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 = 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, -t²-1-(x/B) α f(x; a, B) = Ba 0 x 20 x<0 by x - y and x ≥ 0 by x ≥ 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 1.510 0.624 (d) What are the mean and standard deviation of X? (Round your answers to three decimal places.) mean x days standard deviation ✔ days

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It still says the mean is wrong in part D. 

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,

\[
f(x; \alpha, \beta) = 
\begin{cases} 
\frac{\alpha}{\beta^{\alpha}} (x - \gamma)^{\alpha - 1} e^{-(x/\beta)^\alpha} & x \geq \gamma \\
0 & x < 0 
\end{cases}
\]

by \( x - \gamma \) and \( x \geq 0 \) by \( x \geq \gamma \).

**(a)** Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.)

\[
0.4737 \quad \checkmark
\]

**(b)** Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.)

\[
0.7775 \quad \checkmark
\]

**(c)** What is the 90th percentile of the distribution? (Round your answer to three decimal places.)

\[
2.843 \quad \text{days} \quad \checkmark
\]

**(d)** What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.)

\[
\begin{align*}
\text{mean} & \quad 1.510 \quad \text{x} \\
\text{standard deviation} & \quad 0.624 \quad \text{days} \quad \checkmark
\end{align*}
\]
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, \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^{\alpha}} (x - \gamma)^{\alpha - 1} e^{-(x/\beta)^\alpha} & x \geq \gamma \\ 0 & x < 0 \end{cases} \] by \( x - \gamma \) and \( x \geq 0 \) by \( x \geq \gamma \). **(a)** Calculate \( P(1 < X < 2) \). (Round your answer to four decimal places.) \[ 0.4737 \quad \checkmark \] **(b)** Calculate \( P(X > 1.5) \). (Round your answer to four decimal places.) \[ 0.7775 \quad \checkmark \] **(c)** What is the 90th percentile of the distribution? (Round your answer to three decimal places.) \[ 2.843 \quad \text{days} \quad \checkmark \] **(d)** What are the mean and standard deviation of \( X \)? (Round your answers to three decimal places.) \[ \begin{align*} \text{mean} & \quad 1.510 \quad \text{x} \\ \text{standard deviation} & \quad 0.624 \quad \text{days} \quad \checkmark \end{align*} \]
Expert Solution
Step 1: Given information

Let f be a pdf

f left parenthesis x comma alpha comma beta right parenthesis space equals space alpha over beta to the power of alpha x to the power of alpha minus 1 end exponent e to the power of negative left parenthesis x divided by beta right parenthesis to the power of alpha end exponent ; x>=0

= 0 ; x<0


alpha equals space 2.6
beta equals 1.7
gamma equals 0.5


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It says that 1.496 is also wrong for the mean in part D

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,  
by \(x - \gamma\) and \(x \ge 0\) by \(x \ge \gamma\).]  

\[
f(x; \alpha, \beta) = 
\begin{cases} 
\frac{\alpha}{\beta^\alpha} (x^\alpha) e^{-(x/\beta)^\alpha} & x \ge 0 \\
0 & x < 0 
\end{cases}
\]

(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: 1.496 days ✖️  
- Standard deviation: 0.624 days ✔️  

Note that the calculation for the mean seems to be incorrect.
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, by \(x - \gamma\) and \(x \ge 0\) by \(x \ge \gamma\).] \[ f(x; \alpha, \beta) = \begin{cases} \frac{\alpha}{\beta^\alpha} (x^\alpha) e^{-(x/\beta)^\alpha} & x \ge 0 \\ 0 & x < 0 \end{cases} \] (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: 1.496 days ✖️ - Standard deviation: 0.624 days ✔️ Note that the calculation for the mean seems to be incorrect.
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