If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)^2] equals ___.

GATE 2017 · Engineering Mathematics · Poisson Distribution · medium

Answer: E[(X+2)^2] = 54

1. State Poisson properties: lambda = 5, so E[X] = 5, Var(X) = 5.
2. Expand the expression: Linearity of expectation distributes E over the sum.
3. Compute E[X^2]: E[X^2] = 5 + 5^2 = 5 + 25 = 30.
4. Add all terms: 30 + 20 + 4 = 54.