# Thread: Solving normal distribution using integration and not the table

1. ## 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.

2. ## 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.

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