Problem 43 - Calculus and its applications

Question

The wait times at an urgent care center are exponentially distributed. There is a \(30 \%\) probability that a patient will have to wait up to 1 hr to see a doctor. a) Find \(k,\) and then write the probability density function \(f\) b) Find the probability that a patient will have to wait between \(90 \mathrm{~min}\) and \(3 \mathrm{hr}\) for a doctor.

Step-by-Step Solution

Answer

a) \( k \approx 0.3567, \quad f(x) = 0.3567 e^{-0.3567x} \) b) Probability is \( \approx 0.2445 \).

Step 1: Understanding the Exponential Distribution

When the wait times are exponentially distributed, the probability density function (pdf) is given by \( f(x) = k e^{-kx} \), where \( k \) is the rate parameter. The cumulative distribution function (CDF) for an exponential distribution is \( F(x) = 1 - e^{-kx} \).

Step 2: Use Given Probability to Find k

We know the probability that a patient waits up to 1 hour is \( 0.3 \), or \( P(X \leq 1) = 0.3 \). Using the CDF, this translates to \( 1 - e^{-k \times 1} = 0.3 \). Solving for \( k \), we have \( e^{-k} = 0.7 \). Take the natural logarithm of both sides to find \( k \):\[-k = \ln(0.7) \implies k = -\ln(0.7) \approx 0.3567\].

Step 3: Define the Probability Density Function

Now, plug \( k = 0.3567 \) into the pdf: \( f(x) = 0.3567 e^{-0.3567x} \). This is the probability density function of the wait times.

Step 4: Convert Time Units for Part b

Convert the wait times from minutes to the same units as the pdf. \( 90 \text{ min} = 1.5 \text{ hr} \) and \( 3 \text{ hr} = 3 \text{ hr} \). We are looking for \( P(1.5 \leq X \leq 3) \).

Step 5: Evaluate the Cumulative Distribution Function

Calculate \( P(X \leq 3) \) and \( P(X \leq 1.5) \) using the CDF:\[ P(X \leq 3) = 1 - e^{-0.3567 \times 3} \approx 0.6957 \] \[ P(X \leq 1.5) = 1 - e^{-0.3567 \times 1.5} \approx 0.4512 \].

Step 6: Calculate the Desired Probability

The probability that a patient waits between 1.5 hours and 3 hours is given by the difference:\[ P(1.5 \leq X \leq 3) = P(X \leq 3) - P(X \leq 1.5) = 0.6957 - 0.4512 \approx 0.2445 \].

Key Concepts

Probability Density Function

The Probability Density Function, or PDF, is a fundamental concept when working with continuous probability distributions like the Exponential Distribution. Instead of giving probabilities for individual outcomes (like in a discrete distribution), the PDF describes the relative likelihood of a random variable falling within a particular range of values in a continuous sample space. For the Exponential Distribution, the PDF is expressed as: \[ f(x) = k e^{-kx} \] In this formula, "x" represents the value of the random variable (in our case, wait times), and "k" is the rate parameter, a critical component which we will discuss more in another section. It is key to note that:

• The PDF is always non-negative, ensuring that every possible outcome's likelihood has a valid, non-negative expression.
• Integrating the PDF over the entire space equals 1, reflecting the certainty that some value within the range must occur. In simpler terms, if you add up all the possibilities, they sum to certainty (100%).

Understanding PDFs allows us to visualize and calculate the probabilities for continuous variables such as wait times at an urgent care center. Here, the PDF helps model how likely it is that a patient will wait a specific amount of time to see the doctor.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) complements the PDF by focusing on the probability that a random variable is less than or equal to a certain value. For the Exponential Distribution, the CDF is given by the equation: \[ F(x) = 1 - e^{-kx} \] The CDF answers questions like, "What is the probability that a random event occurs by time x?" This function starts at 0 (for no chance of occurring at the start) and increases to 1, mirroring complete certainty that the event has occurred by the end of the distribution. It essentially sums up probabilities described by the PDF from the start to point x. Here’s a simple breakdown of its features:

• In probabilistic terms, it represents a running total of probabilities, adding up likely outcomes as you progress through the range of x-values.
• Utilizing the CDF helps us answer cumulative questions, which is vital for solving problems related to waiting times where we look for intervals like the number of people who wait less than 1 hour.

This is particularly useful in problems like calculating how many patients may be seen within an hour at a hospital, or determining the likelihood that a patient will need to wait between 1.5 to 3 hours.

Rate Parameter

The Rate Parameter, often denoted as "k", is a key part of the Exponential Distribution. It defines the distribution's rate, affecting the shape and spread of the distribution. In simpler terms, "k" directly influences how "fast" events occur, such as how quickly patients might be seen at a clinic. Finding the right rate parameter is essential for accurately modeling data. In our example, the calculation when using the CDF was used to find "k" like this: Given a 30% probability that a patient waits at most 1 hour, we used: \[ 1 - e^{-k imes 1} = 0.3 \]Solving this equation gave us: \[ k = -\ln(0.7) \approx 0.3567 \] Key points related to the rate parameter include:

• A high value of "k" suggests events happen quickly, indicating a short waiting period.
• A low value of "k" suggests events occur slowly, indicating longer waits.
• Understanding "k" helps in making predictions and decisions. In an emergency department, knowing the correct rate can help manage resources and anticipations for patient flow.

The rate parameter is crucial for capturing the dynamics of wait times in a healthcare setting or any other scenario where time until an event occurs is critical.