Solve Physics Problem: Gaussian Distribution

[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!

 

[EnumaElish]

The problem states the integral value that is "under the bell curve" between x = - 1 and x = 1, for a Gaussian distribution with x_m = 0 and σ_x = 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 x_m - σ_x to x_m + σ_x where x_m = 0 and σ_x = 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 x_m = 0 and σ_x = 1) to any other Gaussian random variable (with an arbitrary x_m \ne 0 and an arbitrary σ_x > 0) by multiplying the standard one with σ_x > 0 then adding x_m \ne 0. For example, X = σ_x Y + x_m where Y is the standard Gaussian and X is any Gaussian, with mean x_m \ne 0 and standard deviation σ_x > 0.

Let f(x;x_m,σ_x) be the Gaussian density with an arbitrary x_m and an arbitrary σ_x > 0. f(x;x_m,σ_x) 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 = σ_x Y + x_m 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.