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    I know how to do part (a), but can't do part (b). Here is the question-

    The number of accidents per week at a factory is a Poisson random variable with parameter 2.

    (a) Find the probability that in any week chosen at random exactly 1 accident occurs.

    (b) The factory is observed for 100 weeks. Determine the expected number of weeks in which 5 or more accidents occur.


    Thanks.
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    (Original post by placenta medicae talpae)
    It's a good idea to start off by letting X be the number of accidents which happen in a week at the factory.

    X~Poisson(2)

    So the probability of X equalling a given x is:

    P(X=x)=\frac{e^{-2} \times 2^x}{x!}

    Okay, really sorry about this: that strange vertical line should be an exclamation mark, the factorial-thing.
    It doesn't seem to want to work atm!

    For that second one, do you have binomial cumulative tables? :teeth:

    Good luck
    That has given me the right answer, thanks for your time!
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    (Original post by AG27)
    That has given me the right answer, thanks for your time!
    Haha, oops, I just deleted the post because I realised you'd said that you had the answer to part (a) and were just looking for advice on part (b)!

    Anyone know how to do CPR? :embarrassed:
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    (Original post by AG27)
    I know how to do part (a), but can't do part (b). Here is the question-

    The number of accidents per week at a factory is a Poisson random variable with parameter 2.

    (a) Find the probability that in any week chosen at random exactly 1 accident occurs.

    (b) The factory is observed for 100 weeks. Determine the expected number of weeks in which 5 or more accidents occur.


    Thanks.
    for b) you can find the probability of 5 or more accidents occur in one week, then from there you can use a binomial distribution: X is the number of weeks in which occurs 5 or more accidents, and X~B(100,p)
    where p is the probability that 5 or more accidents occur in one week.

    then use the formula of for E(X) for the binomial distribution.
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    X is here the number of accidents that occur in one week.

    X~Poisson(2)

    If you can approximate to the binomial, then:
    E(X)=2, which means for the binomial distribution that np is about 2.
    With n being 100, if you then work out p, you can now re-describe X as:

    X~B(n,p)

    For one week, the probability that 5 or more accidents occur will be given by:

    P(X \ge 5)=1-P(X=4)-P(X=3)-P(X=2)-P(X=1)-P(X=0)

    Spoiler:
    Show
    Or if you have skillz with cumulative tables, you could work out P(X \le 4) and take this from 1.

    When you've found the probability that 5 or more accidents occur in one week ...
    Times it by 100 and give a sensible answer to how many weeks !

    Spoiler:
    Show
    The timesing by 100 is because you're finding the expected number, and if we call the P(X \ge 5) thing p, then E(X) is given by np, which is 100p

    Hope you get this! :woo:
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    If you are doing 100p? Than are you just finding E(X) which is just the mean and not the number of weeks? Im teaching myself S2 and I'm intrigued by this question and would like it if someone can explain it to me
 
 
 
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