# Weibull distriution. Problem with paramets and qqPlot

I have 10 years worth of mean wind speed meteorological data (timely 10-minutes data). I want to compare the observed wind speed data against a Weibull distribution on a graph of the probability density distribution. I used function in R:fitdistr(). But the Weibull paramets, which gave me this function is probably wrong? Because when I made qqPlot with this paramets, I had something like this:

Help me please and tell me, what I have been wrong? What should I do?

• You have made the comparison and found your data do not look like they have a Weibull distribution. What that means to you depends on why you were doing this. What was your purpose? – whuber Feb 17 '15 at 22:44
• 1. Some advice: Show much narrower bins in your histogram so you can see more detail. Since you have plenty of data, I'd suggest bins equal to the discrete intervals your data are already binned to. 2. Your data show a heavier right tail than the Weibull that would describe the middle part of the data, and a heavier left tail as well. 3. If your data were close to a Weibull ... what you would do? – Glen_b -Reinstate Monica Feb 18 '15 at 2:05
• I'm writing my Master Thesis at the University about Extreme Wind Speed in Poland. I have a kind of data like I wrote above and I would like to analyse my data by means of Weibul distribution (You know, I want to find the percentiles [0,9; 0,95; 0,999] by menas of qweibull() and I need to do this the paramets of Weibull distribuation). I'm not to good with statistic and I need some help. – Arek Feb 18 '15 at 16:25
• A heavier-tailed distribution seems to be called for; it may be for example that a scale-mixture of Weibulls might be sufficient. – Glen_b -Reinstate Monica Oct 9 '16 at 2:27

You don't need to fit this using a Weibull-distribution. Given $N$ observation, sort the vector of observations and take the 1st, the $N0.9$th,$N0.95$th, etc