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Math Help - p.d.f.

  1. #1
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    p.d.f.

    The p.d.f. of time X to failure of an electronic component is f(x)=2x/1000^{2}*e^{-(x/7)^{2}}, 0<x<\infty

    a) Compute P(X>2000)
    b) Determine the 75th percentile, \pi_{0.75}, of the distribution
    c) Find the 10th and 60th percentiles, \pi_{0.10} and \pi_{0.60}
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  2. #2
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    Hello,
    Quote Originally Posted by antman View Post
    The p.d.f. of time X to failure of an electronic component is f(x)=2x/1000^{2}*e^{-(x/7)^{2}}, 0<x<\infty

    a) Compute P(X>2000)
    By definition of the cumulative density function, this is \int_{2000}^\infty f(x) ~dx
    b) Determine the 75th percentile, \pi_{0.75}, of the distribution
    Find t such that \int_0^t f(x) ~dx=0.75
    c) Find the 10th and 60th percentiles, \pi_{0.10} and \pi_{0.60}
    Same as b)
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    For part a, I got an answer of e^{-4} or 0.0183. I'm not sure how to solve for t in parts b and c. My book doesn't show this way of doing the problem or any other way for that matter. Thank you.
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    For part b, I set the equation up as \int with limits 0\le t \le .75 with f(t)=\frac{2t}{1000^{2}}e^{-(\frac{t}{1000})^{2}} dt but got an answer of 0.000000564, which seems too small to be correct. I tried it also with 75 as my upper limits and got 0.00561, which doesn't seem right either. Can anyone help me figure out what I am doing wrong? Thank you.
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    Quote Originally Posted by antman View Post
    For part b, I set the equation up as \int with limits 0\le t \le .75 with f(t)=\frac{2t}{1000^{2}}e^{-(\frac{t}{1000})^{2}} dt but got an answer of 0.000000564, which seems too small to be correct. I tried it also with 75 as my upper limits and got 0.00561, which doesn't seem right either. Can anyone help me figure out what I am doing wrong? Thank you.
    75th percentile means finding the value of x such that Pr(X < x) < 0.75. So the unknown x is an integral terminal and 0.75 is the value of the integral.
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    When I take the integral, I get the equation -e^{-\frac{1}{1000000}x^{2}}=0.75. Then taking the ln of both sides leaves -\frac{1}{1000000}x^{2}=-.287682. Multiplying both sides by -1000000 gives x^{2}=287682.0725, so x=536.36? Is this how you would solve for x? I don't know what to do with the -e. That's where I got stuck and was looking for an alternative way to solve the problem. Hopefully I have the right idea. Thank you.
    Last edited by antman; April 25th 2009 at 09:13 AM.
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    Quote Originally Posted by antman View Post
    When I take the integral, I get the equation -e^{-\frac{1}{1000000}x^{2}}=0.75. Then taking the ln of both sides leaves -\frac{1}{1000000}x^{2}=-.287682. Multiplying both sides by -1000000 gives x^{2}=287682.0725, so x=536.36? Is this how you would solve for x? I don't know what to do with the -e. That's where I got stuck and was looking for an alternative way to solve the problem. Hopefully I have the right idea. Thank you.
    Check: If you try integrating between 0 and your answer, you should get 0.75 if it's correct. I suggest using technology to make the job easier.
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    By the process of elimination, I got approximately 1177.41 but I don't know how to get there the right way. Does anyone have any ideas? Thank you.
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  9. #9
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    I got it! Thank you for your help. I integrated incorrectly but finally am able to get the answer through integration by hand. Thanks.
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