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

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Official answer: 54 (±0)

Probability and Statistics

For a Poisson distribution with parameter $\lambda = 5$ :

The mean is $E[X] = \lambda = 5$ .
The variance is $\text{Var}(X) = \lambda = 5$ .

The second moment is given by:

E[X^2] = \text{Var}(X) + (E[X])^2 = 5 + 5^2 = 30

Now, expand the expectation expression:

E[(X + 2)^2] = E[X^2 + 4X + 4] = E[X^2] + 4E[X] + 4

Substitute the values:

E[(X + 2)^2] = 30 + 4(5) + 4 = 30 + 20 + 4 = 54

Hence, the answer is 54.

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This Probability and Statistics previous year question is Q48 of the GATE CSE 2017 Set 2 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.