# Gaussian distribution

1. Feb 27, 2008

### houseguest

Hello, I am attaching what was an extra credit question in my physics class which I didn't understand at all. The topic isn't in the book and all the internet searchs I read confuse me. I was hoping someone might give me a walk through.
Thanks!

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Last edited: Feb 27, 2008
2. Feb 27, 2008

### EnumaElish

The problem states the integral value that is "under the bell curve" between x = - 1 and x = 1, for a Gaussian distribution with xm = 0 and σx = 1, which is the numerator of the probability expression (the denominator is $\sqrt{2\pi\sigma^2}$). You should verify this by comparing the two formulas (the one on the left and the one at the top).

Since the integral is defined between -1 to 1, it is defined from xm - σx to xm + σx where xm = 0 and σx = 1.

Let $C(\sigma)=\sqrt{2\pi\sigma^2}$, then the answer is 1.7/C(1), for the standard Gaussian. Since $C(1)=\sqrt{2\pi}$ = 2.5, the probability is 1.7/2.5 = 0.68.

But I can convert the standard Gaussian random variable x (with xm = 0 and σx = 1) to any other Gaussian random variable (with an arbitrary xm $\ne$ 0 and an arbitrary σx > 0) by multiplying the standard one with σx > 0 then adding xm $\ne$ 0. For example, X = σx Y + xm where Y is the standard Gaussian and X is any Gaussian, with mean xm $\ne$ 0 and standard deviation σx > 0.

Let f(x;xmx) be the Gaussian density with an arbitrary xm and an arbitrary σx > 0. f(x;xmx) is identical to the formula on the left margin. The special case of standard Gaussian, f(y;0,1) is identical to the formula at the top of the page.

The conversion X = σx Y + xm does not affect the probability value as long as the bounds of integration (and the C value) are adjusted accordingly. Thus,

$$\int_{x_m-\sigma_x}^{x_m+\sigma_x}f(x;x_m,\sigma_x)dx\left/C(\sigma_x) = \int_{-1}^{1}f(y;0,1)dy\right/C(1) = 1.7/C(1) = 0.68.$$

So the answer is 0.68 not only for the standard Gaussian but for any Gaussian distribution.

Last edited: Feb 28, 2008