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Math Help - Normal distribution word problem...

  1. #1
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    Normal distribution word problem...

    The fluorescent light tube made by the company have lifetimes which are normally distrubted with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life of the tubes, replacing free of charge any tubes that fail to meet this minimum life. If the company wishes to have to replace free only 3% of the tubes sold, find the guaranteed minimum it must set.

    Let L = life, and u = mean life, and o = standard deviation.

    We have:
    L ~ N(2010, 20^2)

    Let Z = (X - 2010) / 20 ... Then Z ~ N(0,1).

    I tried it this way, but it won't work. Not sure why is the following wrong:

    P(L > -0.97) = 1 - \phi (0.97)

    Please let me know the correct way with some explanation as I need to figure it out.
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  2. #2
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    Quote Originally Posted by struck View Post
    The fluorescent light tube made by the company have lifetimes which are normally distrubted with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life of the tubes, replacing free of charge any tubes that fail to meet this minimum life. If the company wishes to have to replace free only 3% of the tubes sold, find the guaranteed minimum it must set.

    Let L = life, and u = mean life, and o = standard deviation.

    We have:
    L ~ N(2010, 20^2)

    Let Z = (X - 2010) / 20 ... Then Z ~ N(0,1).

    I tried it this way, but it won't work. Not sure why is the following wrong:

    P(L > -0.97) = 1 - \phi (0.97)

    Please let me know the correct way with some explanation as I need to figure it out.
    The inverse problem you need to solve is \Pr(L < L_{min}) = 0.03.
    Last edited by mr fantastic; July 18th 2009 at 12:31 AM. Reason: Corrected a minor typo: replaced > with <
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  3. #3
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    I was doing it the way you told me earlier but I failed so I thought I was wrong. So I have spent time figuring it out again and it appears as such that:

    \Pr(L > L_{min}) = 0.03

    Then:

    L_{min} = \phi^{-1} (0.03)
    (L - 2010) / 20 = -1.8808
    L = (-1.8808 * 20) + 2010 = 1972.384


    So the minimum life has to be 1972.384.

    The table I use does not give value for inverse normal distribution for values below 0.5, so it's fair to say that because:
    \phi (1.8808) = 0.97 => \phi (-1.8808) = 0.03

    I guess it's because of the 1 - \phi (z) relationship?

    Thanks.. just a bit confused here.

    Last edited by mr fantastic; July 18th 2009 at 12:34 AM. Reason: Fixed a latex exponent
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  4. #4
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    Hi !

    I have the same problem with this question. So what should be the answer? The book answer is 1970 hours but I only can get 1972.384.. If anyone can get 1970, how did you arrive to that answer?

    Thanks in advance! =)
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    Quote Originally Posted by Cathelyn13 View Post
    Hi !

    I have the same problem with this question. So what should be the answer? The book answer is 1970 hours but I only can get 1972.384.. If anyone can get 1970, how did you arrive to that answer?

    Thanks in advance! =)
    Pr(Z < -1.8808) = 0.03.

    So (L - 2010) / 20 = -1.8808 => L = 1972 (correct to the nearest hour). So you are correct.

    Perhaps the book is rounding to the nearest 10 hours.
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