Question

A car hiring firm has two cars. The demand for cars on each day is distributed as a Posison variate, with mean 1.5. Calculate the proportion of days on which some demand is refused

Answer 1

In a Poisson distribution n = 2

Mean λ = 1.5

x follows a Poisson 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