Question

. In the Journal of Transportation Engineering (June 2005), the number of non-home-based trips per day taken by drivers in Korea was modeled using the Poisson distribution with λ = 1.15. What is the probability that a randomly selected Korean driver will take two or more non-home-based trips per day?

Answer 1

Answer:

So, the probability  that a randomly selected Korean driver will take two or more non-home-based trips per day is P=0.32.

Step-by-step explanation:

We know that the number of non-home-based trips per day taken by drivers in Korea was modeled using the Poisson distribution with λ = 1.15.

We have the Poisson formula:

$P(X=k)=\frac{\lambda^k\cdot e^{-\lambda}}{k!}$

We calculate:

$P(X\geq 2)=1-P(X=0)-P(X=1)\\\\P(X\geq 2)=1-\frac{1.15^0\cdot e^{-1.15}}{0!}-\frac{1.15^1\cdot e^{-1.15}}{1!}\\\\P(X\geq 2)=1-0.32-0.36\\\\P(X\geq 2)=0.32$

So, the probability  that a randomly selected Korean driver will take two or more non-home-based trips per day is P=0.32.