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Math Help - Normal Distribution

  1. #1
    Sep 2010

    Normal Distribution

    I have the following question:

    A machine makes slots of a duralumin forging. The width of the slot is normally distributed with mean of 0.900cm and standard deviation of 0.003cm. The specification limits were given as 0.900 0.007, i.e. forgings outside this interval are
    considered defective.

    (i) What percentage of the forgings will be defective?
    (ii) On average how many forgings will be made until a defective forging is found?
    (iii) How much should we fine-tune the machine (i.e. reduce the standard deviation) in order to reduce the probability of a defective forging to 1%.

    So for (i) is did:


    P( 0.893<= x <= 0.907)
    which gave me a value of 0.9576, so (i) is 1-0.9576 = 0.0424

    (ii) Is this the expected value for the geometric distribution?

    And Im stuck at the final part.
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  2. #2
    MHF Contributor Unknown008's Avatar
    May 2010
    (i) Are you sure for the first part? I'm getting another percentage... (than 4.24%)

    (ii) Yes

    (iii) Use P(x_1 \leq X \leq x_2) = 0.01

    where x_1 = \bar{x} - 3sd

    and x_2 = \bar{x} + 3sd

    And work backwards from there.
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  3. #3
    Sep 2010
    Sorry I think I used the wrong value for part(i), and thanks I understand part (iii) now.
    Last edited by mathsandphysics; November 25th 2010 at 02:24 PM.
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