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Chapter 4: Problem 78

The article "On Assessing the Accuracy of Offshore Wind Turbine Reliability- Based Design Loads from the Environmental Contour Method" (Intl. J. of Offshore and Polar Engr., 2005: 132-140) proposes the Weibull distribution with \(\alpha=1.817\) and \(\beta=.863\) as a model for 1-hour significant wave height ( \(m\) ) at a certain site. a. What is the probability that wave height is at most \(.5 \mathrm{~m}\) ? b. What is the probability that wave height exceeds its mean value by more than one standard deviation? c. What is the median of the wave-height distribution? d. For \(0

Short Answer

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a) Use the CDF: \( F(0.5) = 1 - e^{-(0.5/0.863)^{1.817}} \). b) Exceed probability: \( 1 - F(\mu + \sigma) \). c) Median: \( 0.863(-\ln(0.5))^{1/1.817} \). d) Percentile: \( 0.863(-\ln(1-p))^{1/1.817} \).

Step by step solution

01

Understand the Weibull Distribution

The Weibull distribution is defined by its cumulative distribution function (CDF): \[ F(x) = 1 - e^{-(x/\beta)^\alpha} \] where \( \alpha \) is the shape parameter and \( \beta \) is the scale parameter. Given \( \alpha = 1.817 \) and \( \beta = 0.863 \). We will use this information to solve the different parts of the problem.
02

Probability that Wave Height is at most 0.5 m

To find the probability that the wave height is at most 0.5 m, calculate the CDF at \( x = 0.5 \):\[ F(0.5) = 1 - e^{-(0.5/0.863)^{1.817}} \]Calculate the expression inside the exponent first, then subtract from 1 to find the CDF value.
03

Calculate Mean of the Weibull Distribution

The mean of the Weibull distribution is given by:\[ \mu = \beta \Gamma(1 + 1/\alpha) \]where \( \Gamma \) is the Gamma function. Compute \( \Gamma(1 + 1/1.817) \) and multiply by \( \beta = 0.863 \).
04

Calculate Standard Deviation of the Weibull Distribution

The standard deviation can be calculated using the formula:\[ \sigma = \beta \sqrt{\Gamma(1 + 2/\alpha) - \left(\Gamma(1 + 1/\alpha)\right)^2} \]Compute \( \Gamma(1 + 2/1.817) \) and solve the expression to find \( \sigma \).
05

Probability that Wave Height Exceeds Mean by More Than One Standard Deviation

First, calculate \( \mu + \sigma \) using results from Steps 3 and 4. Then compute the probability that the wave height exceeds this value:\[ P(X > \mu + \sigma) = 1 - F(\mu + \sigma) \]Use the CDF formula from Step 1 to find \( F(\mu + \sigma) \) and subtract from 1.
06

Find the Median of the Distribution

The median of the Weibull distribution satisfies \( F(median) = 0.5 \):\[ 0.5 = 1 - e^{-(\text{median}/\beta)^\alpha} \]Solve for the median:\[ e^{-(\text{median}/\beta)^\alpha} = 0.5 \]\[ -(\text{median}/\beta)^\alpha = \ln(0.5) \]\[ \text{median} = \beta \left(-\ln(0.5)\right)^{1/\alpha} \]
07

General Expression for the 100p-th Percentile

The 100\(p\)-th percentile, \(x_p\), of the Weibull distribution satisfies \( F(x_p) = p \):\[ p = 1 - e^{-(x_p/\beta)^\alpha} \]Solve for \(x_p\):\[ e^{-(x_p/\beta)^\alpha} = 1 - p \]\[ -(x_p/\beta)^\alpha = \ln(1-p) \]\[ x_p = \beta \left(-\ln(1-p)\right)^{1/\alpha} \]

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Key Concepts

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

Probability Calculation
Understanding probability calculations is key to analyzing real-world phenomena, such as wave heights. When using a Weibull distribution to model wave heights, the probability of a wave reaching or staying below a certain height can be calculated using the cumulative distribution function (CDF).

The CDF for a Weibull distribution is given by: \[ F(x) = 1 - e^{-(x/\beta)^\alpha} \] This function helps us determine the probability that a randomly chosen value from the distribution is less than or equal to a given point \( x \). If you want to find the probability that the wave height is at most 0.5 meters, simply plug \( x = 0.5 \) into the equation and solve.

