Question

Every Saturday, at the same time, an individual stands by the side of
a road and tallies the number of cars going by within a 120-minute
window. Based on previous knowledge, she believes that the mean
number of cars going by during this time is exactly 107. Let X represent the appropriate Poisson random variable of the number of cars
passing her position in each Saturday session.
a. What is the probability that more than 100 cars pass her on any
given Saturday?
b. Determine the probability that no cars pass.

Answer 1

The probability distribution of the Poisson random variable X, representing the number of outcomes occurring in a given time interval or specified region denoted by t, is

$$p(x; \lambda t)={e^{-\lambda t}(\lambda t)^x \over x!}$$

where λ is the average number of outcomes per unit time

Given that

$$\lambda ={107 \over 2\ hours}, t=24 \ hours$$

a. What is the probability that more than 100 cars pass her on any given Saturday? 

The $R \ ppois$ function provides the left cumulative probabilities, as in Pr(X ≤ x). 

$$P(X>100)=P(\geq 101)=1-P(X\leq 100)=$$ $$=1-ppois(x=100, lambda=1284)=$$ $$=1-1.938949\times 10^{-405}\approx1$$

b. Determine the probability that no cars pass.

The $R \ dppois$ s function provides the individual Poisson mass function probabilities Pr(X = x) for the Poisson distribution. 

$$P(X=0)=dpois(x=0, lambda=1284)=$$ $$=2.322123\times 10^{-558}\approx0$$