GATE | GATE-CS-2017 (Set 2) | Question 62

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

Note: This question appeared as Numerical Answer Type.
(A) 54
(B) 55
(C) 56
(D) 57


Answer: (A)

Explanation:

Using Linearity of Expectation, we can write, 
E[(X+2)2] = E[X2] + E[4X] + E[4]

In Poisson distribution, mean and variance are same.
Mean is given as 5. So variance should also be 5.
Also, variance = E[X2] – (E[X])2
5 = E[X2] – 25
E[X2] = 30
Thus E[(X+2)2] = 30 + 4*5 + 4 = 54.


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