X = lifetime of transistor

Transcription

1 Kelley Problem 4-57 /3 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 24 weeks and standard deviation 2 weeks. A) What is the probability that a transistor will last between 2 and 24 weeks? B) What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24? Why, or why not? C) What is the 99 th percentile of the lifetime distribution? D) Suppose that the test will be determined after t weeks. What value of t is such that only one-half of % of all transistors would still be operating at termination? X= lifetime of the transistor in weeks E(X)= 24 weeks σ x = 2 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 α and β. X~gamma(α,β) 2 V ( X ) 2 E(X)= αβ β= = = 6 E( X ) 24 V(X)=αβ 2 E( X ) 24 α = = = 4 β 6 Now we can find the solutions: Solution A weeks As shown on pages to solve a gamma distribution problem we have to find the probability that the transistor will last between 2 24 weeks P(2 X 24) X = lifetime of transistor Figure : Shows the area representing the probability that X is between 2 and 24.

2 Kelley Problem /3 The excel formula used to create Figure one is as follows: =gammadist(x,,, False) P(2 X 24) 2 24 = P ( G ) 6 6 = P( 2 G 4 ) = P( G 4) P ( G 2) =. 424 See table A.4 Column α = 4 = =.424 Refer to Figure to get a graphical representation of this value. Solution B What we want to show here is the probability that the transistor will last at most 24 weeks. Which is the probability that Gamma is less than or equal to 6. P(X 24 ) = P( G 6) = P(X 24) X = lifetime of transistor Figure 2: Shows the area representing the probability that X will last at most 24 weeks. For the second part of problem B: P ( X ~ µ ) =.5 (by definition) So ~ µ < 24 because P(X 24 ) =. 567 which is greater than.5 Solution C X.99 = P(X η(.99)) = P( η(.99)) Now go to table A.4 in your book find where 6 α = 4 and trace down until you find.99. Now read across to get the X value, this will be X =.

3 Kelley Problem /3 X so, now you have P ( ) = X = (6)() = P(X 6) = X = lifetime of transistor Figure 3: The area under this curve represents 99% of all transistors which we found out to be 6 transistors. Solution D What we need here is the 99.5 percentile. Again we need to go to table A.4 and trace down from α = 4 until we find.995, then read across to get the needed X value, which will be X =. X So, now we have P( ) = Now we can solve for the 99.5 percentile: X = (6)() = 66 To view a graphical representation see Figure 3 the shaded area should start from 66 and go to the left. A graph wasn t made because this part is so similar to part C.

4 Engr 323 Beautiful HW #9 Kelley Problem 4-94 /3 The paper Determination of the MFT of Positive Photoresists Using the Monte Carlo Method (Photographic Sci. and Engr., 983, pp ) proposes the exponential distribution with parameter λ =.93 as a model for the distribution of a photons free path length (µm) under certain circumstances. Suppose this is the correct model. A) What is the expected path length, and what is the standard deviation of path length. B) What is the probability that path length exceeds 3.? What is the probability that path length is between. and 3.? C) What value is exceeded by only % of all path lengths? X = photons free path length ( m) X~exp( =.93) We were given the random variable X its distribution and parameter. Now we need to find E(X) and σ x. Solution A As shown on pages 7-72 to solve the exponential distribution we have to find the standard deviation and expected variance. E(X) = = =.75µ m σ x = 2 λ.93 λ = = =.75 λ.93 µ m Solution B ere we are trying to find the probability that X is greater than P(X 3) X = photons free path lenght Figure : Shows the area representing the probability that X is greater than 3.

5 Kelley Problem /3 P(X > 3) = - P(X 3) = - F(3) ( 93).93) = -[ e ] =.64 The graph of Figure shows that the area shaded is.64 or about 6%. The excel equation used for the exponential distribution is: =expondist(x,, false) For the next part of the problem we are trying to find the probability that X is distributed exponentially between and 3. See figure 2 to see a graph of the area shaded. P( X 3) = F(3) F() (.93)() =.64 [ e ] = P( X 3) X = photons free path lenght Figure 2: Shows the area representing the probability that X is between and 3. Solution C What we want here is the value expected by only % of all path lengths. Another and more familiar way to look at this problem is to consider it as wanting the 9 th percentile. 9 th percentile -(9) is what we want. Here if we look at page 7 we see that F(x; ) = - -(x), this is true for any x.9 = F(-(9)) = (C) Now solve for C. ln( ) C = = 2.476µ m.93 We can look at figure 3 to see the 9 th percentile.

6 Kelley Problem / The value exceeded by only % of all path lengths or the 9 th percentile X = photons free path lenght