost of an extra number - (4.9,5.0,1000000.0). I doubt it's entirely a co-incidence that the skewness is the third moment and that I seem to need three numbers/dimensions to intuitively visualize a skewed distribution.
That said, there's nothing wrong with it per se - I'm just arguing intuitions and readability here. If you're using it for yourself or your team, go crazy. But I think it would confuse a broad audience.
Thank you for this simple-yet-profound question about the fundamental statistical concepts of mean, median, and mode. There are some wonderful methods /demonstrations available for explaining and grasping an intuitive -- rather than arithmetic -- understanding of these concepts, but unfortunately they are not widely known (or taught in school, to my knowledge).
Mean:
1. Balance Point: Mean as the fulcrum
The best way to understand the concept of mean it to think of it as the balance point on a uniform rod. Imagine a series of data points, such as {1,1,1,3,3,6,7,10}. If each of these points are marked on a uniform rod and equal weights are placed at each point (as shown below) then the fulcrum must be placed at the mean of the data for the rod to balance.
This visual demonstration also leads to an arithmetic interpretation. The arithmetic rationale for this is that in order for the fulcrum to balance, the total negative deviation from the mean (on the left side of the fulcrum) must equal to the total positive deviation from the mean (on the right side). Hence, the mean acts as the balancing point in a distribution.
This visual allows an immediate understanding of the mean as it relates to the distribution of the data points. Other property of the mean that becomes readily apparent from this demonstration is the fact that the mean will always be between the min and the max values in the distribution. Also, the effect of outliers can be easily understood – that a presence of outliers would shift the balancing point, and hence, impact the mean.
2. Redistribution (fair share) value
Another interesting way to understand the mean is to think of it as a redistribution value. This interpretation does require some understanding of the arithmetic behind the calculation of the mean, but it utilizes an anthropomorphic quality – namely, the socialist concept of redistribution – to intuitively grasp the concept of the mean.
The calculation of the mean involves summing up all values in a distribution (set of values) and dividing the sum by the number of data points in the distribution.
$$
\bar{x} = (\sum_{i=1}^n{x_i})/n
$$
One way to understand the rationale behind this calculation is to think of each data point as apples (or some other fungible item). Using the same example as before, we have eight people in our sample: {1,1,1,3,3,6,7,10}. The first person has one apple, the second person has one apple, and so on. Now, if one wants to redistribute the number of apples such that it is “fair” to everyone, you can use the mean of the distribution to do this. In other words, you can give four apples (i.e., the mean value) to everyone for the distribution to be fair/equal. This demonstration provides an intuitive explanation for the formula above: dividing the sum of a distribution by the number of data points is equivalent to partitioning the whole of the distribution equally to all of the data points.
3. Visual Mnemonics
These following visual mnemonics provide the interpretation of the mean in a unique way:
This is a mnemonic for the leveling value interpretation of the mean. The height of the A's crossbar is the mean of the heights of the four letters.
And this is another mnemonic for the balance point interpretation of the mean. The position of the fulcrum is roughly the mean of the positions of the M, E, and doubled N.
Median
Once the interpretation of mean as the balancing point on a rod is understood, the median can be demonstrated by an extension of the same idea: the balancing point on a necklace.
Replace the rod with a string, but keep the data markings and weights. Then at the ends, attach a second string, longer than the first, to form a loop [like a necklace], and drape the loop over a well-lubricated pulley.
Suppose, initially, that the weights are distinct. The pulley and loop balance when the same number of weights are to each side. In other words, the loop ‘balances’ when the median is the lowest point.
Notice that if one of the weights is slid way up the loop creating an outlier, the loop doesn’t move. This demonstrates, physically, the principle that the median is unaffected by outliers.
Mode
The mode is probably the easiest concept to understand as it involves the most basic mathematical operation: counting. The fact that it’s equal to the most frequently occurring data point leads to an acronym: “Most-often Occurring Data Element”.
The mode can also be thought of the most typical value in a set. (Although, a deeper understanding of ‘typical’ would lead to the representative, or average value. However, it’s appropriate to equate ‘typical’ with the mode based on the very literal meaning of the word ‘typical’.)
Sources:
- The Median is a balance point -- Lynch, The College Mathematics Journal (2009)
- Making Statistics Memorable: New Mnemonics and Motivations -- Lesser, Statistical Education, JSM (2011)
- On the Use of Mnemonics for Teaching Statistics -- Lesser, Model Assisted
Statistics and Applications, 6(2), 151-160 (2011)
- What does the mean mean? – Watier, Lamontagne and Chartier, Journal of Statistics Education, Volume 19, Number 2 (2011)
- Typical? Children's and Teachers' Ideas About Average – Russell and Mokros, ICOTS 3 (1990)
OVERALL REFERENCE: http://www.amstat.org/publications/jse/v22n3/lesser.pdf
Best Answer
I disagree with the advice as a flat out rule. (It's not common to all books.)
The issues are more subtle.
If you're actually interested in making inference about the population mean, the sample mean is at least an unbiased estimator of it, and has a number of other advantages. In fact, see the Gauss-Markov theorem - it's best linear unbiased.
If your variables are heavily skew, the problem comes with 'linear' - in some situations, all linear estimators may be bad, so the best of them may still be unattractive, so an estimator of the mean which is not-linear may be better, but it would require knowing something (or even quite a lot) about the distribution. We don't always have that luxury.
If you're not necessarily interested in inference relating to a population mean ("what's a typical age?", say or whether there's a more general location shift from one population to another, which might be phrased in terms of any location, or even of a test of one variable being stochastically larger than another), then casting that in terms of the population mean is either not necessary or likely counterproductive (in the last case).
So I think it comes down to thinking about:
what are your actual questions? Is population mean even a good thing to be asking about in this situation?
what is the best way to answer the question given the situation (skewness in this case)? Is using sample means the best approach to answering our questions of interest?
It may be that you have questions not directly about population means, but nevertheless sample means are a good way to look at those questions (estimating the population median of a waiting time that you assume to be distributed as ab exponential random variable, for example is better estimated as a particular fraction of the sample mean) ... or vice versa - the question might be about population means but sample means might not be the best way to answer that question.