Poisson/exp distribution: event occurs 2x/year on ave; prob of taking >= 9 months?

Kuggoz

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Jun 7, 2018
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Got a little problem with probability and exponential distribution. If something is said to happen twice a year on average, then I need to calculate the probability of it taking at least 9 months to happen, Is the equation 1-e^-2(9-12) or 1-e^-1/2(9-12). Also is it to correct to put it as X~EXP (2) or X~exp (1/2)

I do know exp is about time between the actions and poisson is action/time, but if lambda is given as 2 in the problem, shouldnt it be X~EXP (2) even though i'd use 1/2 in the actual equation?
 

tkhunny

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Got a little problem with probability and exponential distribution. If something is said to happen twice a year on average, then I need to calculate the probability of it taking at least 9 months to happen, Is the equation 1-e^-2(9-12) or 1-e^-1/2(9-12). Also is it to correct to put it as X~EXP (2) or X~exp (1/2)

I do know exp is about time between the actions and poisson is action/time, but if lambda is given as 2 in the problem, shouldnt it be X~EXP (2) even though i'd use 1/2 in the actual equation?
This is the beauty of the Poisson Distribution. It is perfectly scalable.

If you have \(\displaystyle \lambda = 2\;per\;year\), you also have \(\displaystyle \lambda = 1\;per\;six\;months\) and \(\displaystyle \lambda = (1/6)\;per\;month\) and \(\displaystyle \lambda = (3/2)\;per\;nine\;months\).
 
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