Posted by Sean on Sunday, July 19, 2009 at 4:46pm.
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) - Damon, Sunday, July 19, 2009 at 7:00pm
p(k) = e^-(Lt) * (Lt)^k / k!
p(0) = e^-(Lt) * (Lt)^0/0!
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) - Sean, Sunday, July 19, 2009 at 10:55pm
Maybe the book is wrong? Thanks for giving it a shot
Statistics (Poisson) - economyst, Monday, July 20, 2009 at 1:40pm
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) - Sean, Monday, July 20, 2009 at 2:37pm
OK, the book is definitely wrong. I emailed the author. At least I'm understanding the material correctly. Thanks!
Statistics (Poisson) - Sean, Tuesday, July 21, 2009 at 9:47am
BTW, author of book confirmed correction and will fix in next printing.
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