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M

  • 5 years ago

x is exponential random variable f(x) = (1/θ)e^-x/θ given: P(x < 2.40) = .520 find: P(x > 4.36)

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  1. anonymous
    • 5 years ago
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    You need to integrate f(x) from 0 to 2.40 and set that equal to .520. You will have an equation you can solve for theta. Once you find theta integrate the pdf again from 4.36 to infinity, or take one minus the integration from 0 to 4.36 whichever you prefer.

  2. anonymous
    • 5 years ago
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    sorry i dont know why it posted that many times

  3. M
    • 5 years ago
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    i keep getting a negative number for the theta -1.5694 i end up with a negative answer which i don't think it's correct

  4. M
    • 5 years ago
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    make that theta = -3.67

  5. M
    • 5 years ago
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    -e^2.40/theta - [-e^-0/theta] = .520 -e^-2.40/theta + 1 = .520 e^2.40/theta = .48 -2.40/theta = ln(.48) theta = 3.2699 i was trying to ln the limit 0 at first and that's why i was having problem. yay for perseverance :)

  6. M
    • 5 years ago
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    -e^-2.40/theta - [-e^-0/theta] = .520 -e^-2.40/theta + 1 = .520 e^-2.40/theta = .48 -2.40/theta = ln(.48) theta = 3.2699 -e-x/theta from 4.36 to infinity -e^-inf/3.269 - [-e^-4.36/3.269] = 0 + .2635 =.2635

  7. anonymous
    • 5 years ago
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    yep I think that's correct

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