How can we compute this integral? $$\int_{1.96}^{\infty} e^{\frac{x^2}{2}} \ dx$$
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[Here's a tutorial in MathJax](https://math.meta.stackexchange.com/questions/5020/mathjaxbasictutorialandquickreference) – Sahiba Arora Jun 12 '17 at 18:43

This corresponds to a standard normal random variable, and we wish to compute $P(Z > 1.96)$. Tables give this to be about 0.05. – Sean Roberson Jun 12 '17 at 19:08

1i think hes missing a sqrt(1/2pi) right for it to be standard normal? so the answer would be about 0.05*sqrt(2pi) – shoestringfries Jun 12 '17 at 19:12

Hey I think the 0.05 part is not quite correct. Since the pdf above is of a standard normal distribution, and its being integrated from 1.96, the answer should include 0.025, not 0.05. – divya garg Jun 13 '17 at 07:25
2 Answers
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 10.95 = 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.
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I saw the answer to this somewhere as 0.025(2pi)^(1/2). Should one use standard normal distribution or Gamma distribution for the above question? – divya garg Jun 13 '17 at 06:50

1Hey I just checked standard normal distribution pdf, so your method is absolutely correct, just that the final answer is missing a 0.025 which is the area lying to the right of 1.96 under the standard normal curve – divya garg Jun 13 '17 at 07:20

hey sorry for the delay i updated the answer, that was a very good point. – shoestringfries Jun 27 '17 at 15:55
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 $$ \lim _{x \to +\infty}\mbox{erf}(x)=1 $$
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