Approximate mean from .25, .5, .75 percentiles If I have median (.5 percentiles), and the .25 and .75 percentiles, what is the best guess of the mean?
If the .25 and .75 percentiles are in the middle of a probability mass with weight .5 (.25 on each sides) and the median has .5 on each of its sides,
does it makes sense to approximate the average with:
(`.25 pctile` + 2*`median` + `.75 pctile`) / 4

I do not have more granularity as in this case.
I would like to avoid fitting a distribution in some parametric sense as in here
 A: Without imposing some kind of assumption, we can say almost nothing.
It's possible for the population mean to be anything at all - any value on the real line... or possibly infinite, or undefined.
If the underlying distribution is symmetric and unimodal and more or less normalish, your formula would do reasonably well, but symmetry is a pretty strong assumption.
To see that you can get any finite mean at all - even with a sample - consider a sample consisting of five each of the values 1, 2, 3, 4 and 5 and one additional value.
The estimator you propose should always be 3 no matter what the 26th value is. To get any finite mean $m$, choose a value for $m$ and make the additional value $26(m-3) + 3$.
That said, for reasonably symmetric distributions whose mean exists, your suggested estimator (the trimean) often performs quite well as an estimate of the mean; it does well at the normal, and typically better for cases with more of a peak/heavier tail.
If you expect your population distribution to be close to normal, equal weight to the three quartiles is slightly more efficient [1]. If you think it could be substantially more peaked (but still symmetric, like a $t$ or a logistic for example), the trimean will generally be an excellent choice.
[1] Doyle, John R. and Chen, Catherine Huirong, (2009)
"On the efficiency of the Trimean and Q123"
Journal of Statistics and Management Systems,
12:2, pp319-323
A: Without knowing anything about the distribution your percentiles are coming from, I do not think you can tell much about the mean. What you can do is regard the three points you have as your data and calculate the weighted mean, which is what your formula describes.
I would not consider this measure as the mean of your original data though. It might be the best approximation, but the relationship between the mean and the median is heavily dependent on the distribution and the two quartiles don't contribute much to your knowledge.
