Friday

January 30, 2015

January 30, 2015

Posted by **Twg** on Wednesday, August 12, 2009 at 12:01am.

- MATH Prob. -
**drwls**, Wednesday, August 12, 2009 at 7:42amThe distribution should be of the Poisson type. There is a certain expected value "a" of an accident occurring during that interval on a given Friday. You can compute it using

a = 0.84*0 + 0.138*1 + 0.02*2 + 0.01*3 a = 0.208

For a = 0.208, the probability of m=0 accidents according to a Poisson distribution is

P(0) = (a^m e^-a)/m! = e^-.208 = 0.812

P(1) = 0.208*0.812/1 = 0.169

P(2) = 0.208^2*0.812/2 = 0.018

P(3) = 0.208^3*0.812/6 = 0.001

These are in pretty good agreement with your numbers, except for the probability of 3 accidents. Did you mean to write 0.001?

- MATH Prob. -
**Twg**, Wednesday, August 12, 2009 at 7:43pmyou completely lost me when you responded and put 0.138*1 in your reply. May I ask where did you get 0.138?

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