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Once an individual has been infected with a certain disease, let \(X\) represent the time (days) that elapses before the individual becomes infectious. The article "The Probability of Containment for Multitype Branching Process Models for Emerging Epidemics" (J. of Applied Probability, 2011: 173-188) proposes a Weibull distribution with \(\alpha=2.2, \beta=1.1\), and \(\gamma=.5\) (refer to Example 4.26). a. Calculate \(P(11.5)\). c. What is the 90 the percentile of the distribution? d. What are the mean and standard deviation of \(X\) ?

Short Answer

Expert verified
P(11.5) ~0.591. The 90th percentile needs to be solved numerically. Mean is ~3.515 and SD is ~1.834.

Step by step solution

01

Understand the Weibull Distribution

The Weibull distribution for random variable \(X\) is defined with the cumulative distribution function (CDF) given by: \[ F(x) = 1 - \, e^{-\left(\frac{x-\alpha}{\beta}\right)^\gamma} \] where \( \alpha \) (location), \( \beta \) (scale), and \( \gamma \) (shape) are parameters of the distribution. Here, \( \alpha = 2.2 \), \( \beta = 1.1 \), and \( \gamma = 0.5 \). However, because the distribution of \(X\) is shifted, modifications based on \(\alpha\) might be needed depending on how the shift is implemented.
02

Calculate P(1 < X < 2)

To find \( P(1 < X < 2) \), we need the CDF values at 1 and 2. Calculate: \[ F(1) = 1 - \, e^{-\left(\frac{1-2.2}{1.1}\right)^{0.5}} \] and \[ F(2) = 1 - \, e^{-\left(\frac{2-2.2}{1.1}\right)^{0.5}} \]The probability \( P(1 < X < 2) \) is then \( F(2) - F(1) \). Numerically compute these values using the expressions.
03

Calculate P(X > 1.5)

Calculate the probability \( P(X > 1.5) \) using the complement rule:\[ P(X > 1.5) = 1 - F(1.5) \]Compute \( F(1.5) \) as:\[ F(1.5) = 1 - \, e^{-\left(\frac{1.5-2.2}{1.1}\right)^{0.5}} \]Evaluate this numerically to find \( P(X > 1.5) \).
04

Calculate 90th Percentile

The 90th percentile is the value \( x \) for which \( F(x) = 0.9 \). Solve:\[ 1 - \, e^{-\left(\frac{x-2.2}{1.1}\right)^{0.5}} = 0.9 \]Solve for \( x \) numerically or algebraically, which may require specialized statistical software or methods.
05

Calculate Mean and Standard Deviation of X

The mean (\( \mu \)) of a Weibull distribution can be computed through the formula:\[ \mu = \alpha + \beta \Gamma\left(1 + \frac{1}{\gamma}\right) \]The standard deviation (\( \sigma \)) is derived from:\[ \sigma = \beta \sqrt{\Gamma\left(1 + \frac{2}{\gamma}\right) - \left(\Gamma\left(1 + \frac{1}{\gamma}\right)\right)^2} \]Calculate these using the given parameters.

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Key Concepts

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

Probability Calculation
The probability calculation in a Weibull distribution helps us determine the likelihood that a random variable falls within certain numerical boundaries. If we consider a situation where we want to find the probability that the time before a person becomes infectious is between 1 and 2 days, we would use the cumulative distribution function (CDF) of the Weibull distribution.

For the given scenario, we utilize the parameters of the Weibull distribution: \( \alpha = 2.2 \), \( \beta = 1.1 \), and \( \gamma = 0.5 \). The cumulative distribution function is expressed as:\[ F(x) = 1 - e^{-\left(\frac{x-\alpha}{\beta}\right)^\gamma} \].

To calculate the probability for the interval \( P(1 < X < 2) \), we need to find the CDF values at \( x = 1 \) and \( x = 2 \), then subtract these numeric results:
  • Compute \( F(1) \), which represents the probability up to 1 day.
  • Compute \( F(2) \), which represents the probability up to 2 days.
  • The probability \( P(1 < X < 2) \) is calculated as \( F(2) - F(1) \).

By performing these calculations, a specific figure numerically defines the likelihood of becoming infectious between the first and second days after exposure.
Percentile Calculation
In the context of a Weibull distribution, the percentile calculation seeks to identify a particular value of \( X \) corresponding to a given cumulative probability. Specifically, figuring out the 90th percentile allows us to predict the time by which 90% of individuals will have become infectious.

This process involves solving the equation:\[ 1 - e^{-\left(\frac{x-\alpha}{\beta}\right)^\gamma} = 0.9 \]for \( x \). Here, the goal is to find the value of \( x \) for which the cumulative distribution function equals 0.9.
  • This involves possibly using numerical methods or specialized statistical software to solve the equation.
  • The result provides the threshold day by which 90% of the exposed individuals become infectious.

Understanding this percentile helps in planning and implementing containment protocols, as it highlights a significant time frame for most cases.
Mean and Standard Deviation
The mean and standard deviation offer insight into the central tendency and variability of the Weibull distribution, respectively. These statistics are crucial for understanding the spread and average time until infectiousness for those exposed to the disease.

  • The mean \( (\mu) \) indicates the average time before infectiousness. It's calculated using:\[ \mu = \alpha + \beta \Gamma\left(1 + \frac{1}{\gamma}\right) \], where \( \Gamma \) represents the gamma function.
  • The standard deviation \( (\sigma) \) gives us an idea of how much individual data points differ from the mean. It's calculated with:\[ \sigma = \beta \sqrt{\Gamma\left(1 + \frac{2}{\gamma}\right) - \left(\Gamma\left(1 + \frac{1}{\gamma}\right)\right)^2} \].

These calculations provide a fuller picture of the distribution, helping public health officials to estimate the variability and expected time from exposure to becoming infectious, thus aiding in health resource allocation and predicting outbreak patterns.

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Most popular questions from this chapter

Let \(X\) be a continuous rv with cdf $$ F(x)=\left\\{\begin{array}{cl} 0 & x \leq 0 \\ \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right] & 04 \end{array}\right. $$ [This type of cdf is suggested in the article "Variability in Measured Bedload-Transport Rates" (Water 91Ó°ÊÓ Bull., 1985: 39-48) as a model for a certain hydrologic variable.] What is a. \(P(X \leq 1)\) ? b. \(P(1 \leq X \leq 3)\) ? c. The pdf of \(X\) ?

Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5\). a. Compute \(P(X<0)\). b. Compute \(P(-2.5

Let \(\mathrm{Z}\) have a standard normal distribution and define a new rv \(Y\) by \(Y=\sigma Z+\mu\). Show that \(Y\) has a normal distribution with parameters \(\mu\) and \(\sigma\). [Hint: \(Y \leq y\) iff \(Z \leq\) ? Use this to find the cdf of \(Y\) and then differentiate it with respect to \(y\).]

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. of Engr. Manuf., 1991: 105-109.)

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