The time in hours during which an electrical generator is operational is a random variable that follows an exponential tribution. If the average operational time is 160 hours, what is the probability that a generator of this type will be erational between 60 and 160 hours?
The time in hours during which an electrical generator is operational is a random variable that follows an exponential tribution. If the average operational time is 160 hours, what is the probability that a generator of this type will be erational between 60 and 160 hours?
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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`3) Solve the question with the full method.
![### Exponential Distribution and Generator Operational Time
#### Problem Statement:
**Question 3:** The time in hours during which an electrical generator is operational is a random variable that follows an exponential distribution. If the average operational time is 160 hours, what is the probability that a generator of this type will be operational between 60 and 160 hours?
#### Explanation:
This problem involves finding the probability of the operational time of an electrical generator, given that the operational time follows an exponential distribution. An exponential distribution is commonly used to model the time between events in a process where events occur continuously and independently at a constant average rate.
For an exponential distribution, the probability density function (PDF) is given by:
\[ f(x; \lambda) = \lambda e^{-\lambda x} \]
where:
- \( x \) is the time,
- \( \lambda \) is the rate parameter, which is the reciprocal of the average (mean) operational time.
Given in the problem:
- The average operational time (\( \mu \)) is 160 hours.
- Thus, the rate parameter \( \lambda \) is \( \frac{1}{\mu} = \frac{1}{160} \).
We want to find the probability that the generator will be operational between 60 and 160 hours. This involves calculating the cumulative distribution function (CDF) for the exponential distribution.
The CDF for the exponential distribution is given by:
\[ F(x; \lambda) = 1 - e^{-\lambda x} \]
To find the probability that \( X \) is between 60 and 160 hours, we compute:
\[ P(60 \leq X \leq 160) = F(160; \lambda) - F(60; \lambda) \]
Substituting the given values:
\[ F(160; \frac{1}{160}) = 1 - e^{-\frac{160}{160}} = 1 - e^{-1} \]
\[ F(60; \frac{1}{160}) = 1 - e^{-\frac{60}{160}} = 1 - e^{-\frac{3}{8}} \]
Therefore:
\[ P(60 \leq X \leq 160) = \left(1 - e^{-1}\right) - \left(1 - e^{-\frac{3}{8}}\right) \]
\[ P(60 \le](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7824f8d0-65fb-442c-84f7-7ea5547717e1%2Fb7faddf1-a609-49bf-992e-b7ff718f637e%2F6r6su25_processed.png&w=3840&q=75)
Transcribed Image Text:### Exponential Distribution and Generator Operational Time
#### Problem Statement:
**Question 3:** The time in hours during which an electrical generator is operational is a random variable that follows an exponential distribution. If the average operational time is 160 hours, what is the probability that a generator of this type will be operational between 60 and 160 hours?
#### Explanation:
This problem involves finding the probability of the operational time of an electrical generator, given that the operational time follows an exponential distribution. An exponential distribution is commonly used to model the time between events in a process where events occur continuously and independently at a constant average rate.
For an exponential distribution, the probability density function (PDF) is given by:
\[ f(x; \lambda) = \lambda e^{-\lambda x} \]
where:
- \( x \) is the time,
- \( \lambda \) is the rate parameter, which is the reciprocal of the average (mean) operational time.
Given in the problem:
- The average operational time (\( \mu \)) is 160 hours.
- Thus, the rate parameter \( \lambda \) is \( \frac{1}{\mu} = \frac{1}{160} \).
We want to find the probability that the generator will be operational between 60 and 160 hours. This involves calculating the cumulative distribution function (CDF) for the exponential distribution.
The CDF for the exponential distribution is given by:
\[ F(x; \lambda) = 1 - e^{-\lambda x} \]
To find the probability that \( X \) is between 60 and 160 hours, we compute:
\[ P(60 \leq X \leq 160) = F(160; \lambda) - F(60; \lambda) \]
Substituting the given values:
\[ F(160; \frac{1}{160}) = 1 - e^{-\frac{160}{160}} = 1 - e^{-1} \]
\[ F(60; \frac{1}{160}) = 1 - e^{-\frac{60}{160}} = 1 - e^{-\frac{3}{8}} \]
Therefore:
\[ P(60 \leq X \leq 160) = \left(1 - e^{-1}\right) - \left(1 - e^{-\frac{3}{8}}\right) \]
\[ P(60 \le
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