Question

A newly replaced light bulb is used until it burns out and then is replaced immediately by another. The two bulbs are selected randomly from a large lot of bulbs that the manufacturer says have an approximately normal lifetime distribution with a mean of 800 hours and a standard deviation of 100 hours. Describe the distribution of the total lifetime of the two bulbs used in sequence. What is the probability that the sum of the two lifetimes exceeds 1800 hours? What is the probability that the sum of the lifetimes of nine such bulbs selected at random and burned sequentially exceeds 8100 hours? What is the probability that the second light bulb in the sequence burns longer than the first?

Answer 1

Let X is a random variable shows the lifetime of bulb. Here X has normal distribution with parameters as follows:

The total lifetime, T = 2X, of two bulbs will be normally distributed with mean and SD as follows

The z-score for T = 1800 is

The probability that the sum of the two lifetimes exceeds 1800 hours is

P(T > 1800) = P(z > 1) =1 - P(z <=1)= 0.1587

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The total lifetime, T = 9X, of nine bulbs will be normally distributed with mean and SD as follows

The z-score for T = 8100 is

The probability that the sum of the 9 lifetimes exceeds 1800 hours is

P(T > 1800) = P(z > 1) =1 - P(z <=1)= 0.1587

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Let X shows the lifetime of first bulb and Y shows the lifetime of second bulb. So distribution of D=X-Y will be approximately normal with mean and Sd as follows:

and SD:

The z-score for D = 0 is

The probability that the second light bulb in the sequence burns longer than the first, X-Y <0, is