By understanding this calculation, you can apply the same process to find probabilities for any other specified height, thus aiding in various predictive and analytic applications.
Wave Height Data Analysis
Wave height data analysis is an essential aspect of offshore research and environmental studies, especially in assessing offshore wind turbine sites. One of the effective ways to analyze such data is through probability distributions, like the Weibull distribution.

In the context of wave height analysis, the Weibull distribution can model the varying nature of wave heights over a specific period, such as an hour. By adjusting its scale (\( \beta \)) and shape (\( \alpha \)) parameters, it can accommodate different types of data, whether the waves are generally low with occasional spikes, or consistently higher.

Analyzing wave data with a Weibull distribution provides insights into expected conditions, helping to plan and design structures that withstand extreme events, thus ensuring reliability and safety.
Percentiles in Probability Distributions
Percentiles are useful statistical measures that help interpret and understand data distributions. In the context of a Weibull distribution used for modeling wave heights, a percentile is a value below which a given percentage of observations fall.

To calculate the \( 100p \)-th percentile of a Weibull distribution, we use the formula: \[ x_p = \beta \left(-\ln(1-p)\right)^{1/\alpha} \] By applying this formula, you can determine specific wave heights that are expected to be equaled or exceeded by a certain proportion of the observed waves. For instance, the 90th percentile would give us the height that 90% of observed wave heights will fall beneath.

This understanding aids in risk assessment and decision-making processes, especially in fields such as engineering and environmental sciences.
Median Calculation in Statistics
The median is a critical measure in statistics, representing the middle value when a set of numbers is organized from smallest to largest. In a probability distribution such as a Weibull, the median provides the value that divides the probability into two equal halves.

For a Weibull distribution, the median \( m \) satisfies the equation \( F(m) = 0.5 \). Solving for the median involves setting up the equation: \[ 0.5 = 1 - e^{-(m/\beta)^\alpha} \] Then, by manipulating this equation, we find: \[ m = \beta \left(-\ln(0.5)\right)^{1/\alpha} \] This result allows us to understand the typical conditions at a site, providing a central measure that isn't skewed by extreme values as the mean can sometimes be.

Knowing how to calculate and use the median is crucial for understanding the central tendency of environmental data, like wave heights.

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Most popular questions from this chapter

The special case of the gamma distribution in which \(\alpha\) is a positive integer \(n\) is called an Erlang distribution. If we replace \(\beta\) by \(1 / \lambda\) in Expression (4.8), the Erlang pdf is $$ f(x ; \lambda, n)=\left\\{\begin{array}{cc} \frac{\lambda(\lambda x)^{n-1} e^{-\lambda x}}{(n-1) !} & x \geq 0 \\ 0 & x<0 \end{array}\right. $$ It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter \(\lambda\), then the total time \(X\) that elapses before all of the next \(n\) events occur has pdf \(f(x ; \lambda, n)\). a. What is the expected value of \(X\) ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with \(\lambda=.5\), how much time can be expected to elapse before the tenth customer arrives? b. If customer interarrival time is exponentially distributed with \(\lambda=.5\), what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next \(30 \mathrm{~min}\) ? c. The event \(\\{X \leq t\\}\) occurs iff at least \(n\) events occur in the next \(t\) units of time. Use the fact that the number of events occurring in an interval of length \(t\) has a Poisson distribution with parameter \(\lambda t\) to write an expression (involving Poisson probabilities) for the Erlang cdf \(F(t ; \lambda, n)=P(X \leq t)\).

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviation 3 . a. If a specimen is acceptable only if its hardness is between 67 and 75 , what is the probability that a randomly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is \((70-c, 70+c)\), for what value of \(c\) would \(95 \%\) of all specimens have acceptable hardness? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is indepen-dently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.84?

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The paper "Effects of 2,4 -D Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: 1030-1036) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that exactly two of them exceed \(1500 \mu \mathrm{m}\) ?

A family of pdf's that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \(k\) and \(\theta\), both \(>0\), and the pdf is $$ f(x ; k, \theta)=\left\\{\begin{array}{cl} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ a. Sketch the graph of \(f(x ; k, \theta)\). b. Verify that the total area under the graph equals 1 . c. If the rv \(X\) has pdf \(f(x ; k, \theta)\), for any fixed \(b>\theta\), obtain an expression for \(P(X \leq b)\). d. For \(\theta
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