Posted by Twg on Wednesday, August 12, 2009 at 12:01am.
The number of accidents that occur at the intersection of Pine and Linden streets between 3 p.m. and 6 p.m. on Friday afternoons is 0, 1, 2, or 3, with probabilities of 0.84, 0.13, 0.02, and 0.01, respectively. Graph this probability distribution. What is the expected value for the random variable given the number of accidents?

MATH Prob.  drwls, Wednesday, August 12, 2009 at 7:42am
The 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 7:43pm
you 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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