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Math Help - [SOLVED] Normal Distrubtion...

  1. #1
    Member looi76's Avatar
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    Jan 2008
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    [SOLVED] Normal Distrubtion...

    Question:
    The fluorescent light tubes made by the company Well-lit have lifetimes which are normally distrusted with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life. If the company wishes to have to replace free only 3% if the tubes sold, find the guaranteed minimum it must set.

    Attempt:

    \mu = 2010 , \sigma = 20

    How can I get the value of  \phi^{-1} (0.3\%) out of the table?
    Attached Thumbnails Attached Thumbnails [SOLVED] Normal Distrubtion...-normal_distrubtion_table.gif  
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
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    802
    Hi

    Let's call T the random variable associated to the lifetime.
    We're looking for t=\phi^{-1}(3\%) which means P(T\leq t)=\int_{-\infty}^t\frac{1}{\sigma\sqrt{2\pi}}\exp \frac{(x-\mu)^2}{\sigma^2}\,\mathrm{d}x=0.997

    As we know the value taken by the normal distribution for \sigma=1 and \mu=0, we substitute u=\frac{x-\mu}{\sigma} \Rightarrow \mathrm{d}u=\frac{\mathrm{d}x}{\sigma} :
    P(T\leq t)=\int_{-\infty}^{\frac{t-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp u^2\,\mathrm{d}u=0.997

    Using the table, you can find the value of \frac{t-\mu}{\sigma} corresponding and then deduce \phi^{-1}(3\%)=t.
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