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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 IQs of 600 applicants of a certain college are approximately normally distributed with a mean of 115 and a standard deviation of \(12 .\) If the college requires an IQ of at least, \(95,\) how many of these students will be rejected on this basis regardless of their other qualifications?

A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times to be normally distributed. (a) What is the probability that a trip will take at least \(1 / 2\) hour? (b) If the office opens at 9: 00 A.M. and he leaves his house at 8: 45 A.M. daily, what percentage of the time is he late for work? (c) If he leaves the house at 8: 35 A.M. and coffee is served at the office from 8:50 A.M. until 9:00 A.M., what is the probability that he misses coffee? (d) Find the length of time above which we find the slowest \(15 \%\) of the trips. (e) Find the probability that 2 of the next 3 trips will take at least \(1 / 2\) hour.

The life, in years, of a certain type of electrical switch has an exponential distribution with an average life \(\beta=2\). If 100 of these switches are installed in different systems, what is the probability that at most 30 fail during the first year?

In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable \(X\) having a gamma distribution with mean \(\mu=6\) and variance \(a^{2}=12\) (a) Find the values of \(\alpha\) and \(\beta\). (b) Find the probability that on any given day the daily power consumption will exceed 12 million kilowatthours.

A company pays its employees an average wage of \(\$ 15.90\) an hour with a standard deviation of \(\$ 1.50\). If the wages are approximately normally distributed and paid to the nearest cent, (a) what percentage of the workers receive wages between \(\mathrm{S} 13.75\) and \(\mathrm{S} 16.22\) an hour inclusive? (b) the highest \(5 \%\) of the employee hourly wages is greater than what; amount?

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