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How to directly compute an integral which corresponds to the normal distribution

Is there any approximate solution for the following definite integral of normal distribution?

$$\int_{a}^{b} e^{-\frac{(x-\mu)^2}{2s^2}} \ dx$$

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marked as duplicate by Qiaochu Yuan May 30 '11 at 9:00

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See this Wikipedia page for suggestions. –  Stijn May 27 '11 at 7:25
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1 Answer

up vote 5 down vote accepted

You should look into the error function and its approximations

$$\frac{1}{\sqrt{2\pi}\sigma}\int_a^b e^{\frac{-(x-\mu)^2}{2\sigma^2}}\ dx=\frac{1}{2}\left[\text{erf}\left(\frac{b-\mu}{\sqrt{2}\sigma}\right)-\text{erf}\left(\frac{a-\mu}{\sqrt{2}\sigma}\right)\right]$$

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