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

Answer 1

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