Homework Help is Here – Start Your Trial Now!
Math
Probability
Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß= 1.7, and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below, ²²-1-(x/B)a f(x; a, B) = Ba" 0 x20 x < 0 by xy and x 20 by x 2 y.] (a) Calculate P(1 < X < 2). (Round your answer to four decimal places.) 0.4737 (b) Calculate P(X> 1.5). (Round your answer to four decimal places.) 0.7775 (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) 2.843 ✔ days (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) 1.496 X days mean

Once an individual has been infected with a certain disease, let X represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with a = 2.6, ß= 1.7, and y = 0.5. [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter y, called a threshold or location parameter: replace x in the equation below, ²²-1-(x/B)a f(x; a, B) = Ba" 0 x20 x < 0 by xy and x 20 by x 2 y.] (a) Calculate P(1 < X < 2). (Round your answer to four decimal places.) 0.4737 (b) Calculate P(X> 1.5). (Round your answer to four decimal places.) 0.7775 (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) 2.843 ✔ days (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) 1.496 X days mean

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question

I need help finding the mean for part D, I tried 1.510 and got it wrong and I also tried 1.496 and I also got that wrong. 

Once an individual has been infected with a certain disease, let \( X \) represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with \( \alpha = 2.6 \), \( \beta = 1.7 \), and \( \gamma = 0.5 \). [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter \( \gamma \), called a threshold or location parameter: replace \( x \) in the equation below, \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^\alpha} (x)^{\alpha-1} e^{-(x/\beta)^\alpha} & x \ge 0 \\ 0 & x < 0 \end{cases} \] by \( x - \gamma \) and \( x \ge 0 \) by \( x \ge \gamma \).] (a) Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.) - Answer: \( 0.4737 \) (b) Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.) - Answer: \( 0.7775 \) (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) - Answer: \( 2.843 \) days (d) What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.) - Mean: \( 1.496 \) days - Standard deviation: \( 0.624 \) days
Transcribed Image Text:Once an individual has been infected with a certain disease, let \( X \) represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with \( \alpha = 2.6 \), \( \beta = 1.7 \), and \( \gamma = 0.5 \). [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter \( \gamma \), called a threshold or location parameter: replace \( x \) in the equation below, \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^\alpha} (x)^{\alpha-1} e^{-(x/\beta)^\alpha} & x \ge 0 \\ 0 & x < 0 \end{cases} \] by \( x - \gamma \) and \( x \ge 0 \) by \( x \ge \gamma \).] (a) Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.) - Answer: \( 0.4737 \) (b) Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.) - Answer: \( 0.7775 \) (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) - Answer: \( 2.843 \) days (d) What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.) - Mean: \( 1.496 \) days - Standard deviation: \( 0.624 \) days
Once an individual has been infected with a certain disease, let \( X \) represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with \( \alpha = 2.6 \), \( \beta = 1.7 \), and \( \gamma = 0.5 \). [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter \( \gamma \), called a threshold or location parameter: replace \( x \) in the equation below, \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^\alpha}(x - \gamma)^{\alpha - 1}e^{-(x/\beta)^\alpha} & x \geq 0 \\ 0 & x < 0 \end{cases} \] by \( x - \gamma \) and \( x \geq 0 \) by \( x \geq \gamma \).] (a) Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.) \[ 0.4737 \] ✔ (b) Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.) \[ 0.7775 \] ✔ (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) \[ 2.843 \] days ✔ (d) What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.) - Mean: \[ 1.510 \] days ❌ - Standard Deviation: \[ 0.624 \] days ✔
Transcribed Image Text:Once an individual has been infected with a certain disease, let \( X \) represent the time (days) that elapses before the individual becomes infectious. An article proposes a Weibull distribution with \( \alpha = 2.6 \), \( \beta = 1.7 \), and \( \gamma = 0.5 \). [Hint: The two-parameter Weibull distribution can be generalized by introducing a third parameter \( \gamma \), called a threshold or location parameter: replace \( x \) in the equation below, \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^\alpha}(x - \gamma)^{\alpha - 1}e^{-(x/\beta)^\alpha} & x \geq 0 \\ 0 & x < 0 \end{cases} \] by \( x - \gamma \) and \( x \geq 0 \) by \( x \geq \gamma \).] (a) Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.) \[ 0.4737 \] ✔ (b) Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.) \[ 0.7775 \] ✔ (c) What is the 90th percentile of the distribution? (Round your answer to three decimal places.) \[ 2.843 \] days ✔ (d) What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.) - Mean: \[ 1.510 \] days ❌ - Standard Deviation: \[ 0.624 \] days ✔
Expert Solution
Step 1: Given Information

The pdf of the Weibull distribution is: the

f open parentheses x semicolon alpha comma beta close parentheses equals open curly brackets table row cell alpha over beta to the power of alpha x to the power of alpha minus 1 end exponent e to the power of negative open parentheses x divided by beta close parentheses to the power of alpha end exponent space semicolon x greater or equal than 0 end cell row cell 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space semicolon x less than 0 end cell end table close


the values of the parameters are alpha equals 2.6beta equals 1.7gamma equals 0.5.


steps

Step by step

Solved in 3 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer