chisigma said:
Posted on 06 15 2012 on
www.matematicamente.it by the member Edwavit [original in Italian…] and not yet solved…
Hello boys!... a question I can’t solve: X is a Gaussian r.v. with $\mu = 10$ and $\sigma=4$ and $P(X>C)=.05$. Find C…
Setting $\frac{C-\mu}{\sigma}=x$ is...
$\displaystyle P\{X<C\}= \frac{1}{2}\ \{1+ \text{erf}\ (\frac{x}{\sqrt{2}})\} = .95$ (1)
... where...
$\displaystyle \text{erf}\ (t)= \frac{2}{\sqrt{\pi}}\ \int_{0}^{t} e^{- \xi^{2}}\ d \xi$ (2)
... so that from (1) we derive...
$\displaystyle x= \sqrt{2}\ \text{erf}\ ^{-1} (.9)$ (3)
Now of course the problem is the computation of the function $\text{erf}\ ^{-1} (*)$. In...
http://www.mathhelpboards.com/threads/1223-erf
… it has been explained how to find the coefficients of the McLaurin expansion of the function $\text{erf}\ ^{-1} (*)$ and the task defined as ‘tedious but not very difficult’. In …
http://mathworld.wolfram.com/InverseErf.html
… we discover that the ‘tedious task’ has been done by somebody some years ago and the result is the series expansion…
$\displaystyle \text{erf}\ ^{-1} (\frac{2\ x}{\sqrt{\pi}}) = \sum_{n=0}^{\infty} a_{n}\ x^{2n+1}$ (4)
... where...
$\displaystyle a_{n}=\frac{c_{n}}{2n+1}$ (5)
... with $c_{n}$ solution of the difference equation...
$\displaystyle c_{n}= \sum_{k=0}^{n-1} \frac{c_{k}\ c_{n-k-1}}{(k=1)\ (2\ k+1)},\ c_{0}=1$ (6)
The first $a_{n}$ are $a_{0}=1$,$a_{1}= \frac{1}{3}$, $a_{2}= \frac {7}{30}$, $a_{3}= \frac{127}{630}$,... The remaining computation is relatively 'comfortable' and are left to the reader...
Kind regards
$\chi$ $\sigma$
