Math Problem Statement

Let T ∼ Weibull(0.5, 2.400).

Find P(T > 5). (Round the final answer to four decimal places.)

Solution

The probability density function (PDF) of a Weibull distribution is given by:

f(t)=kλ(tλ)k1e(t/λ)kf(t) = \frac{k}{\lambda} \left(\frac{t}{\lambda}\right)^{k-1} e^{-(t/\lambda)^k}

where kk is the shape parameter and λ\lambda is the scale parameter. However, to compute P(T>t)P(T > t), we use the cumulative distribution function (CDF), which for a Weibull distribution is:

F(t)=1e(t/λ)kF(t) = 1 - e^{-(t/\lambda)^k}

The probability P(T>t)P(T > t) is complementary to the CDF:

P(T>t)=1F(t)=e(t/λ)kP(T > t) = 1 - F(t) = e^{-(t/\lambda)^k}

Given k=0.5k = 0.5 and λ=2.4\lambda = 2.4, we want to find P(T>5)P(T > 5):

P(T>5)=e(5/2.4)0.5P(T > 5) = e^{-(5/2.4)^{0.5}}

First, calculate (5/2.4)0.5(5/2.4)^{0.5}:

52.42.0833\frac{5}{2.4} \approx 2.0833

2.08330.51.44332.0833^{0.5} \approx 1.4433

Now, calculate e1.4433e^{-1.4433}:

e1.44330.2360e^{-1.4433} \approx 0.2360

So, the probability is approximately:

P(T>5)0.2360P(T > 5) \approx 0.2360

Final Answer:

P(T>5)0.2360P(T > 5) \approx 0.2360

Would you like more details on this calculation or have any questions?

Here are 5 related questions:

  1. What is the expected value of a Weibull distribution?
  2. How does the shape parameter affect the distribution's behavior?
  3. Can you compute P(T5)P(T \leq 5) for the same distribution?
  4. What is the variance of a Weibull distribution?
  5. How do you find the median of a Weibull distribution?

Tip: The Weibull distribution can model a variety of life data behaviors, especially in reliability and survival analysis. Adjusting the shape parameter can mimic different failure rates.

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Math Problem Analysis

Mathematical Concepts

Weibull Distribution
Probability
Exponential Function

Formulas

Weibull CDF: F(t) = 1 - e^{-(t/λ)^k}
Complementary Probability: P(T > t) = e^{-(t/λ)^k}

Theorems

Weibull Distribution Theorem

Suitable Grade Level

Undergraduate Statistics or Advanced High School (AP Statistics)