Solution Found!
Suppose that the lifetime of a component (in hours), X is
Chapter 4, Problem 164E(choose chapter or problem)
Suppose that the lifetime of a component (in hours), X is modeled with a Weibull distribution with \(\beta=0.5\). and \(\delta=4000\). Determine the following in parts (a) and (b):
(a) \(P(X>3500)\) (b) \(P(X>6000 \mid X>3000)\)
(c) Comment on the probabilities in the previous parts compared to the results for an exponential distribution.
(d) Comment on the role of the parameter β in a lifetime model with the Weibull distribution.
Equation transcription:
Text transcription:
\beta=0.5
\delta=4000
P(X>3500)
P(X>6000 \mid X>3000)
Questions & Answers
QUESTION:
Suppose that the lifetime of a component (in hours), X is modeled with a Weibull distribution with \(\beta=0.5\). and \(\delta=4000\). Determine the following in parts (a) and (b):
(a) \(P(X>3500)\) (b) \(P(X>6000 \mid X>3000)\)
(c) Comment on the probabilities in the previous parts compared to the results for an exponential distribution.
(d) Comment on the role of the parameter β in a lifetime model with the Weibull distribution.
Equation transcription:
Text transcription:
\beta=0.5
\delta=4000
P(X>3500)
P(X>6000 \mid X>3000)
ANSWER:Answer
Step 1 of 4
(a)
Suppose that the lifetime of a component ( in hours) is modeled with a Weibull distribution with
We are asked to find the probability
The random variable with probability density function
is a Weibull random variable with scale parameter and shape parameter
.
The cumulative distribution function of is
………(1)
We can write as,
Using equation (1), we can write,
Hence the probability is