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    https://i.imgur.com/NvSpPRr.jpg
    I just don't understand how I would approach this question.
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    ghostwalker
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    part c btw
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    (Original post by assassinbunny123)
    part c btw
    The samples are independent and the probability that they contain the mean remains the same 90% so its just a binomial distribution with n=4 p=0.9.
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    would of never thought of that. thx
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    (Original post by Dalek1099)
    The samples are independent and the probability that they contain the mean remains the same 90% so its just a binomial distribution with n=4 p=0.9.
    how did you know that samples are independent
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    (Original post by assassinbunny123)
    would of never thought of that. thx
    why wouldn't you be able to do it like just a probability tree.
    with P( 3 time conTaining one not) =0.9x0.9x0.9x0.1 X4( different combos)
    add
    P(4 times containing) o.9x0.9x.9x.9
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    (Original post by assassinbunny123)
    how did you know that samples are independent
    If we're taking separate samples, there's no reason to think they will affect each other.
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    (Original post by assassinbunny123)
    how did you know that samples are independent
    The samples are independent because they come from the same distribution N(mean,variance^2).

    In the real world, most of the time you would be able to make this assumption based around a large enough population and slowly changing distribution but sometimes you wouldn't be able to e.g if you took samples of a considerable size of the population but since the question gives you what the distribution is it can't change. Moreover, the assumption that the distribution remains the same is actually required for the results in this module e.g that the standard error of the mean formula.
 
 
 
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