How to approximate the integral of the standard normal distribution.

Question

So I have this eqn.

$$ f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}} $$

I need to find:

$$ \int\limits_{-1}^1 f(x)dx $$

So I want to use this series to integrate. I know that:

$$ e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$ e^{\frac{-x^2}{2}} =\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!} $$

This is where I get lost. What do I do with the coefficient of e. How do I build this into a series and include the coefficient of e:

$$ \frac {1}{\sqrt{2\pi}}\ $$

Answer 1

Ok I figured it out.

$$ f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}} $$

I need to find:

$$ \int\limits_{-1}^1 f(x)dx $$

So I want to use this series to integrate. I know that:

$$ e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$ e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!} $$

So we can just build up and get the origial.

$$ \frac {1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=\frac {1}{\sqrt{2\pi}} \sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!} $$

Then all we need to do is expand out the series to a few values of n, then we can anti-differentiate the series, and then we can solve using the FTC, where the integral from [a,b] = F(b) - F(a).

Answer 2

Hint: You cannot find an explicit answer for your integration. However the integration of normal distribution can be stated in terms of Q Function. $$ Q(x)=\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}}dt $$ There are various estimates for this function which may be useful. For instance for $x>0$: $$ \frac{x}{(x^2+1)\sqrt{2\pi}}e^{\frac{-x^2}{2}} \leq Q(t)\leq \frac{1}{x\sqrt{2\pi}}e^{\frac{-x^2}{2}} $$

Remark: If you want to continue working with series, it is enough to take the coefficient of the exponential function out of the integration.