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Calculus and Beyond Homework Help
Normal Distribution: PDF of a Normally Distributed Random Variable
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[QUOTE="lanedance, post: 3706246, member: 162674"] Don't worry too much about the constant at the front, this is just to normalise the distribution (though it will be a useful check), the exponential gives it its shape. As sigma has already be taken, to prevent confusion, let's call the standard deviation d so you have [tex] e^{-\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}}[/tex] a normal distribution with standard deviation d, has [tex] e^{-\frac{(x-\mu)^2}{2d^2}}[/tex] equating them we have [tex] \frac{(x-\mu)^2}{2d^2}=\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}[/tex] once you solve for d, you can check it makes sense in the normalisation constant. [/QUOTE]
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Normal Distribution: PDF of a Normally Distributed Random Variable
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