/*! 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 134 . Suppose that \(X\) has a Weibu... [FREE SOLUTION] | 91Ó°ÊÓ

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. Suppose that \(X\) has a Weibull distribution with \(\beta=2\) and \(\delta=8.6 .\) Determine the following: (a) \(P(X<10)\) (b) \(P(X>9)\) (c) \(P(8x)=0.9\)

Short Answer

Expert verified
(a) 0.741, (b) 0.335, (c) 0.227, (d) 2.79

Step by step solution

01

Understanding the Weibull Distribution

The probability density function for a Weibull distribution is given by \( f(x; \beta, \delta) = \frac{\beta}{\delta} \left( \frac{x}{\delta} \right)^{\beta - 1} e^{-\left( \frac{x}{\delta} \right)^\beta} \), where \( \beta \) is the shape parameter and \( \delta \) is the scale parameter. We'll use these parameters in our calculations. The cumulative distribution function (CDF) is \( F(x) = 1 - e^{-\left( \frac{x}{\delta} \right)^\beta} \). For this problem, \( \beta = 2 \) and \( \delta = 8.6 \).
02

Calculate \( P(X < 10) \)

To find \( P(X < 10) \), apply the CDF. \( F(10) = 1 - e^{-\left( \frac{10}{8.6} \right)^2} \). Calculate \( \left( \frac{10}{8.6} \right)^2 \approx 1.352 \). Then, \( F(10) = 1 - e^{-1.352} \approx 1 - 0.259 = 0.741 \). Thus, \( P(X < 10) = 0.741 \).
03

Find \( P(X > 9) \)

We know that \( P(X > 9) = 1 - P(X \le 9) = 1 - F(9) \). Compute \( F(9) = 1 - e^{-\left( \frac{9}{8.6} \right)^2} \). \( \left( \frac{9}{8.6} \right)^2 \approx 1.093 \), so \( F(9) \approx 1 - e^{-1.093} \approx 1 - 0.335 = 0.665 \). Therefore, \( P(X > 9) = 0.335 \).
04

Calculate \( P(8 < X < 11) \)

To find \( P(8 < X < 11) = F(11) - F(8) \), calculate both CDFs. First, \( F(11) = 1 - e^{-\left( \frac{11}{8.6} \right)^2} \approx 1 - e^{-1.637} \approx 1 - 0.194 = 0.806 \). Then, \( F(8) = 1 - e^{-\left( \frac{8}{8.6} \right)^2} \approx 1 - e^{-0.864} \approx 1 - 0.421 = 0.579 \). Thus, \( P(8 < X < 11) = 0.806 - 0.579 = 0.227 \).
05

Determine \( x \) such that \( P(X > x) = 0.9 \)

We seek \( x \) such that \( 1 - F(x) = 0.9 \) or \( F(x) = 0.1 \). Solve \( 1 - e^{-\left( \frac{x}{8.6} \right)^2} = 0.1 \), so \( e^{-\left( \frac{x}{8.6} \right)^2} = 0.9 \). Taking logs, \( -\left( \frac{x}{8.6} \right)^2 = \ln(0.9) \approx -0.105 \). Thus, \( \left( \frac{x}{8.6} \right)^2 = 0.105 \), giving \( \frac{x}{8.6} = \sqrt{0.105} \approx 0.324 \). Finally, \( x \approx 8.6 \times 0.324 = 2.79 \). Therefore, the value of \( x \) is approximately 2.79.

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Key Concepts

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

Probability Density Function
The Probability Density Function (PDF) of a Weibull distribution helps us understand how likely it is for a random variable, say \(X\), to take on a specific value. In mathematical terms, the PDF of a Weibull distribution is given by:
  • \( f(x; \beta, \delta) = \frac{\beta}{\delta} \left( \frac{x}{\delta} \right)^{\beta - 1} e^{-\left( \frac{x}{\delta} \right)^\beta} \)
Here, \(\beta\) is the shape parameter and \(\delta\) is the scale parameter. The PDF provides a way to model the distribution of a random variable, giving us insights into its behavior.
For example, in reliability testing, the Weibull PDF can help predict failure times of products. Its unique formula allows analysts to understand how frequently or at what intervals a particular event occurs. In summary, the PDF is key for examining the probabilities of a variable reaching specific values, thus playing a critical role in probability theory and statistics.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) accumulates the probabilities that a random variable is less than or equal to a particular value. For the Weibull distribution, the CDF is expressed by:
  • \( F(x) = 1 - e^{-\left( \frac{x}{\delta} \right)^\beta} \)
The CDF is immensely helpful because it gives us the probability for a range of values rather than a specific point, thus offering a more comprehensive picture. When dealing with questions like "What's the probability that \(X\) is less than 10?", you would use the Weibull CDF to get the answer.
This function highlights the growth of these probabilities cumulatively, helping analysts understand the likelihood of an event occurring up to a certain point. The CDF thus provides an overall assumption about the distribution of data, aiding in statistical analysis and forecasting.
Shape Parameter
The shape parameter in a Weibull distribution, denoted as \(\beta\), significantly influences the form of the distribution. This parameter determines whether the rate of failure increases or decreases with time.
  • A \(\beta\) less than 1 indicates that the event rates decrease over time, which is often seen in learning and adaptation processes.
  • A \(\beta\) equal to 1 simplifies the Weibull distribution to an exponential distribution, implying a constant failure rate.
  • A \(\beta\) greater than 1 suggests that the failure rate increases over time, a common case in scenarios like aging or wear-out failures in mechanical components.
Understanding the shape parameter is vital for interpreting how likely failures or events are distributed over time, impacting decision-making processes in risk assessment and quality control.
Scale Parameter
The scale parameter, represented as \(\delta\), stretches or compresses the Weibull distribution along the x-axis. This parameter affects the timing of events.
The larger the scale parameter, the longer the expected time until an event occurs. Consequently, the scale parameter serves as a critical indicator in reliability testing and lifetime data analysis.
  • It scales the distribution and influences its width; a larger \(\delta\) results in a broader curve.
  • If \(\delta\) increases, the overall probability of occurrence for earlier times reduces, pushing events further into the future.
Thus, the scale parameter, by altering the spread of the distribution, allows analysts to infer real-world event timings and plan accordingly for scenarios like maintenance schedules and risk mitigation strategies.

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