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Let \(X\) denote the ultimate tensile strength (ksi) at \(-200^{\circ}\) of a randomly selected steel specimen of a certain type that exhibits "cold brittleness" at low temperatures. Suppose that \(X\) has a Weibull distribution with \(\alpha=20\) and \(\beta=100\). a. What is the probability that \(X\) is at most \(105 \mathrm{ksi}\) ? b. If specimen after specimen is selected, what is the long-run proportion having strength values between 100 and \(105 \mathrm{ksi}\) ? c. What is the median of the strength distribution?

Short Answer

Expert verified
a. Approximately 0.675. b. Approximately 0.043. c. Approximately 97.7 ksi.

Step by step solution

01

Understanding the Weibull Distribution

The probability density function (pdf) for a Weibull distribution is given by \( f(x; \alpha, \beta) = \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \) for \( x > 0 \). The cumulative distribution function (cdf) is \( F(x; \alpha, \beta) = 1 - e^{-(x/\beta)^\alpha} \), where \( \alpha \) is the shape parameter and \( \beta \) is the scale parameter.
02

Computing the Probability for a. (X ≤ 105)

To find the probability that \( X \leq 105 \), we use the cumulative distribution function (cdf) formula:\[P(X \leq 105) = F(105; 20, 100) = 1 - e^{-(105/100)^{20}}\]So first, calculate \((105/100)^{20}\) and then use it to find the probability by substituting into the cdf formula.
03

Numerical Solution for a. (Probability)

First, compute the power: \( (105/100)^{20} \approx 1.126825\).Then, calculate the exponent: \( -(105/100)^{20} \approx -1.126825\)Substitute back:\[ 1 - e^{-1.126825} \approx 1 - 0.324652 = 0.675348 \]Thus, the probability that \( X \leq 105 \) ksi is approximately 0.675.
04

Long-Run Proportion Calculation for b. (100 < X ≤ 105)

To find the proportion of specimens with strengths between 100 and 105 ksi, calculate:\[P(100 < X \leq 105) = F(105) - F(100)\]Given that:\[F(100) = 1 - e^{-(100/100)^{20}} = 1 - e^{-1} \approx 1 - 0.367879 = 0.632121\]So:\[P(100 < X \leq 105) = 0.675348 - 0.632121 = 0.043227\]The long-run proportion is approximately 0.043.
05

Calculating the Median for c.

The median of a Weibull distribution is the value \( m \) such that \( F(m) = 0.5 \). Therefore, we solve:\[1 - e^{-(m/100)^{20}} = 0.5\]Rearranging gives:\[e^{-(m/100)^{20}} = 0.5 \quad \Rightarrow \quad -(m/100)^{20} = \ln(0.5)\]\[(m/100)^{20} = -\ln(0.5) \approx 0.693147\]Thus:\[m = 100 \times (0.693147)^{1/20} \approx 100 \times 0.977273 = 97.727\]The median tensile strength is approximately 97.7 ksi.

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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 is a fundamental tool used in statistics to describe probabilities of a random variable. For the Weibull distribution, the pdf is given by: \[ f(x; \alpha, \beta) = \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \]This function helps us understand how the probabilities are distributed over different values of a random variable, in this instance, tensile strength. The pdf is only valid for \(x > 0\), which makes sense since tensile strength cannot be negative.In this context:
  • \(\alpha\) is the shape parameter, which influences the distribution form.
  • \(\beta\) is the scale parameter, which affects the stretch of the distribution along the x-axis.
The pdf graphically shows the likelihood of different tensile strength values occurring and is a continuous curve that integrates to 1 over its range.
Cumulative Distribution Function
The Cumulative Distribution Function (cdf) of the Weibull distribution is closely related to the pdf, but it offers a different perspective. Instead of indicating the probability of a specific outcome, it shows the probability of a random variable being less than or equal to a specific value.For the Weibull distribution, the cdf is given by:\[ F(x; \alpha, \beta) = 1 - e^{-(x/\beta)^\alpha} \]This function provides insights into the cumulative probability up to a particular tensile strength and is especially useful when determining probabilities over intervals.

For example, to find the probability that a specimen's tensile strength is at most 105 ksi, we compute:
  • First, evaluate \((105/100)^{20}\) to get 1.126825.
  • Then calculate \(-e^{-1.126825}\) resulting in approximately -0.324652.
  • Subtract this from 1 to obtain 0.675348.
Therefore, the probability that the tensile strength is 105 ksi or less is approximately 0.675.
Shape Parameter
The Shape Parameter, denoted as \(\alpha\) in the Weibull distribution, is a crucial element that dictates the distribution's behavior and appellation. It fundamentally alters the form of the distribution curve.
  • If \(\alpha < 1\), the distribution is heavily skewed, with a high probability of smaller values.
  • If \(\alpha = 1\), the distribution shapes into an exponential distribution, meaning that the event likelihood doesn't significantly change with increasing x-values.
  • If \(\alpha > 1\), the tail of the distribution becomes heavier, which indicates a greater likelihood of larger values.
For the steel specimen data, \(\alpha\) is set to 20, indicating a strong preference for certain tensile strength levels and affecting how quickly the probabilities increase as tensile strength increases. This particular setting results in smaller variations and a distribution that is heavily weighted towards higher tensile strength values.
Scale Parameter
The Scale Parameter, represented by \(\beta\), plays an equally important role as the shape parameter in defining the Weibull distribution. This parameter dictates the range of the distribution along the x-axis and is critical for understanding the typical values expected.
With a larger \(\beta\), the distribution stretches along the x-axis, intensifying the scale of the values. In our given scenario, \(\beta = 100\), effectively setting the typical tensile strength around 100 ksi. This influences the spread and the rate at which probabilities increase.
The scale parameter acts like a "magnifier," so to speak, for the tensile strength values, allowing us to predict how they will vary across different specimens. For example, it helps us compute the median strength, which rounded out to approximately 97.7 ksi, a figure which would inherently change with different \(\beta\) values. This means the distribution's stretch can profoundly affect statistical measures such as the median, mode, and variance.

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Most popular questions from this chapter

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