Let X have a Weibull distribution with the pdf below. Verify that μ = f(x; a, B) = μ = X. Using the substitution, y = · (²) ª = В x < 0 = µT(1 + ²). [Hint: In the integral for E(X), make the change of variable y = (*)* Now we can simplify as follows. fºx α α Ba -60 (1) = By = pr(1 + ¹) ¹e-(x/B)" dx ¹e-(x/B)" 0 X ]) x 20 Je-y dy Thus, dy = yllªe-y dy - (( =)^²-₁) ². )dx. X , so that x = By¹/a.]
Let X have a Weibull distribution with the pdf below. Verify that μ = f(x; a, B) = μ = X. Using the substitution, y = · (²) ª = В x < 0 = µT(1 + ²). [Hint: In the integral for E(X), make the change of variable y = (*)* Now we can simplify as follows. fºx α α Ba -60 (1) = By = pr(1 + ¹) ¹e-(x/B)" dx ¹e-(x/B)" 0 X ]) x 20 Je-y dy Thus, dy = yllªe-y dy - (( =)^²-₁) ². )dx. X , so that x = By¹/a.]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Let X have a Weibull distribution with the pdf below.
α
{
Ba
f(x; α, ß) =
Verify that μ = BT 1 +
Using the substitution, y =
μ =
=
α
ta
Now we can simplify μ as follows.
- Sªx.
Ba
-6-(0₂ (²)
- (B
1
- M(₁ + ²)
=
1
-1e-(x/B) a
BT(1 + ²). [Hint: In the integral for E(X), make the change of variable y =
= (*) ª.
Χα
-xa-¹e-(x/B)α dx
0
X
ха
ва
Je-y
) [v
x ≥ 0
dy
x < 0
Thus, dy =
y¹/ae-y dy
a
(()*²)
ва
dx.
so that x =
By ¹/α.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddd96ab9-54be-46e6-927a-d9e159267b74%2F8582165b-eba2-4565-97e4-9554a7955524%2Fatnsleb_processed.png&w=3840&q=75)
Transcribed Image Text:Let X have a Weibull distribution with the pdf below.
α
{
Ba
f(x; α, ß) =
Verify that μ = BT 1 +
Using the substitution, y =
μ =
=
α
ta
Now we can simplify μ as follows.
- Sªx.
Ba
-6-(0₂ (²)
- (B
1
- M(₁ + ²)
=
1
-1e-(x/B) a
BT(1 + ²). [Hint: In the integral for E(X), make the change of variable y =
= (*) ª.
Χα
-xa-¹e-(x/B)α dx
0
X
ха
ва
Je-y
) [v
x ≥ 0
dy
x < 0
Thus, dy =
y¹/ae-y dy
a
(()*²)
ва
dx.
so that x =
By ¹/α.]
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