How can the following Gaussian integral be calculated?

Question

How can we compute this integral? $$\int_{1.96}^{\infty} e^{\frac{-x^2}{2}} \ dx$$

Answer 1

Hint:

By definition of the Error function one primitive is:

$$\int e^{-\frac{x^2}{2}} dx=\sqrt{\frac{\pi}{2}}\mbox{erf}\left(\frac{x}{\sqrt{2}}\right) $$

and $$ \lim _{x \to +\infty}\mbox{erf}(x)=1 $$

Answer 2

Note that $$ e^{-\frac{x^2}{2}} $$ is almost the pdf of the standard normal, we are missing a factor of $\frac{1}{\sqrt{2\pi}}$. So we can simply multiply by the factor times its reciprocal so we dont end up messing up the final $$\sqrt{2\pi}\int \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$$.

The x's are the stadard normal random variables, also known as Z scores. From a Z table you can see that Z = 1.96 is about the 95% percentile.

So the missing area is 1-0.95 = 0.05 between Z = -1.96 and Z= 1.96 since the normal table is symmetric. But we only care about the right hand side, from Z= 1.96 to infinity, so the area under this integral $$\int_{1.96}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$$ ends up being 0.025. We need to times this by $$\sqrt{2\pi}$$ to get the final answer which is around 0.063.