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Math Help - Normal distribution help

  1. #1
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    Normal distribution help

    Q: To avoid a complete shutdown of the assembly line if a component fails, major manufacturing companies operate on the principle of preventative maintenance. the lifetime of one component is normally distributed with a mean of 325h and a standard deviation of 20h. How frequently should the component be replaced so that the probability of its failing during operation is less than 0.01?

    A: I don't exactly understand it.

    What exactly do I do? I tried stuff like P(x<0.01) etc but that doesnt work.

    Is it something to do with making it =0.01?
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  2. #2
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    Quote Originally Posted by Johnboe View Post
    Q: To avoid a complete shutdown of the assembly line if a component fails, major manufacturing companies operate on the principle of preventative maintenance. the lifetime of one component is normally distributed with a mean of 325h and a standard deviation of 20h. How frequently should the component be replaced so that the probability of its failing during operation is less than 0.01?

    A: I don't exactly understand it.

    What exactly do I do? I tried stuff like P(x<0.01) etc but that doesnt work.

    Is it something to do with making it =0.01?
    Did you try to solve for a where

    P\left( Z<\frac{a-325}{20}\right)= 0.01
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Did you try to solve for a where

    P\left( Z<\frac{a-325}{20}\right)= 0.01
    so a=325.2 if I am correct?

    No, grr, I don't get it.
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  4. #4
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    Quote Originally Posted by Johnboe View Post
    so a=325.2 if I am correct?

    No, grr, I don't get it.
    Get the z-value z* such that Pr(Z < z*)= 0.01. Now solve z* = (a - 325)/20 for a.
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    Quote Originally Posted by Johnboe View Post
    so a=325.2 if I am correct?
    Think about the curve itself and the case where a = 325

    It gives

    P\left( Z<\frac{325-325}{20}\right)= P(Z<0 )=0.5

    therefore your answer needs to be much less than 325
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  6. #6
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    z*=0.01 @ 0.5040

    0.5040 = a-325 / 20

    10.08+325=a
    a=335.08

    mm, still don't get it
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  7. #7
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    This is the problem.

    Quote Originally Posted by Johnboe View Post
    z*=0.01 @ 0.5040
    z^{*} =0.01 at -2.3263

    Look here Z table - Normal Distribution
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  8. #8
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    Where did you get -2.3263 from?
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  9. #9
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    Quote Originally Posted by Johnboe View Post
    Where did you get -2.3263 from?
    Quote Originally Posted by pickslides View Post
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  10. #10
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    Yes, I looked there... but it just has 2 graphs and I played around and I do not get -2.3263 anywhere.
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  11. #11
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    Quote Originally Posted by Johnboe View Post
    Yes, I looked there... but it just has 2 graphs and I played around and I do not get -2.3263 anywhere.



    Go to Z table - Normal Distribution

    Under the second curve, under "Shadded Area" put 0.01 and tick "Below" A figure will pop up next to "Below"
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  12. #12
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    Which then solves to a=278.474?
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  14. #14
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    Which means: the component should be replaced every 278.5 hours so it fails less than 1%?
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  15. #15
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    Thats how I read it.
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