Understanding Rayleigh distribution I am an aerospace enthusiast. I have obtained wind speed data that's been gathered and published by NCEP/NCAR. It provides wind speed data for every $2.5^\circ$ increment in latitude and longitude for 17 altitude levels. I want to predict the upper bound of the wind speed that a given location and altitude will face with a confidence percentage of 99. Wind speed data is usually modelled as a Rayleigh distribution. How can I find this upper bound?
I have tried something along these lines, please let me know if I'm going about it the right way.
The probability of a wind exceeding $v_i$ for each day ($d$), altitude ($h$), latitude ($\Phi$)
and longitude ($\lambda$) as
$P(d, h, \phi, \lambda)_i = e^{-\frac{\pi}{4}\left(\frac{v^2}{\mu^2}\right)}$
The percent availability (i.e., the percent of the time that the wind is less than $v_i$) is,
$\bar{P}(d,h,\phi,\lambda)_i = 1 - P(d,h,\phi,\lambda)_i \times 100%$
So the desired station keeping probability if set at 99%, we need to find $v_i$ that satisfies,
$0.99 = \frac{\sum_{d=1}^{365} P(d,h,\phi,\lambda)_i}{365}$
i.e., taking the mean of the probability over 365 days.
Is that right?

Wind speed (or the magnitude of any vector with orthogonal components sampled from normal distributions) follow a Rayleigh distribution (or $\chi_2$-distribution). The density and cummulative distribution functions are (from wiki),
\begin{equation}
f(x;\sigma) = \frac{x}{\sigma^2} e^{\frac{-x^2}{(2\sigma^2)}}
\end{equation}
\begin{equation}
F(x;\sigma) = 1 - e^{\frac{-x^2}{(2\sigma^2)}}
\end{equation}
Distribution mean:
\begin{equation}
\mu(X) = \sigma \sqrt{\frac{\pi}{2}}
\end{equation}
We are interested in finding the velocity threshold where the probability of sampled velocity will be less than the threshold is 0.99. That is, we need $v_T$, such that,
\begin{equation*}
Pr(V \leq v_T) = F(v_T;\mu) = 1 - e^{\frac{-\pi}{4}\left(\frac{v_T}{\mu}\right)^2} = 0.99
\end{equation*}
using sampled mean over a year ($\hat{\mu}$),
\begin{equation*}
\hat{\mu} = \frac{\sum_{d=1}^{365}v_{d}}{365}
\end{equation*}
Inversing (with $0.99$ as $p_T$),
\begin{equation*}
v_T = \frac{-2\hat{\mu}}{\sqrt{\pi}} ln(1 - p_T) \approx 5.2 \hat{\mu}
\end{equation*}
Edit:
\begin{equation*}
v_T = \frac{2\hat{\mu}}{\sqrt{\pi}} \sqrt{-\ln(1 - p_T)} \approx 2.4 \hat{\mu}
\end{equation*}
I made an error in coming up with inverse.
Would also suggest you post it on stat stack exchange too. Please link the answer if they provide a more convincing one in your question.
