Question

What are the mean and standard deviation of a probability density function given by `Pr(X=k) = (4^ke^-4)/(k!)` for `k in {0,1,2... oo}`?

Answer 1

`mu=4`, `sigma=2`.

Explanation:

The probability density function given by `P(X=k)=(4^ke^-4)/(k!)` is the Poisson distribution with 4 degrees of freedom. This is because the PDF of the Poisson distribution with `lambda` degrees of freedom is `P(X=k)=(lambda^ke^-lambda)/(k!)`. The mean and variance of this distribution are both `lambda` (proofs can be found at https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution and https://proofwiki.org/wiki/Variance_of_Poisson_Distribution). Since variance is equal to `sigma^2`, we know that `sigma=2`.