Need help solving CDF (Cumulative distribution function)

AI Thread Summary
The discussion revolves around calculating the cumulative distribution function (CDF) by hand, specifically the integral from -1 to 1 of the normal distribution function. The user seeks guidance on how to solve the integral of e^(-z^2/2) and expresses confusion about the substitution method. A helpful response suggests using the error function (erf) to express the result, specifically erf(1/sqrt(2)). The density function associated with this CDF is identified as the normal density function, confirming the user's understanding. The conversation concludes with the user expressing gratitude for the clarification.
invictor
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Hello, I am new. I been looking on the net for a guide how to solve the CDF by hand, i know the answer and I am about to crack this baby but I got stuck...

Im trying to calculate Cumulative distribution function by hand:
\int^{1}_{-1}\frac{1}{2\pi} e^{\frac{-z^{2}}{2}} dz or wolfram alpha: integrate 1/sqrt(2*pi) * e^(-z^2 /2) dz from -1 to 1

Anyway, this is the tricky part, how do this? (I left out the lefthand part above part for easier readability):

\int e^{\frac{-z^{2}}{2}} dz =

u = \frac{-z^{2}}{2}

du = -z dz

\frac{du}{-z} = dz

\int e^{u} \frac{du}{-z} =

then? How do i need to do?. Can any friendly soul here show me step by step how to solve this?

best regrads
invictor
 
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You can define
\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt
and express the result in terms of that, getting \operatorname{erf}(1/\sqrt{2}) as the answer.

You cannot give a (more) exact answer than that (otherwise we wouldn't be needing erf in the first place).
 
Ok so now you have the CDF in terms of the error function erf(/frac{1}{/sqrt{2}}) what is the associating PDF?
 
The density function for the error function is the normal density function, which is the integrand you started with.
 
Thanks i got it now :)
 
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