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Question
If X has Poisson distribution with parameter m = 1, find P[X ≤ 1] [Use `e^-1 = 0.367879`]
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Solution
Given X follows Poisson distribution with parameter m = 1
∴ X ∼ P(1)
p.m.f. is given by
P(X = x) = `(e^-m m^x)/(x!)`
= `(e^-1 1^x)/(x!)`
P(X = x) = `e^-1/(x!)`
Now P[X ≤ l] = P (X = 0) + P (X = 1)
= `e^-1/(0!) + e^-1/(1!)`
= `2e^-1`
= 2(0.367879)
= 0.735758
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If X – P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2) given that e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
`(e^-mm^1)/(1!) = square`
∴ m = `square`
∴ mean = `square` = `square`
Then P(X = 2) = `square` = `square`
