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Question
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poison variate, with mean 1.5. Calculate the proportion of days on which some demand is refused
Solution
In a posion distribution n=2
Mean λ = 1.5
x follows poison distribution
With in P(x) = `("e"^(-lambda) lambda^x)/(x!)`
P(some demand is refused) = P(X > 2)
= `1 - "P"("X" ≤ 2)`
= `1 - ["P"("X" = 0) + "P"("X" = 1) + "P"("X" = 2)]`
= `1 - [("e"^(-1.5) (1.5)^0)/(0!) + ("e"^(-1.5) (1.5)^1)/(1!) + ("e"^(-1.5) (1.5)^2)/(2!)]`
= `1 - "e"^(-1.5) [(1.5)^0/(0!) + (1.5)^1/(1!) +(1.5)^2/(2!)]`
= `1 - "e"^(-1.5) [1 + .5 + 2.25/2]`
= 1 – 0.2231 [1 + 1.5 + 1.125]
= 1 – 0.2231 [3.625]
= 1-0.8087
= 0.1913
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