91Ó°ÊÓ

Let \(X\) denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article "Blade Fatigue Life Assessment with Application to VAWTS" (J. Solar Energy Engrg., 1982: 107-111) proposes the Rayleigh distribution, with pdf $$ f(x ; \theta)=\left\\{\begin{array}{cl} \frac{x}{\theta^{2}} \cdot e^{-x^{2} /\left(2 \theta^{2}\right)} & x>0 \\ 0 & \text { otherwise } \end{array}\right. $$ as a model for the \(X\) distribution. a. Verify that \(f(x ; \theta)\) is a legitimate pdf. b. Suppose \(\theta=100\) (a value suggested by a graph in the article). What is the probability that \(X\) is at most 200 ? Less than 200 ? At least 200 ? c. What is the probability that \(X\) is between 100 and 200 (again assuming \(\theta=100\) )? d. Give an expression for \(P(X \leq x)\).

Step by step solution

01

Verify Legitimate PDF

A probability density function (pdf) is legitimate if it satisfies two conditions: 1) It's non-negative for all values of x, and 2) It integrates to 1 over the entire range of possible values of x. The given pdf is \[ f(x; \theta) = \frac{x}{\theta^{2}} e^{-x^{2} / (2 \theta^{2})}, \quad x > 0 \]We must integrate this from 0 to \(\infty\) and show it equals 1.\[ \int_{0}^{\infty} \frac{x}{\theta^{2}} e^{-x^{2} / (2 \theta^{2})} \, dx = 1 \]Perform a substitution with \( u = \frac{x^{2}}{2 \theta^{2}} \), \( du = \frac{x}{\theta^{2}} \, dx \). This converts the integral to:\[ \int_{0}^{\infty} e^{-u} \, du = 1 \]The integral of \( e^{-u} \) from 0 to \( \infty \) equals 1, satisfying the legitimacy of the pdf.

02

Calculate Probability X ≤ 200

The cumulative distribution function (CDF) for the Rayleigh distribution with parameter \( \theta \) is given by:\[ F(x; \theta) = 1 - e^{-x^{2} / (2 \theta^{2})} \]Substituting \( \theta = 100 \) and \( x = 200 \), we get:\[ F(200; 100) = 1 - e^{-200^{2} / (2 \times 100^{2})} \]\[ F(200; 100) = 1 - e^{-4} \approx 0.9817 \]Thus, the probability that \( X \) is at most 200 is approximately 0.9817.

03

Calculate Probability X < 200

Since continuous probabilities are calculated over intervals and the probability for a single point in a continuous distribution is zero, we have:\[ P(X < 200) = P(X \leq 200) \]Therefore, the probability is also approximately 0.9817.

04

Calculate Probability X ≥ 200

The probability that \( X \) is at least 200 is:\[ P(X \geq 200) = 1 - P(X < 200) = 1 - P(X \leq 200) \]From Step 2, we know \( P(X \leq 200) \approx 0.9817 \), so:\[ P(X \geq 200) = 1 - 0.9817 = 0.0183 \]Thus, the probability that \( X \) is at least 200 is approximately 0.0183.

05

Calculate Probability 100 ≤ X ≤ 200

The probability that \( X \) is between 100 and 200 is:\[ P(100 \leq X \leq 200) = F(200; 100) - F(100; 100) \]From Step 2, \( F(200; 100) \approx 0.9817 \) and:\[ F(100; 100) = 1 - e^{-100^{2} / (2 \times 100^{2})} = 1 - e^{-1} \approx 0.6321 \]Therefore:\[ P(100 \leq X \leq 200) = 0.9817 - 0.6321 = 0.3496 \]Thus, the probability is approximately 0.3496.

06

Expression for P(X ≤ x)

The cumulative distribution function (CDF) for \( X \) is:\[ P(X \leq x) = 1 - e^{-x^{2} / (2 \theta^{2})} \]This expression can be used to determine the probability that \( X \) does not exceed any particular value \( x \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Density Function

In the context of continuous probability distributions, a probability density function (PDF) is a fundamental concept. The PDF allows us to describe the likelihood of a random variable taking a particular range of values within a continuous sample space. For the Rayleigh distribution, the PDF is given by:\[f(x ; \theta) = \frac{x}{\theta^{2}} e^{-x^{2} / (2 \theta^{2})}, \quad x > 0\]The PDF for the Rayleigh distribution must satisfy two conditions to be considered legitimate:

• It must be non-negative for all values of \(x\).
• When integrated over its entire range, from 0 to \(\infty\), the total area under the curve must equal 1, ensuring that the entire probability distribution sums to 1.

A key feature of the Rayleigh distribution is its ability to model phenomena such as vibratory stresses on materials, which are common in engineering fields.

Cumulative Distribution Function

The cumulative distribution function (CDF) offers a way to determine the probability that a random variable takes on a value less than or equal to a certain number. For the Rayleigh distribution, the CDF is given by:\[F(x; \theta) = 1 - e^{-x^{2} / (2 \theta^{2})}\]This function is incredibly useful in practical applications, as it allows engineers and statisticians to calculate probabilities over intervals.
The CDF accumulates probabilities from left to right along the x-axis, providing complete knowledge of the distribution's behavior up to any point \(x\).
In vibratory stress analysis, knowing how likely stress levels will remain below certain thresholds is crucial for assessing potential failure risks in structures, such as wind turbine blades.

Continuous Probability Distributions

Continuous probability distributions describe the probabilities of the possible values of a continuous random variable. These distributions are used when dealing with data that can take on any value within a specified range.
The Rayleigh distribution is a specific type of continuous distribution that is particularly applicable when there is a prevalent "normal" stress level, but random fluctuations from this value occur. This makes it ideal for modeling phenomena where stress and waveforms are commonly observed.
Unlike discrete distributions that handle distinct outcomes, continuous distributions require integration to calculate probabilities over ranges. These distributions have a fundamental role in statistical modeling, where assumptions about the nature of the data guide the selection of an appropriate distribution.

Statistical Modeling

Statistical modeling involves using statistical methods to forecast future outcomes. It is a crucial component of data analysis and decision-making processes. By utilizing mathematical tools like the Rayleigh distribution, engineers and analysts can create models that simulate real-world situations like vibratory stress.
In the case of the Rayleigh distribution, the parameter \(\theta\) significantly influences the spread of the distribution. A larger \(\theta\) suggests a wider spread, meaning higher variability in stress levels.
Models help determine necessary precautions to avoid structural failure or to optimize designs for better performance. By applying these models, organizations can save costs, reduce risks, and enhance operational efficiency.

Vibratory Stress Analysis

Vibratory stress analysis is essential in assessing how mechanical structures respond to varying forces. It evaluates the likelihood of a component experiencing stresses that could cause fatigue or failure.
In applications like wind turbine blades, analyzing vibratory stress with the Rayleigh distribution helps in understanding how often stresses fall within safe limits, and how likely they are to hit levels that could lead to damage.
With insights from vibratory stress analysis, engineers can design components to withstand typical stress fluctuations, ensuring reliability and safety of critical infrastructure.
Utilizing this analysis effectively reduces the risk of unexpected failures, which is crucial for maintaining the longevity and effectiveness of structural components in various engineering applications.