Yes! The formula for weighted mean in the documentation is wrong - but the result is correct. More clearly:

In[70]:= nbpoints = 10;
d = RandomReal[{-1, 1}, nbpoints];
w = Table[1, nbpoints];
Proba = w/Total@w; (* Probability*)
UsualMean = Mean@d
WeightedMean = 
 Mean[WeightedData[d, 
   Proba]] (*Correct, but different from the documentation of Mean!*)
DocMean = (1/nbpoints) Total[
   Proba d](*According to the documentation of Mean for WeightedData *)

Out[74]= -0.268511

Out[75]= -0.268511

Out[76]= -0.0268511

docMean is wrong. As for the variance, compute :

WeightedVar = 
 Variance[
  WeightedData[d, 
   Proba]]  (*  Problematic definition in the documentation, \
for  WeightedData*)
Good = Total[Proba (d - WeightedMean)^2] 
DocVar = (1/(nbpoints - 1)) Total[
   Proba (d - 
       Mean@d)^2](*According to the documentation of Mean for Weighted\
Data *)

WeightedVar =Good, which is correct, I think ; but DocVar is wrong too, and the problem lies in the documentation, not in the computations - a lesser evil !

To add to David's point, data will be treated as population data if wrapped in EmpiricalDistribution[]:

Variance[EmpiricalDistribution@data]

This is documented in Mean[], Variance[], and StandardDeviation[] under "Details":

The data can have the following additional forms and interpretations:... WeightedData — weighted mean, based on the underlying EmpiricalDistribution  »

(This comes a couple of items after the erroneous formulae that have been pointed out already.)

I found this and I am still thinking about it: https://stats.stackexchange.com/questions/51442/weighted-variance-one-more-time

Variance of WeightedData gives the population variance, while Variance on a set of raw observations gives the unbiased sample variance.