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Engineering 323
Kelley
Beautiful HW#9
Problem 4-57
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 12 weeks.
A) What is the probability that a transistor will last between 12 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 99th 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 1% of all transistors would still be operating at
termination?
X= lifetime of the transistor in weeks
E(X)= 24 weeks
σx= 12 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)
V ( X ) 12 2
=
= 6 weeks
E ( X ) 24
E ( X ) 24
α=
=
=4
6
β
E(X)= αβ
β=
V(X)=αβ2
Now we can find the solutions:
Solution A
As shown on pages 167-169 to solve a gamma distribution problem we have to find the
probability that the transistor will last between 12 – 24 weeks.
0.04
f(X)
0.03
0.02
P(12 ∇ X ∇ 24)
0.01
0
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
X = lifetime of transistor
Figure 1: Shows the area representing the probability that X is between 12 and 24.
2/3
Engineering 323
Kelley
Beautiful HW#9
Problem 4-57
The excel formula used to create Figure one is as follows:
=gammadist(X, , , False)
P(12 ≤ X ≤ 24)
12
24
= P( ≤ G ≤ )
6
6
= P( 2 ≤ G ≤ 4 )
See table A.4 Column α = 4
= P( G ≤ 4) − P (G ≤ 2) = .424
= .567-.143 = .424
Refer to Figure 1 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) = .567
0.04
f(X)
0.03
P(X ∇ 24)
0.02
0.01
0
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
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
≤ 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 = 10.
.99 = P(X
η(.99)) = P(
3/3
Engineering 323
Kelley
Beautiful HW#9
Problem 4-57
so, now you have P (
X
≤ 10) = X = (6)(10) = 60
6
0.04
P(X ∇ 60) = .99
f(X)
0.03
0.02
0.01
0
0
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
X = lifetime of transistor
Figure 3: The area under this curve represents 99% of all transistors which we found out to be 60
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 = 11.
X
≤ 11) = .995
6
Now we can solve for the 99.5 percentile: X = (6)(11) = 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.
So, now we have P(
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Engr 323
Kelley
Beautiful HW #9
Problem 4-94
The paper “Determination of the MFT of Positive Photoresists Using the Monte Carlo
Method” (Photographic Sci. and Engr., 1983, pp. 254-260) 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 3.0? What is the probability that path
length is between 1.0 and 3.0?
C) What value is exceeded by only 10% 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 σx.
Solution A
As shown on pages 170-172 to solve the exponential distribution we have to find the
standard deviation and expected variance.
1
1
1
1
1
σx = 2 = =
E(X) = =
= 1.075µ m
= 1.075µm
λ .93
λ .93
λ
Solution B
ere we are trying to find the probability that X is greater than 3.
1
f(X)
0.8
0.6
0.4
P(X∇ 3)
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
X = photons free path lenght
Figure 1: Shows the area representing the probability that X is greater than 3.
5
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Engineering 323
Kelley
Beautiful HW#9
Problem 4-94
P(X > 3) = 1- P(X 3)
= 1- F(3)
= 1-[ 1 − e (93)−.93) ]
= .0614
The graph of Figure 1 shows that the area shaded is .0614 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 1 and 3. See figure 2 to see a graph of the area shaded.
P(1 X 3) = F(3) – F(1)
= .0614 – [ 1 − e − (.93)(1) ]
= .333
1
f(X)
0.8
0.6
P(1 ∇ X ∇ 3)
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
X = photons free path lenght
Figure 2: Shows the area representing the probability that X is between 1 and 3.
Solution C
What we want here is the value expected by only 10% of all path lengths. Another and
more familiar way to look at this problem is to consider it as wanting the 90th percentile.
90th percentile -(90) is what we want.
Here if we look at page 170 we see that F(x; ) = 1-£.90 = F(-(90)) = 1-£
Now solve for C.
ln(−1)
C=
= 2.476 µm
− .93
We can look at figure 3 to see the 90th percentile.
-.93(C)
(x)
, this is true for any x
0
3/3
Engineering 323
Kelley
Beautiful HW#9
Problem 4-94
1
The value exceeded by
only 10% of all path
lengths or the 90th
percentile
f(X)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
X = photons fre e pa th le nght
4.5
5