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Question
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has between 5 and 7 rats inclusive. Given e−5 = 0.0067.
Solution
Let X denote the number of rats per bungalow.
Given, m = 5 and e–5 = 0.0067
∴ X ~ P(m) ≡ X ~ P(5)
The p.m.f. of X is given by
P(X = x) = `("e"^-"m" "m"^x)/(x!)`
∴ P(X = x) = `("e"^-5*(5)^x)/(x!), x` = 0, 1, ..., 5
P(between 5 and 7 rats, inclusive)
= P(5 ≤ X ≤ 7)
= P(X = 5 or X = 6 or X = 7)
= P(X = 5) + P(X = 6) + P(X = 7)
= `("e"^-5 (5)^5)/(5!) + ("e"^-5 (5)^6)/(6!) + ("e"^-5 (5)^7)/(7!)`
= `("e"^-5 (5)^5)/(5!) + ("e"^-5 (5)^6)/(6 xx 5!) + ("e"^-5 (5)^7)/(7 xx 6 xx 5!)`
= `("e"^-5 xx 5^5)/(5!) [1 + 5/6 + 5^2/(7 xx 6)]`
= `("e"^-5 xx 5^5)/(5 xx 4 xx 3 xx 2 xx 1)[1 + 0.833 + 0.595]`
= `(0.0067 xx 5^4)/(24) (2.428)`
= `(0.0067 xx 625 xx 2.428)/(24)`
= `(10.1673)/(24)`
= 0.4236
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