Question

Calculate μ and σ, where σ is the standard deviation, defined by the following.

σ^2 = integral between (−∞,∞) (x-μ)^2 p(x) dx


The smaller the value of σ the more tightly clustered are the values of the random variable X about the mean μ.
p(x) = 1/r e^-x/r on [0, ∞),where r > 0



μ = __________

σ = +/- ________

Answer 1

you have

∫[-∞,+∞] (x-μ)^2 1/r e^-(x/r) dx

This just boils down to

∫x^2 e^(-x) dx
with suitable constants thrown about. Do it with integration by parts twice to get

-(x^2+2x+2)e^(-x)

Now substitute in (x-μ) and x/r and things work out ok.