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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) Use \(F(10)\), (b) Use \(1-F(9)\), (c) Use \(F(11)-F(8)\), and (d) Solve \(e^{-(x/8.6)^2} = 0.9\).

Step by step solution

01

Understanding the Weibull Distribution

The Weibull distribution is a continuous probability distribution often used to model life data and failure times. It is characterized by two parameters: the shape parameter \(\beta\) and the scale parameter \(\delta\). Given \(\beta=2\) and \(\delta=8.6\), we can express the cumulative distribution function (CDF) for a Weibull distributed random variable \(X\) as \(F(x) = 1 - e^{-(x/\delta)^\beta}\).
02

Solve Part (a)

To find \(P(X < 10)\), we calculate the CDF of the Weibull distribution at \(x = 10\):\[F(10) = 1 - e^{-(10/8.6)^2}\].First, compute \((10/8.6)^2\), then calculate the exponential and finally 1 minus the result.
03

Solve Part (b)

\(P(X > 9) = 1 - P(X < 9)\), utilizing the complement rule. Calculate the CDF at \(x = 9\):\[F(9) = 1 - e^{-(9/8.6)^2}\].Thus, \(P(X > 9) = 1 - F(9)\).
04

Solve Part (c)

\(P(8 < X < 11) = F(11) - F(8)\), use the CDF for both values:\[F(11) = 1 - e^{-(11/8.6)^2}\]and\[F(8) = 1 - e^{-(8/8.6)^2}\].Subtract these to get the probability for the range.
05

Solve Part (d)

We need to find the value of \(x\) such that \(P(X > x) = 0.9\), hence \(P(X < x) = 0.1\). Set up the equation:\[1 - e^{-(x/8.6)^2} = 0.1\].Rearrange to solve for \(x\):\[e^{-(x/8.6)^2} = 0.9\].Take the natural logarithm of both sides and solve for \(x\).

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Key Concepts

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

Shape Parameter
The shape parameter, denoted typically as \( \beta \), plays a crucial role in the behavior and properties of the Weibull distribution. It determines how the probability distribution is "shaped" for different scenarios. The value of \( \beta \) can dramatically change how the distribution appears and behaves.
Here's what different values of \( \beta \) imply:
  • If \( \beta = 1 \), the Weibull distribution simplifies to an exponential distribution, often used for modeling and analyzing the life data of products with a constant failure rate.
  • If \( \beta < 1 \), the failure rate decreases over time, indicating that items are less likely to fail as they age. This might be typical in "infant mortality" scenarios where items with initial defects fail early on.
  • If \( \beta > 1 \), the failure rate increases over time, suggesting that items are more likely to fail as they wear out, which is typical of aging products.
With the value of \( \beta = 2 \), as given in the problem, the distribution indicates an increasing failure rate, making it ideal for modeling items that wear out over time. This shape parameter significantly influences the reliability and hazard functions in relation to time.
Scale Parameter
The scale parameter \( \delta \), sometimes referred to as the "characteristic life," is essential for determining the spread or "scale" of the Weibull distribution. This parameter sets the scale of the time or space within which events occur.
A larger scale parameter stretches the distribution, implying that events or failures will occur over a broader range of time. Conversely, a smaller \( \delta \) suggests that events are more concentrated within a smaller time frame.
Given in the exercise, the scale parameter is \( \delta = 8.6 \). This means that 63.2% of the population will likely have failed by the time \( t = \delta \) due to the nature of the Weibull distribution.
The scale parameter \( \delta \) can also be used to adjust the "horizontal" position of the distribution:
  • It helps in shifting the graph of the Weibull distribution along the time axis.
  • Determines the median or average life period in many applications.
  • Plays a pivotal role in reliability analysis, helping in predicting the lifespan of products and materials.
This makes \( \delta \) a powerful tool for engineers and researchers in optimizing designs and predicting product lifetimes.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) is a foundational concept in probability and statistics, providing the probability that a continuous random variable, such as one following the Weibull distribution, will take on a value less than or equal to a given point. In essence, it helps to understand how the values of a random variable are distributed.
For a Weibull distribution with parameters \( \beta \) and \( \delta \), the CDF is expressed as:\[ F(x) = 1 - e^{-(x/\delta)^\beta} \]Breaking down this formula:
  • \( e^{-(x/\delta)^\beta} \) represents the probability that the random variable exceeds \( x \).
  • The subtraction from 1 converts this to the probability that the variable is \( \leq x \).
Here’s why the CDF is crucial:
  • It is a non-decreasing function; as \( x \) increases, \( F(x) \) either increases or stays constant.
  • Provides a comprehensive understanding of the probability distribution by accounting for all possible values up to \( x \).
  • Used to compute probabilities and perform statistical inference or hypothesis testing on a dataset.
  • In practical terms, tools like the CDF help in decision-making processes, particularly in quality control and risk management scenarios.
The Weibull CDF thereby becomes a valuable tool in reliability analysis, providing insights into the likelihood of different outcomes and helping in the efficient design and testing of products.

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Most popular questions from this chapter

An article in the Journal of Geophysical Research ["Spatial and Temporal Distributions of U.S. of Winds and Wind Power at 80 m Derived from Measurements" (2003, vol. 108)] considered wind speed at stations throughout the United States. A Weibull distribution can be used to model the distribution of wind speeds at a given location. Every location is characterized by a particular shape and scale parameter. For a station at Amarillo, Texas, the mean wind speed at \(80 \mathrm{~m}\) (the hub height of large wind turbines) in 2000 was \(10.3 \mathrm{~m} / \mathrm{s}\) with a standard deviation of \(4.9 \mathrm{~m} / \mathrm{s} .\) Determine the shape and scale parameters of a Weibull distribution with these properties.

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