Respuesta :
Answer: a) 4.6798, and b) 19.8%.
Step-by-step explanation:
Since we have given that
P(n) = [tex]\dfrac{15}{120}=0.125[/tex]
As we know the poisson process, we get that
[tex]P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda[/tex]
So, for exactly one car would be
P(n=1) is given by
[tex]=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599[/tex]
Hence, our required probability is 0.2599.
a. Approximate the number of these intervals in which exactly one car arrives
Number of these intervals in which exactly one car arrives is given by
[tex]0.2599\times 18=4.6798[/tex]
We will find the traffic flow q such that
[tex]P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr[/tex]
b. Estimate the percentage of time headways that will be 14 seconds or greater.
so, it becomes,
[tex]P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%[/tex]
Hence, a) 4.6798, and b) 19.8%.
