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August 27th, 2014, 07:57 PM   #1
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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!
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August 28th, 2014, 06:14 AM   #2
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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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