Asked by Sean

Cars arrive at a toll booth according to a Poisson process with mean 80 cars per hour.

How long can the attendant's phone call last if the probability is at least .4 that no cars arrive during the call.

The book's answer is 23 seconds. That doesn't match mine and I was hoping to understand why...

p(y) = λ^y * e^-λ / y!
p(0) = e^-λ

p(0) > 0.4
e^-λ > 0.4
-λ > ln 0.4
λ < -ln 0.4

If x = call length (in seconds)
λ = x * 80 / 3600

x * 80/3600 < -ln 0.4
x < -3600/80 * ln 0.4
x < 41.2 seconds

Can anyone point out how to get 23 seconds or what I did wrong?
Answered by Damon
p(k) = e^-(Lt) * (Lt)^k / k!

p(0) = e^-(Lt) * (Lt)^0/0!
so
p(0) = e^-(Lt)

.4 = e^-(80t/3600)
ln .4 = - .022222 t
-.9163 = -.022222 t
t = 41.23
Beats me, I am having whatever trouble that you are having with it.
Answered by Sean
Maybe the book is wrong? Thanks for giving it a shot
Answered by economyst
You and Damon, I believe are both right. 23 seconds would correspond a 40% probability that at least one car showed up.
(1-.4) = e^(80t/3600)
ln(.6) = -.0222222t
-.5108 = -.0222222t
t = 22.99
Answered by Sean
OK, the book is definitely wrong. I emailed the author. At least I'm understanding the material correctly. Thanks!
Answered by Sean
BTW, author of book confirmed correction and will fix in next printing.
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