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 | Select- $20, since P(X ) = .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 0.5% of all transistors would still be operating at termination? (Round your answer to the nearest whole number.) t= weeks

Explanation / Answer

Solution:
Let X denote the life time of the transistor in weeks and follows a gamma distribution with parameters(, )
From the given information, E(X) = 20 weeks and x = 10 weeks
Meean of the gamma distribution is, E(X) = -----(1)
Variance of the gamma distribution is, V(X) = ^2
From equation(2),
^2 = V(X)
() = V(X)
= V(X)/E(X) (Since, = E(X))
= (x)^2/E(X) (Since, V(X) = (x)^2)
= (10)^2/20
= 100/20 = 5
Now consider equation(1) ,
E(X) =
20 = (5) (Since, E(X) = 20 and = 5)
=> = 20/5 = 4
Hence, the required parameters is, = 4 and = 5.

a) Find the probability that a transistors will last between 10 and 20 weeks.
P(10 < X < 20) = P(X 20) -P(X 10)
= F(x/, ) - F(x/, )
= F(20/5, 4) - F(10/5, 4)
= F(4, 4) - F(2,4) = 0.567 -0.143 (From Gamma table values)
= 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 (From Gamma table values)
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 which is greater than the 0.5.

c) Find the 99th percentile 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 = 5 with x = 10 to get the 99th percentile.
99th percentile = x = 10 * 5 = 50
Hence, the required lifetime is, 50 weeks

d) Determine the number of weeks until only 0.5 remian working, find the value in the incomplete gamma function 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 wih x = 11 to get the 99.5th percentile.
t = x = 11 * 5 = 55
Hence, the required lifetime is, t = 55 weeks

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