The time to failure, t, in hours, of a machine is often exponentially distributed with a probability density function f(t)=ke-kt, 0≤t<∞o, where k=- and a is a the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is 80 hr. What is the probability that a failure will occur in 52 hr or less? The probability is (Round to four decimal places as needed.)
The time to failure, t, in hours, of a machine is often exponentially distributed with a probability density function f(t)=ke-kt, 0≤t<∞o, where k=- and a is a the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is 80 hr. What is the probability that a failure will occur in 52 hr or less? The probability is (Round to four decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Time to Failure Analysis of a Machine using an Exponential Distribution
The time to failure, \( t \), in hours, of a machine is often exponentially distributed with a probability density function given by:
\[ f(t) = k e^{-kt}, \quad 0 \leq t < \infty \]
where \( k = \frac{1}{a} \) and \( a \) is the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is 80 hours.
**Question:**
What is the probability that a failure will occur in 52 hours or less?
To calculate the probability, we use the exponential distribution's cumulative distribution function (CDF):
\[ F(t) = 1 - e^{-kt} \]
Given:
- Average time \( a = 80 \) hours
- \( k = \frac{1}{a} = \frac{1}{80} \)
- Time \( t = 52 \) hours
Plugging in the given values into the CDF:
\[ F(52) = 1 - e^{-\frac{1}{80} \times 52} \]
**Solution:**
The probability is \(\boxed{\phantom{00}}\).
*Round to four decimal places as needed.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc56b3bff-26b7-4f28-9b8d-5c65e8bb1b58%2F4ce1e7d3-31a9-45c4-ad89-a8dd08cca76f%2F1uwqfjd_processed.png&w=3840&q=75)
Transcribed Image Text:### Time to Failure Analysis of a Machine using an Exponential Distribution
The time to failure, \( t \), in hours, of a machine is often exponentially distributed with a probability density function given by:
\[ f(t) = k e^{-kt}, \quad 0 \leq t < \infty \]
where \( k = \frac{1}{a} \) and \( a \) is the average amount of time that will pass before a failure occurs. Suppose that the average amount of time that will pass before a failure occurs is 80 hours.
**Question:**
What is the probability that a failure will occur in 52 hours or less?
To calculate the probability, we use the exponential distribution's cumulative distribution function (CDF):
\[ F(t) = 1 - e^{-kt} \]
Given:
- Average time \( a = 80 \) hours
- \( k = \frac{1}{a} = \frac{1}{80} \)
- Time \( t = 52 \) hours
Plugging in the given values into the CDF:
\[ F(52) = 1 - e^{-\frac{1}{80} \times 52} \]
**Solution:**
The probability is \(\boxed{\phantom{00}}\).
*Round to four decimal places as needed.*
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