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Im stuck in this statistics problem

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data

    the weight of items produced by a production line is normally distributed with a mean of 12 ounces and a standard deviation of 2 ounces.
    Suppose that quality control requires the weight of items to be within 8 and 16 ounces. You select 7 items at random (each item is independent). What is the probability that 3 items will fulfill quality control requirements.

    What I have to the moment is:
    X~n(12,2)
    P(x<16) - P(x<8)
    z=(16-12)/2
    z=2

    z= (8-12)/2
    z=-2

    p(z<2)-P(z<-2)
    =.9772-.0228
    =.9544

    p(x>20)
    z=(20-12)/2
    z=4
    P(z>4) = 1

    Those where other requirements of the problem. I'm stuck at the supposing, cant decide if it is binomial or not, and how to start the procedure
     
  2. jcsd
  3. Aug 25, 2010 #2

    presbyope

    User Avatar
    Gold Member

    So you calculated the probability of success for each trial to be 0.9544? That takes care of the supposing part. Now figure out what the 3 out of 7 means.

    .. personally I can't tell. Is it asking for _exactly_ 3 out of 7?
     
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