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Statistics, probability ! Help!

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data

    The number of defects in a sample of 20 parts is recorded for quality control purposes; over the last year the number of defects and their occurrence rate has been : 0 defects, 82%; 1 defect, 13%; 2 defects, 4%; 3 defects, 1%. Find the expected number of defects in the next sample of 20 parts.

    2. Relevant equations

    (work shown below)

    3. The attempt at a solution

    1*.13+2*.04+3*.01=.24
    20*.24=4.8 the expected number of defects in the next sample of 20 parts.
     
  2. jcsd
  3. Jul 6, 2011 #2

    tiny-tim

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    Hi Calculator14! :smile:
    hmm … you don't know when to stop, do you? :redface:

    How can the expected number be between 4 and 5 when the given probabilities only go up to 3 ??

    What do you think the 0.24 is ? :wink:
     
  4. Jul 6, 2011 #3
    OHHHHHHHH!!! So I think I went a little too far with this, haha! My apologies tiny-tim, thank you for pointing out my mistake!! I believe my answer should be .24??
     
  5. Jul 7, 2011 #4

    tiny-tim

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    he he! :biggrin:

    yes :smile:
     
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