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    • Thread Starter

    Struggling with this question;

    Lightbulb lifetime is normally distributed; mean=1100 hrs, standard deviation=80hrs

    A newly installed fitting takes 6 bulbs, lifetimes are independent of each other.

    The probability that the light fitting can run for t hrs without the bulbs failing is 0.95.

    At first i was thinking of using the Binomial where: 6C6 x t^6 = 0.95. But this wouldn't work because t is the number of hours not the probability. Then I tried finding the z value by using t as x but was not sure if the mean and standard deviation values would stay the same with 6 bulbs?
    • Thread Starter


    I think you do this:

    Say that Phi (the symbol for area on the normal tables) = 0.95, as this is the probability.
    And then find the corresponding z value, which is between 1.64 and 1.65, and as it turns out, it's directly in the middle, so z = 1.645. But, as it says 0.95 probability of the light bulb not failing, change z to -1.645.

    z = (x-mu)/sigma ... ((value-mean)/standard deviation)
    so, -1.645 = (x-1100)/80
    Rearrange to get x = 968.4 The light bulb can run for 968.4 hours with a 0.95 probability of not being needed to be replaced.

    That's how I would do it anyway.
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