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?

Question

drwls · Answer

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?

Twg · Answer

you completely lost me when you responded and put 0.138*1 in your reply. May I ask where did you get 0.138?