Saturday

August 23, 2014

August 23, 2014

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?

**Related Questions**

math - The number of accidents that occur at the intersection of Pine and Linden...

Math - The number of accidents that occur at the intersection of Pine and Linden...

Math - The number of accidents that occur at the intersection of Pine and Linden...

MATH - The number of accidents that occur at the intersection of Pine and Linden...

Math - The number of accidents that occur at the intersection of Pine and Linden...

math - The number of accidents that occur at the intersection of Pine and Linden...

math/graphing - The number of accidents that occur at the intersection of Pine ...

math157 - The number of accidents that occur at the intersection of Pine and ...

statistics - Suppose that the number of accidents occurring in an industrial ...

Statistics - I have a problem I need some help on. I answered some. For many ...