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Math Help - Integrating the Standard Normal Distribution

  1. #1
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    Integrating the Standard Normal Distribution

    hey i was trying to integrate the standard normal distribution to find where t wil equal one half of the distribution

    \int_{-\infty}^{t} \frac{1}{\sqrt{2\pi}}e^{(-x^{2})/2} dx = \frac{1}{2}
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    At t=0. Because from -oo to +oo the area is 1. So half way will give you 1/2.
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    Quote Originally Posted by ThePerfectHacker View Post
    At t=0. Because from -oo to +oo the area is 1. So half way will give you 1/2.
    true, but how do you integrate that... and evaluate it at negative infinity... i tried integrating it in maple, and it included the error function
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    Quote Originally Posted by maroon_tiger View Post
    true, but how do you integrate that... and evaluate it at negative infinity... i tried integrating it in maple, and it included the error function
    You cannot integrate it. It is not doable from the know functions. You need to know the properties of the Normal distribution, one of those properties is that the integral from -oo to +oo is 1.
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    don't you have one of these http://sweb.cz/business.statistics/normal.jpg ?
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    Quote Originally Posted by ThePerfectHacker View Post
    You cannot integrate it. It is not doable from the know functions. You need to know the properties of the Normal distribution, one of those properties is that the integral from -oo to +oo is 1.
    thanks, i never knew about the "know functions". integrating from -oo to +oo and getting 1 makes sense..
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    Quote Originally Posted by bobak View Post
    don't you have one of these http://sweb.cz/business.statistics/normal.jpg ?
    yeah, the main challenge was to see if i could throughly integrate that function and then evaluate it....
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    Quote Originally Posted by maroon_tiger View Post
    yeah, the main challenge was to see if i could throughly integrate that function and then evaluate it....
    Yes, life is certainly full of challenges.

    One of my main challenges is to see if I can flap my arms and fly. The probability of success is slightly greater for me than for you (1/googolplex versus 0).
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