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David24 · #2 November 24th, 2008, 05:03 AM

Quote:

Originally Posted by Simo

Dear all,
I need to know if is it possible to find the mean of a normal distribution knowing the ratio of the area under the "bell" curve on the left and on the right of a given point.
To be more precise I need to solve the following equation
$\frac{1}{\sigma \sqrt{2\pi }}\int_{x=\bar{X}}^{+\infty }\exp \left( \frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx=\frac{k}{\sigma \sqrt{2\pi }}\int_{x=-\infty}^{\bar{X} }\exp \left( \frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx$
where I know the ratio $k$ , the variance $\sigma ^{2}$ and the point $\bar{X}$ is fixed. i need to find the mean of the distribution $\mu$ that achieve the ratio of $k$ between the two portion of area.

thanks in advance.

hey mate, just so Im reading your your question correctly, you have an integral equation of the form

int(f(x)) = k*inf(g(x)) and you want to solve for k?

Regards,

David