/*! 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 VAW'S" (J. of Solar Energy Engr:, 1982: 107-111) proposes the Rayleigh distribution, with pdf $$ f(x ; \theta)=\left\\{\begin{array}{cc} \frac{x}{\theta^{2}} \cdot e^{-x^{2} /\left(2 \theta^{2}\right)} & x>0 \\ 0 & \text { otherwise } \end{array}\right. $$ 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 valid 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

For a function to be a legitimate probability density function (pdf), it must satisfy two conditions:1. \( f(x; \theta) \geq 0 \) for all \( x \).2. The integral over all possible values of \( x \) must equal 1: \( \int_{0}^{\infty} f(x; \theta) \, dx = 1 \).Since \( f(x; \theta) = \frac{x}{\theta^{2}} \cdot e^{-x^{2}/(2 \theta^{2})} \) is clearly non-negative for \( x > 0 \), the first condition is satisfied.To verify the second condition, calculate the integral:\[\int_{0}^{\infty} \frac{x}{\theta^{2}} \cdot e^{-x^{2}/(2 \theta^{2})} \, dx\]Using the substitution \( u = \frac{x^{2}}{2 \theta^{2}} \), \( du = \frac{x}{\theta^{2}} \, dx \), the integral becomes \( \int_{0}^{\infty} e^{-u} \, du = 1 \), meeting the second condition.
02

Calculate P(X ≤ 200)

To find the probability that \( X \) is at most 200, we need to compute:\[P(X \leq 200) = \int_{0}^{200} \frac{x}{100^{2}} \cdot e^{-x^{2}/(2 \times 100^{2})} \, dx\]This calculation often uses the cumulative distribution function (CDF) for the Rayleigh distribution, which is:\[F(x) = 1 - e^{-x^{2}/(2 \theta^{2})}\]Substitute \( x = 200 \) and \( \theta = 100 \):\[F(200) = 1 - e^{-200^{2}/(2 \times 100^{2})} = 1 - e^{-2} \approx 0.8647\]Thus, \( P(X \leq 200) \approx 0.8647 \).
03

Calculate P(X < 200)

Since the Rayleigh distribution is a continuous distribution, \( P(X \leq 200) \) and \( P(X < 200) \) are essentially the same:\[P(X < 200) = P(X \leq 200) \approx 0.8647\]Thus, \( P(X < 200) \) is also approximately 0.8647.
04

Calculate P(X ≥ 200)

Use the complement rule:\[P(X \geq 200) = 1 - P(X < 200) = 1 - 0.8647 = 0.1353.\]Thus, \( P(X \geq 200) \approx 0.1353 \).
05

Calculate P(100 ≤ X ≤ 200)

The probability that \( X \) is between 100 and 200 is:\[P(100 \leq X \leq 200) = F(200) - F(100)\]Substitute \( \theta = 100 \):\( F(100) = 1 - e^{-100^{2}/(2 \times 100^{2})} = 1 - e^{-0.5} \approx 0.3935 \)Then calculate the difference:\[P(100 \leq X \leq 200) = 0.8647 - 0.3935 = 0.4712.\]Therefore, the probability is approximately 0.4712.
06

Expression for P(X ≤ x)

The cumulative distribution function (CDF) for the Rayleigh distribution gives us the expression for \( P(X \leq x) \):\[P(X \leq x) = F(x) = 1 - e^{-x^{2}/(2 \theta^{2})}\]This expression can be used with any specific value of \( x \) and \( \theta \).

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Key Concepts

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

Probability Density Function (PDF)
The Probability Density Function (PDF) is a fundamental concept in continuous probability distributions. For a continuous random variable, the PDF, represented as \( f(x) \), describes the likelihood of the random variable taking on a particular value. In a graphical sense, it's the function used to specify the probability of the variable within a certain range of values.

