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.
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 = +/- ________
1 answer
1 answer