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March 30, 2015

March 30, 2015

Posted by **Sean** on Sunday, July 19, 2009 at 4:46pm.

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 7:00pmp(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) -
**Sean**, Sunday, July 19, 2009 at 10:55pmMaybe the book is wrong? Thanks for giving it a shot

- Statistics (Poisson) -
**economyst**, Monday, July 20, 2009 at 1:40pmYou 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 2:37pmOK, 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 9:47amBTW, author of book confirmed correction and will fix in next printing.

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