# Thread: A random variable is gaussian

1. ## A random variable is gaussian

A random variable is gaussian with $\displaystyle u_{x}=0$ and $\displaystyle \omega_{x}=1$.
a) What is the probability of |X| > 2?

So, I plugged the mean and variance into the gaussian and for this function I got

$\displaystyle 1-2\times\int^2_0\frac{1}{\sqrt{2\pi}}e^{\frac{-(x)^2}{2}}$

But I had trouble integrating the function.
and that's why I was asking about integrating $\displaystyle e^{x^2}$
I know I can use the Emperical Rule for standard deviations, and so I know the answer is 5%, but I want to know how to do it through integration of the pdf.

Is this really to hard to do by hand?

2. Originally Posted by Bucephalus
A random variable is gaussian with $\displaystyle u_{x}=0$ and $\displaystyle \omega_{x}=1$.
a) What is the probability of |X| > 2?

So, I plugged the mean and variance into the gaussian and for this function I got

$\displaystyle 1-2\times\int^2_0\frac{1}{\sqrt{2\pi}}e^{\frac{-(x)^2}{2}}$

But I had trouble integrating the function.
and that's why I was asking about integrating $\displaystyle e^{x^2}$
I know I can use the Emperical Rule for standard deviations, and so I know the answer is 5%, but I want to know how to do it through integration of the pdf.

Is this really to hard to do by hand?
What does $\displaystyle \omega_{x}=1$ mean? That the variance is one?

Are you just trying to find $\displaystyle P(|Z|>2)$, where Z is a st normal rv?

Using http://www.danielsoper.com/statcalc/calc21.aspx I get .0455001.

You cannot integrate $\displaystyle e^{-x^2}$, but I show in my classes that it's easy to approximate via its taylor series.
It converges quickly and since it's an alternating series you can estimate the error easily.

3. ## Greekk alphabet malfunction

yeah you're right, I meant the variance $\displaystyle \sigma$

Sorry about that. That was an awesome response.
I haven't read that link yet. I will take a look now.

4. Originally Posted by matheagle
What does $\displaystyle \omega_{x}=1$ mean? That the variance is one?

Are you just trying to find $\displaystyle P(|Z|>2)$, where Z is a st normal rv?

Using Free Two Tailed Area Under the Standard Normal Curve Calculator I get .0455001.

You cannot integrate $\displaystyle e^{-x^2}$, but I show in my classes that it's easy to approximate via its taylor series.
It converges quickly and since it's an alternating series you can estimate the error easily.
I will look into this taylor series approximation later on. I have an exam tomorrow so I have to keep moving.
If I still can't work it out I will ask you again.
Thanks MathEagle.

5. Originally Posted by Bucephalus
A random variable is gaussian with $\displaystyle u_{x}=0$ and $\displaystyle \omega_{x}=1$.
a) What is the probability of |X| > 2?

So, I plugged the mean and variance into the gaussian and for this function I got

$\displaystyle 1-2\times\int^2_0\frac{1}{\sqrt{2\pi}}e^{\frac{-(x)^2}{2}}$

But I had trouble integrating the function.
and that's why I was asking about integrating $\displaystyle e^{x^2}$
I know I can use the Emperical Rule for standard deviations, and so I know the answer is 5%, but I want to know how to do it through integration of the pdf.

Is this really to hard to do by hand?
Alas, it can't be done. See this page: Integration of Nonelementary Functions

6. Expanding $\displaystyle e^{-x^2/2}$ and integrating term by term is really easy and an excellent exercise. It is the best use of Taylor I've seen and it makes students appreciate series. It converges quickly so I recommend spending 5 minutes and doing this.