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Laboratory tests show that the lives of light bulbs are normally distributed with a mean of 750 hours and a standard deviation of 75 hours. Find the probability that a randomly selected light bulb will last between 825 and 900 hours

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Answer:

P(825 to 900 hours) = 0.4772-0.3413=0.1359

Explanation:

We are given that the life-span of the bulbs are normally distributed.

First we need to find the z-values, where

z(x)=(x-mean)/standard deviation

=(x-750)/75

therefore

z(825)=(825-750)/75=1

z(900)=(900-750)/75=2

Now we need to find the probabilities of the given z-values, from a probability table or from software.

P(z=1) = 0.3413

P(z=2)=0.4772

So probability that the life-span falls between 825 and 900 hours would be P(825 to 900 hour) = 0.4772-0.3413=0.1359

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