Estimate percentile of mean from other percentiles From a data set, I know the mean, P10, P50 and P90.
How do I estimate the cumulative probability of the mean?
The reason is that my data set may have extreme events, pushing the mean value to a high percentile. 

It always hold that P50 < mean < P90. So I would like to have a better estimate of the "percentile" of the mean.For now I'm just using linear interpolation and the result is ~60%. But in this region a linear interpolation will give a lower value. 
 A: You can't do this without some strong assumptions about what the extreme tail can do.
You could have two different data sets (or indeed distributions), one where the mean is below the 10th percentile and one where the mean is above the 90th percentile, yet the three percentiles you mention could be the same for both. 
For example consider these two samples:
Sample 1:
40  96  97  98  99 100 101 102 103 104 105

Sample 2:
95  96  97  98  99 100 101 102 103 104 160

According to the quantile function in R (with default definition of sample quantiles), both these samples have the same values for the three quantiles you mentioned:
> quantile(sample1,p=c(.1,.5,.9))
10% 50% 90% 
 96 100 104 
> quantile(sample2,p=c(.1,.5,.9))
10% 50% 90% 
 96 100 104 

But the mean is below the 10th percentile in the first sample, and above the 90th percentile in the second:
> mean(sample1)
[1] 95
> mean(sample2)
[1] 105

With larger samples you could move the mean deeper, past more tail quantiles; you could put the mean below the first percentile or above the 99th percentile, for example.

Which is to say, if all you know is those three percentiles, you really can't pin the percentile of the mean down much at all -- it must lie above the minimum and below the maximum, but it could be above or below any other observation.
A: I believe the question can be summarized as
$$F[x]\equiv\Pr[X<x],\,F[x_{10},x_{50},x_{90}]=[0.1,0.5,0.9],\,x_{50}<\bar{x}<x_{90}\implies{F}[\bar{x}]=?$$
As noted by many responses, there is not enough information here to solve the problem uniquely.
However building on the OP suggestion of linear interpolation, a simple approach would be to just extend the order of the polynomial interpolation. So for example we might assume
$$F[t]=a+bt+ct^2 \,,\,t=\frac{x-x_{50}}{x_{90}-x_{50}}$$
where the coefficients must satisfy the constraints $F_0=0.5$ and $F_1=0.9$. From the definition of a CDF and PDF, we must also have $F^\prime[t]\geq{0}$.
Together these constraints give
$$a=F_0\,,\,c=F_1-F_0-b\,,\,0\leq{b}\leq{2}(F_1-F_0)$$
so we have
$$t^2\leq\frac{F[t]-F_0}{F_1-F_0}\leq{t}(2-t)$$
Note that $t\in[0,1]$, and in the linear case the expression in the middle reduces to $t$, so this includes the OP's linear interpolation solution as a special case.
If the OP was intended to be on the "tail", in the sense that $c\leq{0}$ so the PDF is decreasing over the interval, then the lower bound above can be increased from $t^2$ up to $t$ (while the upper bound stays the same, at $t(2-t)$). This is how I interpret the OP comment "in this region a linear interpolation will give a lower value".
Again I will end with the caveat that this is not a general solution, as there is no reason in general that a quadratic approximation will be appropriate for an arbitrary CDF of the $[x_{50},x_{90}]$ interval!
A: I am not aware of a function that does it. However you should be able to simply count. Some (python'ish) pseudo code:
len(data[data.value <= data.value.mean()]) / len(data.value)

A: Think I have the answer, from the definition of the mean using probabilities:
$$\overline x = \sum x_i p_i$$
Now I have mean, P10, P50 and P90:
$\overline x, x_{p10}, x_{p50}, x_{p90}$.
They are respectively associated to the cumulative probabilities: 
$q_{\overline x}, 0.1, 0.5, 0.9$.
And the best we can assume with this small set is that their probabilities are respectively (I know from my data that mean is larger than P50, or $\overline x > x_{p50} \to q_{\overline x} > 0.5$): 
$q_{\overline x}-0.5, 0.1, 0.4, 0.9-q_{\overline x}$
So:
$$\overline x = 0.1  x_{p10} + 0.4 x_{p50} + (q_{\overline x} - 0.5) \overline x + 
(0.9 - q_{\overline x})x_{p90}$$
$$\cdots$$
$$q_{\overline x}=\frac{0.1  x_{p10} + 0.4 x_{p50}+0.9 x_{p90}-1.5 \overline x}{x_{p90}-\overline x}$$
