If X~Po(λ) and E(X^2) = 2, find λ.
That's pretty straight forward, isn't it? The Poisson distribution is given by $\displaystyle P(k)= \frac{\lambda^k}{k! e^k}$ where k is a positive integer. So the Expected value of $\displaystyle X^2$ is $\displaystyle \sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2$.
What must $\displaystyle \lambda$ be in order that that equation be true?
Can also try something like this.
Since X~Po(λ), we know two things:
1) E(X) = λ
2) Var(X) = λ = E(X^2) - E(X)^2
combining these two gives:
λ = E(X^2) - λ^2
E(X^2) = λ + λ^2
If this is equal to 2 then its the same as solving the quadratic λ + λ^2 = 2