Answer:
0.2905
Step-by-step explanation:
Given that vehicle arrivals at a certain intersection from 9-10 am have a Poisson distribution = 70/hr.
Also given that A Berkeley TE student counts arrivals at the intersection during this time period for a number of days.
To find the probability for 75 or more vehicles
Let X be the no of vehicles
P(k events ) = [tex]\frac{70^k e^{-70} }{k!}[/tex]
Hence here required probability=[tex]P(X\geq 75)[/tex]
=[tex]1-\Sigma_{i=1}^{i=74} e^{-70} (\frac{\lambda ^i}{i!} )[/tex]
Reqd prob = [tex]P(X\geq 75)=0.29048[/tex]