Key characteristics of a PDF include:
  • Non-negativity: \( f(x) \geq 0 \) for all \( x \), ensuring probabilities aren't negative.
  • Normalization: The total area under the curve of the PDF across its entire range equals one, i.e., \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \). This property ensures that the sum of all possible probabilities is 1.
In the Rayleigh distribution, the PDF is \( f(x; \theta) = \frac{x}{\theta^2} \cdot e^{-x^2 / (2\theta^2)} \) for \( x > 0 \). This formula models the vibratory stress on a wind turbine blade in our given problem. The parameter \( \theta \) is known as the scale parameter, influencing the shape of the distribution curve.

To verify a legitimate PDF, the conditions above must be checked, as demonstrated in the solution by ensuring the PDF integrates to 1 over its defined range.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is an integral part of understanding continuous probability distributions. It provides the probability that a random variable takes on a value less than or equal to a specific point. In essence, the CDF is the sum of probabilities up to a maximum value.

For the Rayleigh distribution, the CDF is given by:

\[ F(x) = 1 - e^{-x^2/(2\theta^2)} \]

This function smoothly captures the cumulative probability from the lower bound to any point \( x \).
  • This equation allows us to calculate probabilities up to any value of \( x \).
  • For example, \( P(X \leq 200) \) results in 0.8647, representing an 86.47% chance that stress values are 200 psi or less.
When working with continuous distributions like the Rayleigh, \( P(X \leq x) \) and \( P(X < x) \) are effectively the same due to the continuous nature, eliminating the gap between less than and equal to.

The CDF's insight deepens our understanding of distribution behavior beyond the PDF's specifics, capturing the full range of probabilities in a single function.
Continuous Probability Distribution
Continuous Probability Distributions deal with random variables that can take on an infinite number of values within a given range. Unlike discrete distributions with countable outcomes, continuous distributions allow for any value along an interval on the number line.

Key aspects of continuous distributions include:
  • Probability Density Function (PDF): Used to define the relative likelihood of a variable within a range of values.
  • Continuous nature: No gaps exist between possible outcomes, as the variable can assume any value in a specified continuum.
In our context, the Rayleigh distribution is a prime example of a continuous probability distribution, often used in scenarios where phenomena are influenced by random fluctuations around a known average. The calculation of probabilities in continuous distributions involves integrating the PDF over the variable's desired range to determine likelihoods.

For instance, when determining probabilities like \( P(100 \leq X \leq 200) \) for stress values, integration of the PDF allows us to pinpoint the chance that the random variable \( X \) falls within a specified interval, leveraging the continuity of possible outcomes.

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Most popular questions from this chapter

The automatic opening device of a military cargo parachute has been designed to open when the parachute is \(200 \mathrm{~m}\) above the ground. Suppose opening altitude actually has a normal distribution with mean value \(200 \mathrm{~m}\) and standard deviation \(30 \mathrm{~m}\). Equipment damage will occur if the parachute opens at an altitude of less than \(100 \mathrm{~m}\). What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

Based on data from a dart-throwing experiment, the article "Shooting Darts" (Chance, Summer 1997, 16-19) proposed that the horizontal and vertical errors from aiming at a point target should be independent of one another, each with a normal distribution having mean 0 and variance \(\sigma^{2}\). It can then be shown that the pdf of the distance \(V\) from the target to the landing point is $$ f(v)=\frac{v}{\sigma^{2}} \cdot e^{-v^{2} / 2 \sigma^{2}} \quad v>0 $$ a. This pdf is a member of what family introduced in this chapter? b. If \(\sigma=20 \mathrm{~mm}\) (close to the value suggested in the paper), what is the probability that a dart will land within \(25 \mathrm{~mm}\) (roughly \(1 \mathrm{in}\).) of the target?

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 \(F(t ; \lambda, n)=P(X \leq t)\).

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean \(\mu\), the actual temperature of the medium, and standard deviation \(\sigma\). What would the value of \(\sigma\) have to be to ensure that \(95 \%\) of all readings are within \(.1^{\circ}\) of \(\mu\) ?

Consider babies born in the "normal" range of \(37-43\) weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\). [The article "Are Babies Normal?"' (The American Statistician, 1999: 298-302) analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.] a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? Is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? d. How would you characterize the most extreme .1\% of all birth weights? e. If \(X\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0\) ), then \(Y=a X\) also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

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