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xEnOnn Group Title

In a city, the daily number of traffic accidents follows a Poisson distribution with \[\lambda=9\]. What is the probability that the total number of traffic accidents exceeds 1000 in 100 days?

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    @Zarkon can we change to \(\lambda = 900\)?

    • 2 years ago
  2. xEnOnn Group Title
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    One of my solutions is to have: \[\lambda=9*100\] Because the interval considered in this case is 100 days. Then I calculate for: \[P(X>1000)\]. But the problem with having Lambda as 900 is that it is too huge. P(X>1000)=1.

    • 2 years ago
  3. satellite73 Group Title
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    how do you calculate \(P(x\geq 1000)\)? using poisson? usually "greater than ..." means calculate everything up to it and subtract from one. must be another formula for this. i will try to look it up

    • 2 years ago
  4. xEnOnn Group Title
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    Precisely. Here's what I have done: \[P(X>1000)=1-\sum _{ i=0 }^{ 1000 }{ poisson(i,\quad \lambda =900) } \] However, I can't calculate possion up to 1000. It is too huge. Not even my calculator can do it. I used an online calculator to do it but the problem is I believe this question shouldn't require me to do such calculations that require a powerful calculator like a computer.

    • 2 years ago
  5. xEnOnn Group Title
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    And the value returned from Mathematica for calculating the sum from 0 to 1000 of the poisson function is 0.9995093672673, which is almost 1. So, I think my answer is still wrong.

    • 2 years ago
  6. satellite73 Group Title
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    yeah i think there is another way of computing \(P(x\geq k)\) for poisson using variance, but it might require some table i am trying to look it up in a third book

    • 2 years ago
  7. xEnOnn Group Title
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    ohh thanks! I didn't know there is another way to compute poisson using variance. But if we need a table, would a table list up to 900? The variance for poisson is the same as its expectation, which is Lambda.

    • 2 years ago
  8. satellite73 Group Title
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    yeah each of these damned books has poisson all over the place still looking.

    • 2 years ago
  9. xEnOnn Group Title
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    haha.... okay sure... thanks! i'm still trying too.

    • 2 years ago
  10. Zarkon Group Title
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    use a normal approximation

    • 2 years ago
  11. satellite73 Group Title
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    i have an idea, probably a bad one you expect 900 accidents, and want to know what is the probability you get over 1000 reduce all by a factor of 100 and change to you expect 9 accidents, what is the probability you get more than 10 this seems to be done by tables, one of which i found on line, where it will give you this

    • 2 years ago
  12. satellite73 Group Title
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    @Zarkon so we need a table for that too right?

    • 2 years ago
  13. Zarkon Group Title
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    My calculator can do it

    • 2 years ago
  14. xEnOnn Group Title
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    I have tried this idea too. The probability is around 0.7 something. There is a formula for poisson.

    • 2 years ago
  15. xEnOnn Group Title
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    But with this idea, the interval considered for the probability computed would be for each day, not 100 days any more?

    • 2 years ago
  16. satellite73 Group Title
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    i found this http://www.wku.edu/~david.neal/statistics/discrete/poisson.html which tells you how to do it with a calculator, but if my idea is right, then i found a table that says for \(\lambda =9\) \(P(x<10)=0.70599\)

    • 2 years ago
  17. satellite73 Group Title
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    if it is valid to reduce all by 100, and it seems like it might be, then your answer would be \(1-.70599\) but again this is relying on a table

    • 2 years ago
  18. satellite73 Group Title
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    sorry but every other example i can find sums up all the others and subtracts from one, as you did at first

    • 2 years ago
  19. xEnOnn Group Title
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    yea, it seems pretty valid. thanks! :) My only concern is if this would be computing for an interval of 1 day instead of 100 days? Because usually, I remember I would adjust the value of Lambda according to the required interval specified in the problem. But in this case, adjusting the Lambda seems too huge. Adjusting the interval value gives a more believable value but casts doubts on the interval that probability value computed for.

