The probability distribution function for a normal distribution is

f(x) = 1/(σ*sqrt(2*π))*e^[-(x-μ)^2/(2*σ^2)]

How do I derive this?

To derive the probability distribution function (PDF) for a normal distribution, also known as the Gaussian distribution, you can use the method of calculus. Here's a step-by-step explanation of how to derive it:

Step 1: Start by assuming a general form for the PDF:

f(x) = C * e^[-(x-μ)^2/(2*σ^2)]

where C is a constant to be determined.

Step 2: Normalize the PDF by ensuring that the area under the curve is equal to 1. To achieve this, integrate the PDF over the entire range of x and set it equal to 1:

∫ f(x) dx = 1

Step 3: Evaluate the integral on the left side:

∫[C * e^[-(x-μ)^2/(2*σ^2)] ] dx

The integral of this Gaussian function is not straightforward to compute analytically. However, by making use of certain properties, we can simplify the calculation.

Step 4: Perform a u-substitution to facilitate the integration. Let u = (x-μ)/σ, then du = dx/σ:

∫[C * e^[-(x-μ)^2/(2*σ^2)] ] dx = C * σ * ∫ e^[-u^2/2] du

Step 5: Notice that the resulting integral is the standard normal distribution (with mean 0 and standard deviation 1). Its integral over the entire range is a known value equal to √(2π), which allows us to determine the constant C:

C * σ * ∫ e^[-u^2/2] du = 1
C * σ * √(2π) = 1

Therefore, C = 1 / (σ * √(2π)).

Step 6: Substitute the value of C back into the general form of the PDF:

f(x) = 1 / (σ * √(2π)) * e^[-(x-μ)^2/(2*σ^2)]

And there you have it! With the constant determined, we derive the probability distribution function for the normal distribution.

It's worth noting that this is a standard method for deriving the PDF of a normal distribution. However, it requires some familiarity with calculus and mathematical properties to carry out the integration and simplification.