Problem 135 - Applied Statistics and Probability for Engineers

Question

An article in Vaccine ["Modeling the Effects of Influenza Vaccination of Health Care Workers in Hospital Departments" (2009, Vol.27(44), pp. \(6261-6267\) ) ] considered the immunization of healthcare workers to reduce the hazard rate of influenza virus infection for patients in regular hospital departments. In this analysis, each patient's length of stay in the department is taken as exponentially distributed with a mean of 7.0 days. (a) What is the probability that a patient stays in hospital for less than 5.5 days? (b) What is the probability that a patient stays in hospital for more than 10.0 days if the patient has currently stayed for 7.0 days? (c) Determine the mean length of stay such that the probability is 0.9 that a patient stays in the hospital less than 6.0 days.

Step-by-Step Solution

Answer

(a) 0.5442, (b) 0.6517, (c) 2.605 days

Step 1: Identify the Distribution

The problem states that each patient's length of stay is exponentially distributed. Thus, the probability density function is given by \[ f(x) = \frac{1}{\theta} e^{-x/\theta} \] where \( \theta \) is the mean of the distribution.

Step 2: Determine Parameter \( \theta \)

The mean length of stay \( \theta \) is given as 7.0 days. Therefore, \( \theta = 7.0 \).

Step 3: Calculate Part (a)

To find the probability of a stay less than 5.5 days, use the cumulative distribution function (CDF):\[ P(X < 5.5) = 1 - e^{-5.5/7.0} \]Evaluate this expression to find the probability.

Step 4: Calculate Part (a) Result

Compute the expression:\[ P(X < 5.5) = 1 - e^{-5.5/7.0} \approx 1 - e^{-0.7857} \approx 1 - 0.4558 = 0.5442 \]

Step 5: Calculate Part (b) Using Memoryless Property

Because the exponential distribution is memoryless, the probability a patient stays more than 10 days given they have already stayed 7 days is\[ P(X > 10 \, | \, X > 7) = P(X > 3) = e^{-3/7} \]

Step 6: Calculate Part (b) Result

Compute \( P(X > 3) \):\[ P(X > 3) = e^{-3/7} \approx e^{-0.4286} \approx 0.6517 \]

Step 7: Calculate Part (c)

To find the mean that results in a 0.9 probability of a stay less than 6 days, solve for \( \theta \) in:\[ P(X < 6) = 0.9 = 1 - e^{-6/\theta} \]Rearrange to solve for \( \theta \):\[ e^{-6/\theta} = 0.1 \]\[ -\frac{6}{\theta} = \ln(0.1) \]\[ \theta = -\frac{6}{\ln(0.1)} \]

Step 8: Calculate Part (c) Result

Compute \( \theta \):\[ \theta = -\frac{6}{\ln(0.1)} \approx \frac{6}{2.3026} \approx 2.605 \]

Key Concepts

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function (CDF) is a critical concept in probability and statistics, especially when dealing with continuous random variables like the exponential distribution. It serves as a tool to calculate the probability that a random variable takes on a value less than or equal to a specific value.
For the exponential distribution, the CDF is given by:

• \( F(x) = 1 - e^{-x/\theta} \)

Where:

• \( \theta \) is the mean or average rate of the distribution.
• \( e \) is the base of the natural logarithm.

This function helps us understand probabilities in various scenarios. For instance, to find the probability that a patient stays in the hospital for less than 5.5 days, we use the exponential CDF as shown in the original problem. By substituting 5.5 for \( x \) and 7 for \( \theta \), we estimate the cumulative probability.
The CDF gives us a straightforward way to anticipate the behavior of an exponentially distributed variable over a given time or range. This is crucial in practical applications like healthcare, where understanding patient flow and stay durations can inform hospital resource management.

Memoryless Property

The memoryless property is a unique and defining feature of the exponential distribution. It is characterized by the fact that the probability of an event occurring in the future is independent of how much time has already elapsed. This means that past data or events do not affect the probability of future occurrences. The mathematical expression for the memoryless property is:

• \( P(X > s + t \mid X > s) = P(X > t) \)

Where:

• \( s \) and \( t \) are time periods.

This can be somewhat counter-intuitive at first. Still, it's essential in fields involving waiting times and reliability, such as telecommunications and healthcare. In the original exercise, the memoryless property was used to calculate the probability of a stay longer than 10 days, given a 7-day stay. By reframing the question, using memoryless behavior, the problem simplifies down to calculating the probability of a future duration, starting anew. This focuses analyses more on the current situation than the overall history.
Such insights help in strategically planning for unpredictable scenarios without over-dependence on past data.

Probability Density Function (PDF)

The Probability Density Function (PDF) is an essential concept that describes the likelihood of a continuous random variable to take on a particular value. For the exponential distribution, the PDF is given by:

• \( f(x) = \frac{1}{\theta} e^{-x/\theta} \)

Here, \( \theta \) represents the mean time between events, while \( e \) denotes the base of the natural log.
This function isn't about exact probabilities for individual points but rather helps to depict how probability is distributed over continuous intervals. The area under the curve of the PDF across a range of values provides the actual probabilities, commonly visualized with graphs.
In the original exercise, understanding the PDF helps identify how patient hospital stays fit into a broader pattern and mean duration, calculated as 7.0 days. By analyzing how the density function changes, healthcare providers can estimate how hospital resources will be used over time.
In summary, the PDF is a fundamental tool for modeling and predicting patterns in data, providing a deeper understanding of how random variables behave under certain conditions.