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Poisson Distribution of Accidents

  • Thread starter bitty
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  • #1
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Homework Statement


In New York in the last 3 years there were 55 driving accidents. Assume all days are alike. What is the approximate probability that "in the next 3 years there will be at least 2 days with more than one accident".


Homework Equations


Poisson approximation


The Attempt at a Solution


P[in the next 3 years there will be at least 2 days with more than one accident]=
1-P[in the next 3 years there will be 0 days with more than one accident]-P[in the next 3 years there will be 1 day with more than one accident]

P[in the next 3 years there will be 0 days with more than one accident] is simple to calculate: 1095!/(1040!*1095^55), w/ 3 years=1095 days.

But I have no idea how to calculate P[in the next 3 years there will be 1 day with more than one accident].
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


In New York in the last 3 years there were 55 driving accidents. Assume all days are alike. What is the approximate probability that "in the next 3 years there will be at least 2 days with more than one accident".


Homework Equations


Poisson approximation


The Attempt at a Solution


P[in the next 3 years there will be at least 2 days with more than one accident]=
1-P[in the next 3 years there will be 0 days with more than one accident]-P[in the next 3 years there will be 1 day with more than one accident]

P[in the next 3 years there will be 0 days with more than one accident] is simple to calculate: 1095!/(1040!*1095^55), w/ 3 years=1095 days.

But I have no idea how to calculate P[in the next 3 years there will be 1 day with more than one accident].
If you assume that the daily number of accidents is a Poisson random variable with some mean r, what is the expected number of accidents 3 years = 1095 days? Based on the only data available, how would you estimate the value of r?

Now assuming the value of r you obtained above, what is p2 = P{>= 2 accidents in any single day}? Take it from there.

RGV
 

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