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Derive the mean and variance of the Weibull distribution.

Short Answer

Expert verified
The mean of a Weibull distribution is given by \( a* \Gamma(1+1/b) \) and the variance is given by \( a^2 * (\Gamma(1+2/b) - [ \Gamma(1+1/b) ]^2 ) \), where \( \Gamma \) is the gamma function.

Step by step solution

01

Write down the Probability Density Function (PDF) of a Weibull distribution

The PDF of a Weibull distribution is given by \( f(x; a, b) = \frac{b}{a} \left(\frac{x}{a}\right)^{b-1} e^{-(x/a)^{b}} \) for \( x \geq 0 \),and \( b > 0 \) where \( a > 0 \) is the scale parameter, and \( b > 0 \) is the shape parameter.
02

Calculate the mean of the Weibull distribution

The mean (expected value) of a Weibull distribution is given by \( E[X] = a* \Gamma(1+1/b) \), where \( \Gamma \) is the gamma function.
03

Calculate the variance of the Weibull distribution

The variance of a Weibull distribution is given by \( Var[X] = a^2 * (\Gamma(1+2/b) - [ \Gamma(1+1/b) ]^2 ) \), where \( \Gamma \) is the gamma function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function (PDF)
Understanding the Probability Density Function (PDF) is fundamental when studying continuous random variables like the Weibull distribution. The PDF describes how the probability of the variable is distributed over a range of values. For the Weibull distribution, the PDF is particularly useful in reliability engineering and failure analysis because it models the time until a particular event, such as the failure of a product, occurs.

The PDF for the Weibull distribution is expressed as \( f(x; a, b) = \frac{b}{a} \left(\frac{x}{a}\right)^{b-1} e^{-(x/a)^{b}} \) for \( x \geq 0 \), and \( b > 0 \) where \( a > 0 \) is the scale parameter that stretches or shrinks the distribution, and \( b > 0 \) is the shape parameter that makes the distribution curve steeper or flatter. This function can tell us the likelihood of a failure occurring at a specific time, poking at critical insights for planning and decision making.
Expected Value
The expected value, often termed as the mean, is a critical concept which represents the average or central value of a random variable's probability distribution. In the context of the Weibull distribution, the expected value can predict the 'average failure time', making it a vital tool in areas like quality control and risk management.

For any given Weibull distribution, the expected value is calculated using the formula: \( E[X] = a * \Gamma(1 + 1/b) \). Here, 'a' is the scale parameter, 'b' is the shape parameter, and \( \Gamma \) denotes the gamma function—a complex function extending the factorial function to real and complex numbers. It's essential to grasp that a higher expected value could suggest a longer life expectancy of a product or system.
Gamma Function
The gamma function \( \Gamma(z) \) is an advanced mathematical concept that extends the factorial to complex numbers. It is particularly significant in calculating the expected value and variance for distributions such as Weibull.

Conventionally, the factorial of a positive integer \( n \) is the product of all positive integers up to \( n \). However, the gamma function allows factorials for non-integer values, calculated as \( \Gamma(z) = \int_0^{\infty} x^{z-1}e^{-x} dx \), for real or complex \( z \), where the real part of \( z \) is positive. In the realm of Weibull distribution, this function is indispensable since it helps compute moments like the expected value and variance, providing deeper insight into the behavior of the distribution.
Variance
Variance is a measure of how much the values of a random variable spread out or differ from the mean (expected value). In the case of the Weibull distribution, variance gauges the dispersion of failure times around the average failure time and is a square of the standard deviation.

The variance for the Weibull distribution is found using the formula: \( Var[X] = a^2 * (\Gamma(1+2/b) - [ \Gamma(1+1/b) ]^2 ) \). Here, the gamma function plays a crucial part again, as it calculates the necessary moments of the distribution. Recognizing the variance is essential for quality analysts and reliability engineers to understand the reliability and consistency of the products or systems they work with. It enables them to anticipate the range of potential outcomes and adjust their strategies accordingly.

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Most popular questions from this chapter

The heights of 1000 students are normally distributed with a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. Assuming that the heights are recorded to the nearest half-centimeter, how many of these students would you expect to have heights (a) less than 160.0 centimeters? (b) between 171.5 and 182.0 centimeters inclusive? (c) equal to 175.0 centimeters? (d) greater than or equal to 188.0 centimeters?

In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08 . Assume that the distribution of percent purity was approximately normal. (a) What percentage of the purity values would you expect to be between 99.5 and \(99.7 ?\) (b) What purity value would you expect to exceed exactly \(5 \%\) of the population?

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Statistics released by the National Highway Traffic Safety Administration and the National Safety Council show that on an average weekend night, 1 out of every 10 drivers on the road is drunk. If 400 drivers are randomly checked next Saturday night, what is the probability that the number of drunk drivers will be (a) less than \(32 ?\) (b) more than \(49 ?\) (c) at least 35 but less than \(47 ?\)

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