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Question
The distribution of the number of road accidents per day in a city is poisson with mean 4. Find the number of days out of 100 days when there will be atleast 2 accidents
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Solution
In a possion distribution
Mean λ = 4
n = 100
x follows possion distribution with
P(x) = `("e"^(-lambda) lamda^x)/(x!)`
= `("e"^-4 (4)^x)/(x!)`
P(atleast 2 accidents)
= P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4) + …………
= 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= `1 - [("e"^-4(4)^0)/(0!) + ("e"^-4(4)^1)/(1!)]`
= 1 – e-4[l + 4]
= 1 – 0.0183(5)
= 1 – 0.0915
= 0.9085
= Out of 100 days there will be atleast 2 accidents
= n × P(X ≥ 2)
= 100 × 0.9085
= 90.85
= 91 days .......(approximately)
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