Question

would like some help solving this question. I know what formula to use but not sure how to plug in the info.

Z 2.1)C Problem 3 Suppose that the City of Chicago is evaluating the expected traffic congestion if construction reduces the lane on Peterson Avenue (Eastbound) from two lanes to one. During rush hour, the time between arrivals at the Pulaski-Peterson intersection heading east is determined to be 5 seconds. If a single lane of Peterson can handle only one car every 8 seconds, what percentage of time will there be congestion? Probab:ty Po isson Noimal 2 lcres 1cer eurs se.ccnts are:1araur 6 Seenrds

Answer 1

Let X is the inter time between arrivals of cars at the intersection     
here X follows exponential distribution with λ = 1/5 cars per second     
λ = 1/5 cars/second     
λ = 8/5 cars / 8 seconds     
Let Y is the number of cars arriving in 8 seconds     
here, Y follows Poisson distribution with λ = 8/5 cars/ 8 seconds     
There will be congestion if there are more than 1 cars in 8 seconds in the lane     
we need to find out P(Y > 1)     

P(Y > 1) = 1 - P(Y ≤ 1)     

=1-[P(Y=0)+P(Y=1)]
poisson formula : (e^-λ ) (λ ^x) / x! (here:λ =μ)
               = 1- [(e^-(8/5)) ((8/5)⁰) / 0! +(e^-(8/5)) ((8/5)¹) / 1! ]
               = 1-[0.20190+0.32303]

=0.47507

=47.51%