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# Normal Distribution S2 watch

1. 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?
2. bump
3. 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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Updated: March 20, 2013
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