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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 at most 3 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(atmost 3 accident) = P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= `[("e"^-4(4)^0)/(0!) + "e"^-4 (4)^1/(1!) + ("e"^-4(4)^2)/(2!) + ("e"^-4(4)^3)/(3!)]`
= `"e"^-4 [4^0/(0!) + 4^1/(1!) + 4^2/(2!) + 4^3/(3!)]`
= `0.0183 [1 + 4 + 16/2 + 64/6]`
= `0.0183 [1 + 4 8 + 32/3]`
= `0.0183[1 + 2/3]`
= `0.0183[(9 + 32)/3]`
= `0.0061[71]`
= 0.4331
Out of 100 days there will be at most 3 acccident = n × P(X ≤ 3)
= 100 × 0.4331
= 43.31
= 43 days ........(approximately)
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