/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Let \(X\) denote the vibratory s... [FREE SOLUTION] | 91Ó°ÊÓ

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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. of Solar Energy Engr., 1982: 107-111) proposes the Rayleigh distribution, with pdf $$ f(x ; \theta)=\left\\{\begin{array}{cc} \frac{x}{\theta^{2}} \cdot e^{-x^{2} \cdot\left(2 \theta^{2}\right)} & x>0 \\ 0 & \text { otherwise } \end{array}\right. $$ 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 \(P(X \leq x)\).

Short Answer

Expert verified
a. It's a legitimate pdf. b. P(X ≤ 200) ≈ 0.8647, P(X < 200) ≈ 0.8647, P(X ≥ 200) ≈ 0.1353. c. P(100 ≤ X ≤ 200) ≈ 0.4712. d. P(X ≤ x) = 1 - e^(-x²/(2θ²)).

Step by step solution

01

Verify the PDF Condition

To verify that \(f(x; \theta)\) is a legitimate probability density function (pdf), we need to check two conditions: the function must be non-negative, and the integral over its entire range must equal 1. The pdf is given as \( f(x; \theta) = \frac{x}{\theta^2} \cdot e^{-\frac{x^2}{2\theta^2}} \) for \(x > 0\). It's clearly non-negative for \(x > 0\).Next, integrate \(f(x; \theta)\) over \((0, \infty)\): \[ \int_{0}^{\infty} \frac{x}{\theta^2} \cdot e^{-\frac{x^2}{2\theta^2}} \, dx \]Use substitution \(u = \frac{x^2}{2\theta^2}\), then \(du = \frac{x}{\theta^2} \, dx\), making the integral become \[ \int_{0}^{\infty} e^{-u} \, du = 1 \]Thus, \(f(x; \theta)\) integrates to 1, verifying it's a legitimate pdf.
02

Compute P(X ≤ 200)

We need the cumulative distribution function (CDF) to find \(P(X \leq 200)\). For the Rayleigh distribution, \(P(X \leq x) = 1 - e^{-\frac{x^2}{2\theta^2}}\).Substituting \(x = 200\) and \(\theta = 100\), we get: \[P(X \leq 200) = 1 - e^{-\frac{200^2}{2 \cdot 100^2}} = 1 - e^{-2}\]Calculate: \[P(X \leq 200) \approx 1 - e^{-2} \approx 0.8647\]
03

Compute P(X < 200)

Since \(X\) is a continuous random variable, \(P(X < 200) = P(X \leq 200)\). Thus,\[P(X < 200) = P(X \leq 200) \approx 0.8647\]
04

Compute P(X ≥ 200)

Use the complement rule:\[P(X \geq 200) = 1 - P(X < 200)\]Using the result from Step 3, we find:\[P(X \geq 200) = 1 - 0.8647 \approx 0.1353\]
05

Compute P(100 ≤ X ≤ 200)

This probability can be found using the CDF:\[P(100 \leq X \leq 200) = P(X \leq 200) - P(X < 100)\]Using \(P(X \leq x) = 1 - e^{-\frac{x^2}{2\theta^2}}\), we get:\[P(X \leq 100) = 1 - e^{-\frac{100^2}{2 \cdot 100^2}} = 1 - e^{-0.5} \approx 0.3935\]Thus,\[P(100 \leq X \leq 200) = 0.8647 - 0.3935 \approx 0.4712\]
06

General Expression for CDF P(X ≤ x)

The cumulative distribution function for a Rayleigh distributed variable is:\[P(X \leq x) = 1 - e^{-\frac{x^2}{2\theta^2}}\]

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Key Concepts

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

Probability Density Function
In the realm of probability and statistics, a probability density function (pdf) is crucial for continuous random variables like the one described in this problem. A pdf, denoted as \( f(x) \), is a function that characterizes the likelihood of a continuous random variable taking on a specific value. Although individual probabilities are zero in continuous distributions, you can find the probability that the variable falls within a particular range by integrating the pdf over that range.

For a function to qualify as a legitimate pdf, it must satisfy two conditions:
  • The function \( f(x) \) must be non-negative for all values of \( x \).
  • The total area under the curve of \( f(x) \) over the entire space must equal 1. This ensures that there is a 100% probability the variable will take a value within the defined range.
In the given Rayleigh distribution pdf, these conditions are met when \( x > 0 \), as verified through integration.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a continuous random variable provides a complete description of the distribution's probability behavior. Unlike the pdf, which gives a density at a point, the CDF, denoted as \( P(X \leq x) \), illustrates the probability that the random variable \( X \) is less than or equal to a particular value \( x \).

The CDF is expressed as an increasing function ranging between 0 and 1. For the Rayleigh distribution stated in this exercise, the CDF is derived as:\[ P(X \leq x) = 1 - e^{-\frac{x^2}{2\theta^2}} \]This formula provides the probability of the vibratory stress on the wind turbine blade being less than or equal to a certain value when \( \theta \) is specified.

The CDF is a useful tool in calculating probabilities over intervals, segments, and in determining values like median or percentiles.
Continuous Random Variable
A continuous random variable is a variable that can take an infinite number of possible values within a given range. Unlike discrete variables, which have specific outcomes, continuous variables are represented by intervals on the real number line.

Examples include heights, weights, or, as in the exercise, the vibratory stress \( X \) on a wind turbine under certain conditions.

Because the variable can take any value in a continuum:
  • The probability of it assuming a specific singular value is zero. This contrasts with discrete variables where you can assign actual probabilities to individual outcomes.
  • Probabilities for continuous random variables are found using areas under the probability density function and cumulative distribution function.
The concept of continuity helps in understanding how probabilities are distributed over a scale, making it essential in various practical and theoretical applications.
Integration in Statistics
Integration is a key mathematical tool in statistics, especially when dealing with continuous random variables. This process allows us to find probabilities, expectations, variances, and other useful statistics.
  • For probability density functions (pdf), integration helps determine whether the total probability is 1 over all possible values.
  • In cumulative distribution functions (CDFs), integrating the pdf gives the probability that the continuous variable is less than or equal to a given value.
In the Rayleigh distribution provided in the exercise, integration was used to verify that the area under the curve of the pdf equals 1, confirming its legitimacy. Techniques such as substitution simplify this process, making it possible to evaluate integrals that define essential properties and probabilities for continuous distributions.

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Most popular questions from this chapter

Consider an rv \(X\) with mean \(\mu\) and standard deviation \(\sigma\), and let \(g(X)\) be a specified function of \(X\). The first-order Taylor series approximation to \(g(X)\) in the neighborhood of \(\mu\) is $$ g(X) \Rightarrow g(\mu)+g^{\prime}(\mu) \cdot(X-\mu) $$ The right-hand side of this equation is a linear function of \(X\). If the distribution of \(X\) is concentrated in an interval over which \(g(\cdot)\) is approximately linear [e.g., \(\sqrt{x}\) is approximately linear in \((1,2)\) ], then the equation yields approximations to \(E(g(X))\) and \(V(g(X))\). a. Give expressions for these approximations. [Hint: Use rules of expected value and variance for a linear function \(a X+b .]\) b. If the voltage \(v\) across a medium is fixed but current \(l\) is random, then resistance will also be a random variable related to \(I\) by \(R=v / I\). If \(\mu_{l}=20\) and \(\sigma_{l}=.5\), calculate approximations to \(\mu_{R}\) and \(\sigma_{R}\).

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