Suppose that when a transistor of a certain type is subjected to an accelerated

Question

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean 20 weeks and standard deviation 10 weeks. (a) What is the probability that a transistor will last between 10 and 20 weeks? (Round your answer to three decimal places.) (b) What is the probability that a transistor will last at most 20 weeks? (Round your answer to three decimal places.) Is the median of the lifetime distribution less than 20? Why or why not? The median is less than 20, since P(X p) = .5. (c) What is the 99th percentile of the lifetime distribution? (Round your answer to the nearest whole number.) d) Suppose the test will actually be terminated after t weeks what value of t is such that only your answer to the nearest whole number.) 5% of all rans stors would still be operating at termination? Round weeks You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Solution:

X= lifetime of the transistor in weeks

E(X)= 20 weeks x= 10 weeks

We were given the random variable X and its distribution and the expected value and variance. Before we can find the problems answers, we need to find the parameters and .

Mean of the gamma distribution is, E(X) =   ---------->(1)

Variance of the gamma distribution is, V(X) =  2 ------>(2)

From equation (2),  

2 = V(X)
() = V(X)/E(X)   
= V(X)/E(X) (Since, = E(X))
= (x)2/E(X) (Since, V(X) = (x)2)
= 102/20
= 100/20
= 5

Now, consider equation (1),
E(X) =
20 = (5)
=> = 20/5 = 4
Hence, the required parameter is, = 4 and = 5.

a) Find the probability that a transistors will last between 10 and 20 weeks

P(10 < X < 20) = P( X 32) -P( X 16)

= F(x/, ) -F(x/, )

= F(20/5,4)-F(10/5,4)

= F(4,4)-F(2,4)

= 0.567-0.143 = 0.424

Hence, the required probability is, 0.424

b) Find the probability that a transistor will last at most 20 weeks.

P(X 20) = F(20/5, 4) = F(4,4)

= 0.567

Hence, the required probability is 0.567

Now find the median of the life time distribution less than 20.

P(X -bar) = 0.5 From the given information, so -bar < 20 because, P(X 20) = 0.567

c) Find the 99th percentike of the life time distribution.

F(x; ) = 99%

F(x :4) = 0.99

From the gamma tabulated values at the = 4 and the corresponding 0.99 percent value to x is, 10.

Now, multiply by = 8 with x = 10 to get the 99th percentile .

99th percentile = x

= 10 * 8 = 80

Hence, the required lifetime is, 80 weeks.

d) Determine the number of weeks until only 0.5 remain working, find the value in the incomeplete gamma fucnction with = 4 such that

F(x; 4) = 1-0.5%

F(x; 4) = 1- 0.005

F(x; 4) = 0.995

Now, find the 99,5 percentile.

From the gamma tabulated values at the = 4 and the corresponding 0.995 percent value to x is, 11.

Now, multiply by = 8 with x = 11 to get the 99.5th percentile.

t = x = 11*8 = 88

Hence, the required lifetime is, t = 88 weeks.

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