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August 27th, 2014, 06:57 PM  #1 
Newbie Joined: Aug 2014 From: Australia Posts: 1 Thanks: 0  Poisson Distribution
Hi Everyone. Thanks in advance for any help. I'm having some major issues trying to calculate a Poisson distribution. the question I have is: A particularly dangerous intersection follows a Poisson Distribution for the number of accidents in any given day. The probability of having at least 2 accidents is 0.99. What is the value of lambda What I've done so far follows. I feel as though I'm missing some vital rule that I should be applying. P(X => 2) = 0.99 = 1 (P(X=1) + P(X = 0) the Poisson Distribution formula is P(X = k) = ((e^m)(m^k))/k! so P(X => 2) = 0.99 = 1 (((e^m)(m^0)/0!) + (((e^m)(m^1)/1!) 0.01= (e^m) + (e^m)(m) 0.01 = (e^m)(1+m) This is where I get stuck. How do I now calculate what M is? Thanks again! 
August 28th, 2014, 05:14 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Use Newton's method or bisection? There is a function, Lambert's W, that can solve this in 'closed form' but it's not generally taught at that level. I'd just use numerical methods here. If you try $\lambda=6$ you'll find it's too small, and if you use $\lambda=7$ you'll find it's too big, so the answer lies somewhere between the two. 

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