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Math Help - HELP: Normal distribution probability problem

  1. #1
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    HELP: Normal distribution probability problem

    I have the following problem:

    A certain small freight elevator has a max. capacity C, which is Normally distributed, with mean 400 lbs., and standard deviation 4 lbs.

    The weight of the boxes being loaded into the elevator is a R.V., with mean 30 lbs., and standard deviation 0.3 lbs.

    How many boxes may be loaded into this elevator before the probability of disaster exceeds 0.2?

    I have some idea of how to get started, but I can't seem to find the right formula. Can any one help? Thanks in advance, any help is greatly appreciated.

    Mikhail
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  2. #2
    MHF Contributor matheagle's Avatar
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    P(disaster)=P\left(C<\sum_{i=1}^nB_i\right)

    C is your capacity rv
    Bi is the rv associated with the weight of the ith box.
    n is the fixed number of boxes.
    IF you have independence between the C and the boxes, then

    C-\sum_{i=1}^nB_i is a normal random variable and you want

    P\left(C-\sum_{i=1}^nB_i<0\right)=.2

    YOU can find the 20th percentile of a standard normal and the distribution of

    W=C-\sum_{i=1}^nB_i
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  3. #3
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    Thank you!
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