Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†). (Round your answer to three decimal places.) (a) What is the probability that the number of drivers will be at most 18?

The number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20.

The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.
The variance of Poisson distribution is equal to the mean of Poisson distribution.

a) P(number of drivers will be at most 18)

Formula:

[tex]P(X =k) = \displaystyle\frac{\mu^k e^{-\mu}}{k!}\\\\ \mu \text{ is the mean of the distribution}[/tex]

[tex]P( x \leq 18) =P(x=0) + P(x =1) + P(x = 2) + ... + P(x = 18)\\\\= \displaystyle\frac{20^0 e^{-20}}{0!} + \displaystyle\frac{20^1 e^{-20}}{1!} +...+ \displaystyle\frac{20^{18} e^{-20}}{18!}\\\\ = 0.381[/tex]

Thus, 0.381 is the probability that the number of drivers will be at most 18.