Statistics (Poisson)

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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?

• Statistics (Poisson) -

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.

• Statistics (Poisson) -

Maybe the book is wrong? Thanks for giving it a shot

• Statistics (Poisson) -

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

• Statistics (Poisson) -

OK, the book is definitely wrong. I emailed the author. At least I'm understanding the material correctly. Thanks!

• Statistics (Poisson) -

BTW, author of book confirmed correction and will fix in next printing.

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