Dax Posted April 26, 2014 Report Share Posted April 26, 2014 Hey Guys, could someone tell me how to do the Poisson Distribution Question in Section A of Paper 2 TZ1 2013 May?Thanks! Reply Link to post Share on other sites More sharing options...
Negotiation Posted April 26, 2014 Report Share Posted April 26, 2014 (edited) Doesn't the mark scheme lay it out nicely? It's formatted exactly the way my answer is. What are you finding tricky? Edited April 26, 2014 by Negotiation Reply Link to post Share on other sites More sharing options...
rinik Posted April 27, 2014 Report Share Posted April 27, 2014 (edited) a)In T minutes there will be T/4 cars and hence the parameter you use is 0.25TP(X<=3)= P(0)+P(1)+P(2)+P(3)=0.6Substitute 0.25T as the parameter in the equation and solve using a gdc.for example P(0)= (0.25T)^0*e^(-0.25T)/0!do this for all of the probabilities and you get an equation with one unknown, Tb)A ferry can carry a max of 3 cars. That means that a total of 6 can arrive because if more arrive than it will take another 10 minutes to transport all of the cars. This means that at least 3 have to arrive in the first 10 minutes because if 4 or more cars arrive in the second 10 minutes you will need another 10 to transport all of the cars. The new parameter is 10/4=2.5There are 4 possibilities No more than 3 cars arrive in the first 10 minutes and no more than 3 cars arrive in the next 10 minutes 4 cars arrive in the first 10 min and no more than 2 arrive in the next 105 cars arrive in the first 10 minutes and no more arrive than 1 in the next 106 arrive in the first 10 minutes and no more than 0 arrive in the next 10 minutes.no more than= cumulative distributionJust multiply the values for each situation and add up all 4 situations and you get the required probability. Use poissonpdf and poissoncdf or this will take you forever.Edit: you can also do this with a lot longer method. Using the same parameter and the fact that there can be a max of 6 cars you can calculate the probabilities individually for everything and sum them upP(0)P(0)+P(1)P(0)+P(2)P(0)+P(3)P(0)+P(3)P(1)..............................P(6)P(0).Try expanding the cumulative probabilities in the previous method and you will see that both expressions are the sameI am writing this to show you that the probabilities when less than 6 cars arrive are present in the first method Edited April 27, 2014 by rinik Reply Link to post Share on other sites More sharing options...
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