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Math Help - finding percentile rank of a lightbulb that failed at 9000 hours

  1. #1
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    finding the percentile rank of a lightbulb that failed at 9000 hours

    I'm trying to find the approximate percentile rank of a lightbulb that failed at
    9000 hours (based on the information from the problem below), but I'm not sure how to solve this...

    A consumer product testing lab tested 400 13-watt compact

    fluorescent light bulbs by turning them all on at once, leaving them
    on until they burned out, and recording the time-to-failure of each
    bulb. The test revealed a symmetric and mound-shaped
    distribution of times-to-failure with mean 7800 hours and standard
    deviation 1200 hours. The number of lightbulbs that lasted between 5400 and 9000 hours were about 326 lightbulbs.

    This is my guess:

    percentile = 7800/9000 = about 86%, since 9000 hours is 1 standard deviation from the mean
    Last edited by nyago; December 5th 2012 at 10:29 AM.
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  2. #2
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    Re: finding percentile rank of a lightbulb that failed at 9000 hours

    Hey nyago.

    The thing you will have to do is use the information about the distribution to obtain the probability.

    To look at failures before 9000 hours, we need to look at the probability where P(X < 9000) where X is the distribution for failures.

    It says the distribution is symmetric and mound shaped which suggests a hint to use the normal distribution. It also gives you the mean and standard deviation.

    So given X ~ Normal(7800,1200^2) [mean,variance] can you find P(X < 9000)?
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  3. #3
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    Re: finding percentile rank of a lightbulb that failed at 9000 hours

    I'm not sure if we've gone over normal distribution in our class, just z-scores, percentile rank and the empirical rule.... I got this answer from someone, is this right way to find it? ,
    9000 hours is exactly one standard deviation away from the mean (7800+1200=9000) so is approximately 34.1% + 50 % = 84.1%
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  4. #4
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    Re: finding percentile rank of a lightbulb that failed at 9000 hours

    You are spot on.

    Using R we get the probability to be:
    > pnorm(1,0,1)
    [1] 0.8413447
    Thanks from nyago
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