/*! 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 80 Let \(X\) be the number of mater... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article +'Methodology for Probabilistic Life Prediction of Multiple-Anomaly Materials" (Amer. Inst. of Aeronautics and Astronautics \(J .\), 2006: 787-793) proposes a Poisson distribution for \(X\). Suppose that \(\mu=4\). a. Compute both \(P(X \leq 4)\) and \(P(X<4)\). b. Compute \(P(4 \leq X \leq 8)\). c. Compute \(P(8 \leq X)\). d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

Short Answer

Expert verified
a) \(P(X \leq 4)\approx 0.6288\), \(P(X < 4)\approx 0.4335\); b) \(P(4 \leq X \leq 8)\approx 0.7275\); c) \(P(8 \leq X)\approx 0.0419\); d) \(P(4 < X \leq 6)\approx 0.3425\).

Step by step solution

01

Understand the Problem and Identify Given Parameters

This exercise involves a Poisson distribution for the number of anomalies \(X\) with a given mean \(\mu = 4\). We're tasked with computing several probabilities related to this distribution.
02

Probability Definitions for Poisson Distribution

For a Poisson distribution with mean \(\mu\), the probability of observing \(k\) events is given by \(P(X = k) = \frac{\mu^k e^{-\mu}}{k!}\). We need to calculate probabilities such as \(P(X \leq 4)\) and \(P(X < 4)\).
03

Compute \(P(X \leq 4)\)

Calculate \(P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)\). Use the formula for each \(k\) from 0 to 4.
04

Compute \(P(X < 4)\)

This is the probability that \(X\) is less than 4, which equals \(P(X=0) + P(X=1) + P(X=2) + P(X=3)\).
05

Compute \(P(4 \leq X \leq 8)\)

This requires finding \(P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)\) using the Poisson formula for each value.
06

Compute \(P(8 \leq X)\)

Use the complement rule: \(P(X \geq 8) = 1 - P(X < 8)\). Calculate \(P(X < 8) = P(0) + P(1) + ... + P(7)\) and subtract from 1.
07

Probability of Exceeding Mean by No More Than One Standard Deviation

For a Poisson distribution, the standard deviation is \(\sigma = \sqrt{\mu} = 2\). Thus, calculate \(P(4 < X \leq 6)\) as the number of anomalies exceeds its mean by no more than one standard deviation.
08

Final Calculation and Result

Calculate and sum each probability from the formulas in the previous steps to get the final probabilities for each part of the question.

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 Calculations
In the context of the Poisson distribution, probability calculations involve finding the likelihood of observing a specific number of events (in this case, material anomalies) within a given time frame or spatial area. The Poisson distribution is appropriate for modeling events that happen independently and with a known constant mean rate. Here, we use the formula \( P(X = k) = \frac{\mu^k e^{-\mu}}{k!} \) where \( \mu \) is the mean number of events, \( e \) is the base of the natural logarithm, and \( k! \) is the factorial of \( k \).

To calculate \( P(X \leq 4) \), we add the probabilities of events from 0 up to 4: \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \). Similarly, for \( P(X < 4) \), we sum the probabilities from 0 to 3. The key part of solving these calculations is understanding the cumulative nature of probability sums in Poisson distribution.
  • For probabilities up to or less than a value, sum up all probabilities to that point.
  • Utilize the Poisson formula for each event number to compute individual probabilities.
Standard Deviation
Standard deviation is a measure of the spread or dispersion of a set of values. In a Poisson distribution, the standard deviation is defined as the square root of the mean \( \mu \). This gives us the formula for standard deviation as \( \sigma = \sqrt{\mu} \). When \( \mu = 4 \), the standard deviation is \( \sigma = 2 \).

The standard deviation tells us how much the number of anomalies might vary from the mean. It helps in determining the range in which the anomalies are likely to fall.

In exercises like calculating the probability that the number of anomalies exceeds its mean by no more than one standard deviation, we need to evaluate \( P(4 < X \leq 6) \). This is the range from \( 4 + 1 \times \sigma \), showing how probability and standard deviation interplay to give an understanding of variability.
  • Standard deviation is crucial in understanding the spread of data in Poisson distributions.
  • The "no more than one standard deviation" logic is commonly used to set bounds for probability calculations.
Mean Value
The mean value in the context of the Poisson distribution is symbolized by \( \mu \) and indicates the expected number of occurrences in a given interval or region. For this exercise, \( \mu = 4 \) which means, on average, 4 anomalies are expected.

Mean value is central to the Poisson formula, since it shapes the probability distribution. By analyzing the mean, one can predict how the number of occurrences will concentrate around this average.\

In applications, knowing the mean allows you to quickly assess the probability of a number of events occurring close to this mean, often using the properties of Poisson and standard deviation. For instance, in determining probabilities for a range like \( 4 \leq X \leq 8 \), understanding the mean helps ground these calculations by setting expectations based on average occurrence.
  • The mean value provides a basis for calculation in Poisson distribution models.
  • It serves as the axis around which probabilities are centered.
  • Poisson distribution assumes the intervals of event occurrence are independent.
Probabilistic Life Prediction
Probabilistic life prediction is a technique in reliability engineering and risk assessment used to estimate the lifespan of a component, such as a gas-turbine disk, by considering variability and random events. Specifically for materials with anomalies, this approach calculates the likelihood of failure and longevity over time.

Using distributions like Poisson, engineers estimate how frequently certain failures or anomalies may occur, providing a statistical basis for planning inspections, maintenance, and guarantees. By understanding the distribution, the mean value, and standard deviation, the extent to which anomalies deviate from expected norms can be measured, enabling data-driven life predictions.

For such predictions, effective understanding of probability calculations is essential. Using statistical models:
  • Methodical predictions can help reduce risks and prevent catastrophic failures.
  • Probabilistic methods incorporate variability in materials and environments.
  • This approach offers a strategic vision of life expectancy in engineering designs.
By evaluating predictions through probability distributions, material anomalies can be better managed and anticipated, allowing for more resilient and reliable design practices.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article 'Should You Report That FenderBender?" (Consumer Reports, Sept. 2013: 15) reported that 7 in 10 auto accidents involve a single vehicle (the article recommended always reporting to the insurance company an accident involving multiple vehicles). Suppose 15 accidents are randomly selected. Use Appendix Table A.l to answer each of the following questions. a. What is the probability that at most 4 involve a single vehicle? b. What is the probability that exactly 4 involve a single vehicle? c. What is the probability that exactly 6 involve multiple vehicles? d. What is the probability that between 2 and 4 , inclusive, involve a single vehicle? e. What is the probability that at least 2 involve a single vehicle? f. What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles?

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let \(X\) be the number among the six who are taking the test for the first time. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X=\) the number of months between successive payments. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<1 \\ .30 & 1 \leq x<3 \\ .40 & 3 \leq x<4 \\ .45 & 4 \leq x<6 \\ .60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{cases} $$ a. What is the pmf of \(X\) ? b. Using just the cdf, compute \(P(3 \leq X \leq 6)\) and \(P(4 \leq X)\).

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is \begin{tabular}{l|cccc} \(y\) & 0 & 1 & 2 & 3 \\ \hline\(p(y)\) & \(.60\) & \(.25\) & \(.10\) & \(.05\) \end{tabular} a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

The \(n\) candidates for a job have been ranked \(1,2,3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)= \begin{cases}1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise }\end{cases} $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6 .]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.