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Thread: Solving normal distribution using integration and not the table

  1. #1
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    Solving normal distribution using integration and not the table

    The question is: If a normal distribution with mean mu and variance > 0 has 46th percentile equal to 20*sigma, then what is mu in terms of standard deviation? I know the answer should be mu = 20.1*sigma by solving using the standard normal table, but I need to solve it using integration.
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  2. #2
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    Re: Solving normal distribution using integration and not the table

    "Forty sixth percentile 20 sigma" means that \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{20\sigma} e^{-\frac{(x- \mu)^2}{2\sigma^2}}dx= 0.46. You want to solve that for \mu in terms of \sigma.

    What do you mean by "solve it by integration"? You can use the usual substitution to reduce to the "standard normal distribution" and then look the result up in a table but the whole reason there is a table is that e^{-z^2} cannot be integrated in terms of an elementary function, the best you can do is integrate numerically.
    Last edited by HallsofIvy; Apr 6th 2017 at 03:22 AM.
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