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 August 27th, 2014, 07:57 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, 06:14 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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### poisson distribution of e-0.99

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