Let X have a Weibull distribution with the pdf below. f(x; a, B) = μ = Using the substitution, y = · (‡)“ - = X. Verify that μ = (1+¹). [Hint: In the integral for E(X), make the change of variable y = (*). Now we can simplify as follows. x²-1e-(x/p)a - 6*([ α 0 -xa-¹e-(x/B) dx = pr(1 + ²) e-Y dy x 20 x < 0 ])"y¹ªey dy Thus, dy = dx. , so that x = By¹/a.]
Let X have a Weibull distribution with the pdf below. f(x; a, B) = μ = Using the substitution, y = · (‡)“ - = X. Verify that μ = (1+¹). [Hint: In the integral for E(X), make the change of variable y = (*). Now we can simplify as follows. x²-1e-(x/p)a - 6*([ α 0 -xa-¹e-(x/B) dx = pr(1 + ²) e-Y dy x 20 x < 0 ])"y¹ªey dy Thus, dy = dx. , so that x = By¹/a.]
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter3: The Derivative
Section3.CR: Chapter 3 Review
Problem 7CR
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![Let X have a Weibull distribution with the pdf below.
α
f(x; α, ß):
#
=
ва
Verify that μ = Br1 +
Using the substitution, y =
μ =
X.
1
· pr(1 + ¹). [Hint: In the integral for E(X), make the change of variable y =
X
(ỗ)
Now we can simplify μ as follows.
· 6²×
/0
= 6*([
- (
=
α
-xa-¹e-(x/B) α x ≥ 0
Ba
1
= A ( ₁ + ²)
0
X
ха
(*)* - *.
=
¹e-(x/B) a dx
Je-y dy
).6²x
x < 0
y¹/αe-y dy
Thus, dy =
dx.
α
so that x =
By¹/a.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e2cacc5-c358-4323-b5fe-70f1523c28bd%2F71a637d9-1833-4748-bf86-68bec6547290%2Fsf82fz_processed.png&w=3840&q=75)
Transcribed Image Text:Let X have a Weibull distribution with the pdf below.
α
f(x; α, ß):
#
=
ва
Verify that μ = Br1 +
Using the substitution, y =
μ =
X.
1
· pr(1 + ¹). [Hint: In the integral for E(X), make the change of variable y =
X
(ỗ)
Now we can simplify μ as follows.
· 6²×
/0
= 6*([
- (
=
α
-xa-¹e-(x/B) α x ≥ 0
Ba
1
= A ( ₁ + ²)
0
X
ха
(*)* - *.
=
¹e-(x/B) a dx
Je-y dy
).6²x
x < 0
y¹/αe-y dy
Thus, dy =
dx.
α
so that x =
By¹/a.]
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