A Normally distributed random variable with mean [mu] has a probability density function given by
[gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).
Its standard deviation is given by:
*Sorry I'm still learning how to use Latex
A Normally distributed random variable with mean [mu] has a probability density function given by
[gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).
Its standard deviation is given by:
*Sorry I'm still learning how to use Latex
Sorry but I can't make sense of the expression you have posted for the pdf.
If it's a normal distribution then it has the usual form found here: Normal distribution - Wikipedia, the free encyclopedia
So unless you can re-post a clearer expression I suggest you read the link, fit your pdf to the normal pdf given in the link and then read off what the variance and hence standard deviation is.
Is it $\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}}$ ?
That can't be right. It has to be either $\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y^2 (x - \mu)^2}{2 \sigma}}$ or $\displaystyle \frac{\sqrt{y}}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}}$
Assuming the former expression, then I would have thought that the comparison with the pdf in the link I gave you is straightforward:
$\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} = \frac{1}{\sigma_X \sqrt{2 \pi}}$ therefore $\displaystyle \sigma_X = \, ....$
If it's the latter expression, then you do a similar thing.