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Help with a Stats question! (Poisson Distribution) watch

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    I have been given this question;
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    Well from the information you know that Po(6) per week.

    a) Is basically asking what is P(x≥2) You can look up P(x≤1) in the formula book then do 1-P(x≤1) to get P(x≥2)

    sorry b was completely wrong.
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    (Original post by Dj2013)
    ...
    If you have cumulative tables for the Poisson distriubtion with parameter=6 available, then it's just a question of looking up in the tables.

    1) At lease two are sold means you're looking for P(X>=2)

    Considering the complementary event, this will equal 1-P(X<=1) and you can lookup the latter.

    2) Just scan down the table for lambda = 6, until you get a value >=0.99. The first value you come to that satisfies that criterion will correspond to the minimum number you require. Viz.

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    Thank you guys,
    I have the table h
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    (Original post by Dj2013)
    Thank you guys,
    I have the table here in front of me. I am unsure as to how I should be reading it?

    Mine is set out like this;


    Lambda----0.2-----0.4----0.6-----0.8 and so on
    r = 0
    1
    2
    3
    Then there are value going across.

    So if i am looking for P(X>=2) Po(6) would it be 0.0620?
    The values listed in the table are the cumulative distribution values

    0.0620 is P(X<=2) when lambda = 6.

    So, you're in the right column.

    But you need P(X<=1) and whatever value it is, you need to subtract it from 1 to get the answer to the first part.
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    Thanks again mate,
    So the answer I have is 0.9826.

    Please tell me this is right!
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    (Original post by Dj2013)
    Thanks again mate,
    So the answer I have is 0.9826.

    Please tell me this is right!
    Yep, that's right!
 
 
 
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Updated: November 17, 2013

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