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Simo · #12 November 24th, 2008, 10:07 AM

doh...

ok, I've made another mistake...
the equation that I've actually solved is

$\frac{1}{\sigma \sqrt{2\pi }}\int_{-\infty }^{x=\bar{X}}\exp \left( -\frac{\left( x-\mu \right)^{2}}{2\sigma ^{2}}\right) dx = \frac{K}{\sigma \sqrt{2\pi }}\int_{x=\bar{X}}^{+\infty }\exp \left( -\frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx$

instead of

$\frac{1}{\sigma \sqrt{2\pi }}\int_{x=\bar{X}}^{+\infty }\exp \left( -\frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx=\frac{K}{\sigma \sqrt{2\pi }}\int_{-\infty }^{x=\bar{X}}\exp \left( -\frac{\left( x-\mu \right)^{2}}{2\sigma ^{2}}\right) dx$

that was the first one posted and pasted in the post with the scratch of the solution...I've switched the integrals

sorry