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Thread: Probability

  1. #1
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    Probability

    Got this wrong on my assignment, please help?



    The average life of a light bulb is 500 hours. Find

    a)the probability that a bulb will last more than 1,000 hours
    b)the probability that a bulb will last less than 100 hours
    c)the median life
    d)the probability that a bulb will last exactly 500 hours
    e)the 95 Percentile


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    Distribution, work you did that was wrong. . .etc.
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  3. #3
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    Quote Originally Posted by lisa1984wilson View Post
    Got this wrong on my assignment, please help?



    The average life of a light bulb is 500 hours. Find

    a)the probability that a bulb will last more than 1,000 hours
    b)the probability that a bulb will last less than 100 hours
    c)the median life
    d)the probability that a bulb will last exactly 500 hours
    e)the 95 Percentile

    Is it known what the distribution is that the life of a bulb follows? is the variance or standard deviation known? Without knowing the whole question it's impossible to provide help.
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    That is the full question, and this was is below it.



    (Note: From the table on page 722 of your textbook, e^ ( -3.0) is almost 0.05, so the 95 percentile life would be -3.0/0.002 = 1,500 hours)
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    If it's an exponential distribution with an AVERAGE (mu) lifetime of 500 hours, then from your notes you should remember that:

    $\displaystyle \mu_x=\frac{1}{\lambda}$

    Solving for lambda will allow you to get your parameter lambda.

    To find a through e you have to remember that $\displaystyle P(X\leq x)$ is represented by $\displaystyle 1-e^{x\lambda}$, and for $\displaystyle P(X> x)$ is represented by $\displaystyle 1-P(X\leq x)$.

    Try to work with that and see where you get, or at least show us the work you did that was determined to be wrong. To help you get started:

    A. $\displaystyle P(X>1000)=1-P(X\leq 1000)$. Why? Because to find the probability that a component lasts longer than 1000 hours, we only need to find the compliment of it lasting SHORTER than 1000 hours. So:
    $\displaystyle P(X>1000)=1-[1-e^{1000*-0.002}]$
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  6. #6
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    thanks I figured it out!
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