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An article in the Journal of the Indian Geophysical Union titled "Weibull and Gamma Distributions for Wave Parameter Predictions" \((2005,\) Vol. \(9,\) pp. \(55-64)\) described the use of the Weibull distribution to model ocean wave heights. Assume that the mean wave height at the observation station is \(2.5 \mathrm{~m}\) and the shape parameter equals \(2 .\) Determine the standard deviation of wave height.

Short Answer

Expert verified
The standard deviation of wave height is approximately \(0.9524 \mathrm{~m}\).

Step by step solution

01

Understand the Weibull Distribution

The Weibull distribution is a continuous probability distribution used to model reliability and lifetime data, among other things. It is defined by the scale parameter \( \lambda \) (lambda) and the shape parameter \( k \). In this problem, the shape parameter \( k \) is given as \( 2 \), and the challenge is to find the standard deviation given the mean wave height, \( 2.5 \mathrm{~m} \).
02

Mean Formula of Weibull Distribution

For a Weibull distribution, the mean \( \mu \) can be expressed as: \[ \mu = \lambda \Gamma \left( 1 + \frac{1}{k} \right), \] where \( \Gamma \) is the Gamma function. You're provided with the mean, \( \mu = 2.5 \mathrm{~m} \), and \( k = 2 \).
03

Solve for Scale Parameter \( \lambda \)

Using the mean formula, substitute the values to solve for \( \lambda \):\[ 2.5 = \lambda \Gamma\left( 1 + \frac{1}{2} \right). \]The Gamma function \( \Gamma\left( 1.5 \right)\) equals \( \frac{\sqrt{\pi}}{2} \). Therefore:\[ 2.5 = \lambda \frac{\sqrt{\pi}}{2}. \]Solve for \( \lambda \):\[ \lambda = \frac{2.5 \times 2}{\sqrt{\pi}}. \]
04

Standard Deviation Formula for Weibull Distribution

The formula for the standard deviation \( \sigma \) of a Weibull distribution is:\[ \sigma = \lambda \sqrt{\Gamma\left( 1 + \frac{2}{k} \right) - \left( \Gamma\left( 1 + \frac{1}{k} \right) \right)^2}. \]Given \( k = 2 \), evaluate \( \Gamma\left(1 + \frac{2}{2}\right) = \Gamma(2) = 1 \), and \( \Gamma\left( 1.5 \right) = \frac{\sqrt{\pi}}{2} \).
05

Substitute Values and Calculate Standard Deviation

Substitute the known values for \( \lambda \) and the Gamma functions into the standard deviation formula:\[\sigma = \left(\frac{2.5 \times 2}{\sqrt{\pi}}\right) \sqrt{1 - \left(\frac{\sqrt{\pi}}{2}\right)^2},\]and simplify the expression to get the standard deviation.

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Key Concepts

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

Understanding Probability Distributions
Probability distributions are essential in statistics and are used to describe the likelihood of different outcomes. The Weibull distribution is a particular type of probability distribution that is often applied in fields like reliability engineering and failure analysis.
In essence, a probability distribution provides a mathematical function that gives the probabilities of occurrence of different possible outcomes for an experiment. Some common characteristics of probability distributions include:
  • Mean: Represents the expected value or average of the distribution. It's the point around which all the data points tend to cluster, and it serves as a measure of the central tendency.
  • Variance and Standard Deviation: While the variance measures the spread of the distribution by averaging the squared deviations from the mean, the standard deviation is the square root of the variance, offering a more intuitive measure of data spread.
  • Shape: Often defined by parameters, the shape of a distribution can vary from symmetrical to skewed.
For the Weibull distribution, the primary characteristic is that it can model various types of data depending on its parameters, making it versatile for different applications.
Standard Deviation of Weibull Distribution
The standard deviation is a vital statistic that offers insight into the variance or 'spread' of data points in a dataset. For the Weibull distribution, determining the standard deviation involves several calculations, particularly because Weibull's equation uses parameters like the scale parameter (\( \lambda \)) and the shape parameter (\( k \)).
The formula to calculate the standard deviation \( \sigma \) of the Weibull distribution includes the Gamma functions. It is expressed as:
  • \[ \sigma = \lambda \sqrt{\Gamma\left( 1 + \frac{2}{k} \right) - \left( \Gamma\left( 1 + \frac{1}{k} \right) \right)^2} \]
To find \( \sigma \), you must first solve for the scale parameter \( \lambda \), which involves using the given mean and the shape parameter. The formula shows that \( \Gamma \) (the Gamma function) plays a key role. Once you've calculated \( \lambda \) and applied the appropriate Gamma function values, you can easily compute the standard deviation to understand the dispersal of wave heights.
The standard deviation provides an understanding of how much wave heights, for instance, deviate from the average (mean) height. A higher standard deviation indicates more variability in wave heights, while a lower value points to consistency.
The Role of Shape Parameter in Weibull Distribution
The shape parameter (\( k \)) in the Weibull distribution is crucial as it determines the distribution's form, allowing Weibull to model a wide range of data types. Each value of \( k \) gives the distribution distinct characteristics, which are pivotal in applications like predicting lifetimes of products or modeling natural phenomena.
Here's how different values of \( k \) influence the Weibull distribution:
  • \( k < 1 \): The distribution is highly skewed and decreases rapidly, often used to model infant mortality or items failing early.
  • \( k = 1 \): The Weibull distribution simplifies to an exponential distribution, commonly used for scenarios with constant failure rates over time.
  • \( k > 1 \): It becomes less skewed and more bell-shaped, modeling wear-out failures, where items are more likely to fail as time progresses.
In the given problem, where \( k = 2 \), the distribution falls into the 'wear-out' category, suggesting that wave heights are expected to have increased variability over time.
The shape parameter is essential in tailoring the Weibull distribution to accurately reflect real-world phenomena, ensuring that predictions and analyses are meaningful and relevant.

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