(7 points) Cars arrive at a toll both according to a Poisson process with mean 80 cars per hour. If the attendant makes a two-minute phone call, what is the probability that at least 1 car arrives during the call

The probability that at least 1 car arrives during the call is 0.9306

Step-by-step explanation:

Cars arriving according to Poisson process - 80 Cars per hour

If the attendant makes a 2 minute phone call, then effective λ = 80/60 * 2 = 2.66666667 = 2.67   X ≅ Poisson (λ = 2.67)

Now, we find the probability: P(X≥1)

P(X≥1) = 1 - p(x < 1)

P(X≥1) = 1 - p(x=0)

P(X≥1) = 1 - [ (e^-λ) * λ^0] / 0!

P(X≥1) = 1 - e^-2.67

P(X≥1) = 1 - 0.06945

P(X≥1) = 0.93055

P(X≥1) = 0.9306

Thus, the probability that at least 1 car arrives during the call is 0.9306.

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

answer: -1

step-by-step explanation: