Solve Physics Problem: Gaussian Distribution

In summary, the conversation discusses a physics problem involving a Gaussian distribution and a probability expression. The integral value between -1 and 1 is equal to 1.7 divided by the square root of 2*pi*sigma^2. The standard Gaussian can be converted to any other Gaussian by multiplying by sigma and adding xm. This conversion does not affect the probability value as long as the bounds of integration and the C value are adjusted accordingly. Thus, the answer of 0.68 is not only for the standard Gaussian but for any Gaussian distribution.
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houseguest
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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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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 [itex]\sqrt{2\pi\sigma^2}[/itex]). 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 [itex]C(\sigma)=\sqrt{2\pi\sigma^2}[/itex], then the answer is 1.7/C(1), for the standard Gaussian. Since [itex]C(1)=\sqrt{2\pi}[/itex] = 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 [itex]\ne[/itex] 0 and an arbitrary σx > 0) by multiplying the standard one with σx > 0 then adding xm [itex]\ne[/itex] 0. For example, X = σx Y + xm where Y is the standard Gaussian and X is any Gaussian, with mean xm [itex]\ne[/itex] 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,

[tex]\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.[/tex]

So the answer is 0.68 not only for the standard Gaussian but for any Gaussian distribution.
 
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FAQ: Solve Physics Problem: Gaussian Distribution

What is a Gaussian distribution in physics?

A Gaussian distribution is a type of probability distribution that is commonly used to model various physical phenomena. It is also known as a normal distribution and is characterized by a bell-shaped curve.

How is a Gaussian distribution used to solve physics problems?

Gaussian distributions are used in physics to describe the behavior of random variables that are subject to various types of noise or error. They are particularly useful in statistical mechanics, where they can be used to model the distribution of particles in a gas or the fluctuations of energy in a system.

What is the equation for a Gaussian distribution?

The equation for a Gaussian distribution is y = (1/σ√(2π)) * e^(-((x-μ)^2)/(2σ^2)), where μ is the mean and σ is the standard deviation.

How is the central limit theorem related to Gaussian distributions?

The central limit theorem states that when independent random variables are added together, their sum tends to follow a Gaussian distribution. This is why Gaussian distributions are so commonly used in physics, as many physical systems involve the interaction of multiple variables.

Are there any limitations to using a Gaussian distribution to solve physics problems?

While Gaussian distributions are widely used and quite versatile, they do have some limitations. They are only appropriate for continuous variables, and they assume that the variables are normally distributed. In certain cases, other types of distributions may be more appropriate for modeling physical phenomena.

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