Question 7844 – Engineering-Mathematics
December 10, 2023Question 7915 – Engineering-Mathematics
December 10, 2023Question 8010 – Engineering-Mathematics
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)2] equals _________.
Correct Answer: A
Question 28 Explanation:
In Poisson distribution:
Mean = Variance
E(X) = E(X2) – (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Mean = Variance
E(X) = E(X2) – (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
54
55
56
57
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