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    hi,
    can someone help with the following....

    the mean number of faulty bolts produced by a machine per hour is 0.3. the machine runs non stop for 8hrs. wat is the probability that more than 4 bolts are faulty?

    Thank you
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    X
    = Number of faulty bolts in 8 hours
    ~ Po(2.4)

    P(X > 4)
    = 1 - P(X <= 4)
    = 1 - e^(-2.4) (1 + 2.4 + 2.4^2/2! + 2.4^3/3! + 2.4^4/4!)
    = 0.096
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    (Original post by Vijay1)
    Jonny's answer would be an estimate of the binomial by using the poisson distribution.
    There can be more than one faulty bolt in an hour. (So there can be more than eight in eight hours.)

    Number of faulty bolts in a hour ~ Po(0.3)
    Number of faulty bolts in eight hours ~ Po(2.4)
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    (Original post by Jonny W)
    There can be more than one faulty bolt in an hour. (So there can be more than eight in eight hours.)

    Number of faulty bolts in a hour ~ Po(0.3)
    Number of faulty bolts in eight hours ~ Po(2.4)
    Oh I see, :rolleyes:
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    (Original post by Vijay1)
    Yes but why use the poisson when the chance is only of a probabilty 0.3 of faulty bolt and not faulty 0.7.
    Since there is only two possiblities you would use the binomial.
    The question DIDN'T say

    "In each hour, the machine produces a faulty bolt with probabilty 0.3 and no faultly bolts with probability 0.7".
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    (Original post by Jonny W)
    The question DIDN'T say

    "In each hour, the machine produces a faulty bolt with probabilty 0.3 and no faultly bolts with probability 0.7".
    Yes I misread, your right.
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    Ariel68 posted this question two weeks ago in another thread. She got the same (right and wrong) responses as she has had here!

    http://www.thestudentroom.co.uk/t120950.html
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    I know it did. but my teacher checked it and he said it was wrong!?
    How sure are you about your conclusion?
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    (Original post by Ariel68)
    I know it did. but my teacher checked it and he said it was wrong!?
    How sure are you about your conclusion?
    99.9%
 
 
 
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