    • 2 years ago
  20. xEnOnn Group Title
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    *casts doubts on the interval that the probability value is computed for.

    • 2 years ago
  21. satellite73 Group Title
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    i wish i knew. my only answer is that this would be an approximation, and perhaps a bad one. sorry i cannot be more help, stumped on this one, but i would go with reducing and computing the approximation

    • 2 years ago
  22. Zarkon Group Title
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    do you know what the answer is supposed to be?

    • 2 years ago
  23. xEnOnn Group Title
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    Thanks a lot, @satellite73. :)

    • 2 years ago
  24. satellite73 Group Title
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    yw, sorry it was not better

    • 2 years ago
  25. xEnOnn Group Title
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    @Zarkon, unfortunately, I don't know the correct answer for this question. I am not given. :(

    • 2 years ago
  26. Zarkon Group Title
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    what don't you like about the answer Mathematica gave you?

    • 2 years ago
  27. xEnOnn Group Title
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    Because I won't have Mathematica in the exam. :( This question suppose to be an exam question.

    • 2 years ago
  28. Zarkon Group Title
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    will you have tables?

    • 2 years ago
  29. Zarkon Group Title
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    will you have a normal table?

    • 2 years ago
  30. xEnOnn Group Title
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    yea, I will have normal tables. Can I approximate poisson with normal?

    • 2 years ago
  31. xEnOnn Group Title
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    I'm only aware that I could approximate binomial with normal and poisson. But I don't know if I could approx for poisson with normal.

    • 2 years ago
  32. Zarkon Group Title
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    yes...quite accurately with continuity corrections

    • 2 years ago
  33. Zarkon Group Title
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    \[P(X>1000)=P(X>1000.5)\] \[=P\left(\frac{X-900}{\sqrt{900}}>\frac{1000.5-900}{\sqrt{900}}\right)\] \[=P\left(Z>\frac{1000.5-900}{\sqrt{900}}\right)\]

    • 2 years ago
  34. xEnOnn Group Title
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    But with this, I have to make an assumption that the original data follows a normal distribution? Otherwise, how can I be sure that the data is close to a normal distribution?

    • 2 years ago
  35. Zarkon Group Title
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    No... here you essentially have the sum of 100 iid random variables. By the CLT it will be approx normal

    • 2 years ago
  36. Zarkon Group Title
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    that is how you get \(\lambda=9\cdot 100\) if \(X_1,X_2,\cdots X_n\) are poisson(\(\lambda\)) then \[\sum_{k=1}^{n}X_k\sim\text{poisson}(\lambda\cdot n)\]

    • 2 years ago
  37. xEnOnn Group Title
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    ohhh.....yeah!! This sounds very logical. And it looks like a very workable hand method.!

    • 2 years ago
  38. Zarkon Group Title
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    good

    • 2 years ago
  39. Zarkon Group Title
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    sorry i didn't jump into the conversation earlier...I'm grading now...man I hate to grade.

    • 2 years ago
  40. xEnOnn Group Title
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    It is through CLT, that this whole poisson becomes close to normal. And therefore, the approximation can take place. It is not exact but close because of the CLT, n more than 30. Thank you so much for your help! :D

    • 2 years ago
  41. xEnOnn Group Title
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    haha....thanks! it's okay. I am very happy to learn something from you. And, grading? You mean like grading students' assignment?

    • 2 years ago
  42. Zarkon Group Title
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    Complex analysis exams

    • 2 years ago
  43. xEnOnn Group Title
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    @satellite73 Zarkon gave a pretty good explanation. Just thought you may be interested to learn about it too.

    • 2 years ago
  44. xEnOnn Group Title
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    oh wow... Sounds like a difficult subject. haha... thanks for your help!

    • 2 years ago
  45. satellite73 Group Title
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    grading is the worst. guess what i am doing?

    • 2 years ago
  46. Zarkon Group Title
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    it is the worst.....worst part of the job.. I hope you are not grading :)

    • 2 years ago
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