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Question
If X has a Poisson distribution with variance 2, find P (X = 4)
[Use e-2 = 0.1353]
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Solution
Variance = 2 ∴ m = 2
X ~ P (2)
P (X = 4) = `("e"^-2 2^4)/(4!)`
`= 0.1353 xx 16/24`
`= 0.2706/3 = 0.0902`
P (X = 4) = 0.0902
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Solution: X : Follows Poisson distribution
∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2
∴ P(X = 1) = `square` P(X = 2).
`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,
`"e"^-"m" = square "e"^-"m" "m"/2`, m ≠ 0
∴ m = `square`
In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.
(Given that e-0.2 = 0.8187)
Solution:
Here, m = `square` and X − P(m) with parameter m.
The p.m.f. X is:
P(X = x) = `(e^(−m).m^x)/(x!), x = 0, 1, 2,...`
P(X ≥ 3) = 1 − P(X < 3)
= 1 − [`square + square + square`]
= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`
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Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^square"m"^1)/(1!) = ("e"^"-m""m"^2)/square`
∴ m = `square`
∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`
