##### Help solve this problem??

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A certain brand of light bulb has a life, before burn-out that is normally distributed with a mean of 3000 hours and a standard deviation of 200 hours.

If one such light bulb is randomly selected and used in a warehouse where it will be left on continuously. What is the probability that it will last for at least 120 days (4 months)? Note: 120 days of continual use is 120*24=2880 hours.

Jun 3rd, 2015

So then we're looking for the probability that it will last 2880 or MORE.

First let's find the Z value

Z = 2880 - 3000 / 200 = -120 / 200 = -0.6

According to the Z table of standard  normal distribution, -0.6 corresponds to a probability of 0.2743

But that's for a bulb lasting 2880hrs or LESS

For 2880hrs or MORE we have to subtract from 1

1 - 0.2743 = 0.7257

Jun 3rd, 2015

Thank you so so much!! I didn't realize that is how to approach it...

So if six light bulbs are used in a warehouse where each light bulb would be left on continuously. What is the probability that more than half of the bulbs will last for at least 120 days? And the bulbs are independent of other bulbs?

Jun 3rd, 2015

Then you use the formula nCr(Ps)^r(Pf)^n-r

where n is the number of trials

r is the number of successes

Ps is the probability of success

Pf is the probability of failure

Jun 3rd, 2015

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Jun 3rd, 2015
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Jun 3rd, 2015
May 24th, 2017
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