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Magazine Chapter 4: Problem 125
Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the mean and variance of \(X\)
Short Answer
Expert verified
The mean is 12000 hours and the variance is 36144000000 hours虏.
Step by step solution
01
Identify the Weibull distribution parameters
The Weibull distribution is defined by two parameters: the shape parameter \(\beta\) and the scale parameter \(\delta\). In this case, we are given \(\beta = 0.2\) and \(\delta = 100\). These are the values we will use to find the mean and variance.
02
Formula for the mean of a Weibull distribution
The mean (also called the expected value) \(E(X)\) for a Weibull distribution is given by the formula: \[ E(X) = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right) \]where \(\Gamma(z)\) is the gamma function.
03
Calculate the gamma function for the mean
For our parameters, we need to evaluate \(\Gamma\left(1 + \frac{1}{0.2}\right) = \Gamma(6)\). Since \(\Gamma(6) = (6-1)! = 5! = 120\), this gives us the value for \(\Gamma(6)\).
04
Compute the mean of the Weibull
Substitute the values into the mean formula:\[ E(X) = 100 \cdot 120 = 12000 \text{ hours} \]This is the expected mean value for \(X\).
05
Formula for the variance of a Weibull distribution
The variance \(Var(X)\) for a Weibull distribution is given by the formula:\[ Var(X) = \delta^2 \cdot \left(\Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2\right) \]
06
Calculate the gamma functions for variance
First, calculate \(\Gamma\left(1 + \frac{2}{0.2}\right) = \Gamma(11)\) and \(\Gamma(6)^2\) from previous steps:- \(\Gamma(11) = (11-1)! = 10! = 3628800\)- \(\Gamma(6)^2 = 120^2 = 14400\)
07
Compute the variance of the Weibull
Substitute the values into the variance formula:\[Var(X) = 100^2 \cdot (3628800 - 14400) = 10000 \cdot 3614400 = 36144000000 \text{ hours}^2\]This is the variance of \(X\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Weibull Distribution Mean
The mean of a Weibull distribution is a measure of the central tendency for data that is modeled using this distribution. For the Weibull distribution, the mean is calculated using the formula: \[ E(X) = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right) \] where \( \delta \) is the scale parameter and \( \beta \) is the shape parameter. The term \( \Gamma(z) \) stands for the gamma function, which is a special function that extends the concept of factorials to non-integer values.
Some interesting points about the Weibull mean include:
Some interesting points about the Weibull mean include:
- It tells us the 'average' time until an event (like failure) happens, depending on the scale of the distribution.
- Given \( \beta = 0.2 \) and \( \delta = 100 \), we computed \( E(X) = 100 \cdot \Gamma(6) = 12000 \text{ hours} \).
- The mean leverages the gamma function to handle the continuous nature of this distribution efficiently.
Weibull Distribution Variance
Variance in the context of the Weibull distribution provides a measure of spread. It indicates how much the values of the distribution deviate from the mean. The formula for finding the variance of a Weibull distributed random variable \( X \) is: \[ Var(X) = \delta^2 \cdot \left(\Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2\right) \] where \( \delta \) is the scale parameter and \( \beta \) is the shape parameter.
Some key aspects of the Weibull variance are:
Some key aspects of the Weibull variance are:
- It gives insight into the variability and reliability of the data modeled by the Weibull distribution.
- In our example, \( Var(X) = 100^2 \cdot (3628800 - 14400) = 36144000000 \text{ hours}^2 \).
- The variance takes into account higher powers of the beta parameter and multiple uses of the gamma function.
Gamma Function
The gamma function \( \Gamma(z) \) is a complex mathematical function that generalizes the concept of factorial to non-integer numbers. It plays a significant role in various probability distributions, including the Weibull distribution. The gamma function is defined for positive reals and is given by: \[ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \, dt \] For integer values, it's related to factorials: \( \Gamma(n) = (n-1)! \).
A few essential points about the gamma function are:
A few essential points about the gamma function are:
- It appears frequently in the context of continuous probability distributions.
- In the Weibull distribution, it helps calculate important metrics like mean and variance by extending the concept of multiplication beyond integers.
- The gamma function has recurrent relation properties like \( \Gamma(z+1) = z\Gamma(z) \).
Probability Distributions
Probability distributions are mathematical functions that describe the likelihood of different outcomes in a random experiment. They are fundamental in statistics and are used to model various types of data. A crucial concept is that they help in predicting the behavior of systems under uncertainty.
Key aspects of probability distributions include:
Key aspects of probability distributions include:
- Discrete distributions, like the binomial distribution, describe scenarios with finite or countable outcomes.
- Continuous distributions, such as the Weibull distribution, model data that can take an infinite number of possible outcomes within a certain range.
- Each distribution has its own set of parameters, such as mean and variance, that help in understanding its characteristics and how they relate to real-world data.
