Problem 12 - Calculus Concepts

Question

ER Arrivals The time \(t,\) in minutes, between successive arrivals at an emergency room of a certain city hospital on Saturday nights can be described by the function \(E(t)=0.2 e^{-0.2 t}\) when \(t \geq 0 .\) Two patients arrive at the emergency room every 10 minutes. a. Calculate the probability that successive arrivals are between 20 and 30 minutes apart. b. Calculate the probability that 10 minutes or less elapses between successive arrivals. c. Calculate the probability that successive arrivals will be more than 15 minutes apart.

Step-by-Step Solution

Answer

a. \(0.01582\), b. \(0.8647\), c. \(0.0498\)

Step 1: Understand the density function

We are given the function \( E(t) = 0.2 e^{-0.2 t} \), which describes the probability density for the time \( t \) between arrivals. This means the time between arrivals follows an exponential distribution with rate parameter \( \lambda = 0.2 \).

Step 2: Determine the cumulative distribution function (CDF)

The cumulative distribution function \( F(t) \) of an exponential distribution with rate \( \lambda \) is given by \( F(t) = 1 - e^{-\lambda t} \). Substituting \( \lambda = 0.2 \), we have \( F(t) = 1 - e^{-0.2t} \).

Step 3: Probability of arrivals between 20 and 30 minutes apart

Calculate \( P(20 < t < 30) \) using the CDF: \[ P(20 < t < 30) = F(30) - F(20) \]. Substitute the CDF values: \( F(30) = 1 - e^{-0.2 \times 30} \) and \( F(20) = 1 - e^{-0.2 \times 20} \).

Step 4: Compute F(30) and F(20)

Calculate \( F(30) = 1 - e^{-6} \) and \( F(20) = 1 - e^{-4} \). Use a calculator to find: \( F(30) \approx 1 - 0.00248 \) and \( F(20) \approx 1 - 0.0183 \).

Step 5: Substitute and find the probability

Now we have \( P(20 < t < 30) = (1 - 0.00248) - (1 - 0.0183) \). This simplifies to \( 0.01582 \).

Step 6: Probability of 10 minutes or less

Use the CDF to find \( P(t \leq 10) = F(10) = 1 - e^{-0.2 \times 10} \). Calculate \( F(10) = 1 - e^{-2} \approx 1 - 0.1353 \). Thus, \( P(t \leq 10) \approx 0.8647 \).

Step 7: Probability of more than 15 minutes

To find \( P(t > 15) \), use the complementary probability: \( P(t > 15) = 1 - F(15) = e^{-0.2 \times 15} \). Compute \( e^{-3} \approx 0.0498 \).

Key Concepts

Probability Density Function

The probability density function (PDF) is a crucial concept in understanding exponential distributions. For this exercise, the function given is \( E(t) = 0.2 e^{-0.2 t} \). Here, \( E(t) \) represents how the time \( t \) between emergency room arrivals is distributed. The PDF is used to find the likelihood of a random variable—in our case, time \( t \)—taking on a particular value.

In an exponential distribution, the rate parameter \( \lambda \) dictates how quickly events occur, with larger values indicating more frequent occurrences. For our scenario, the rate \( \lambda = 0.2 \) means arrivals at the hospital follow a particular pace, and this rate helps inform calculations of probabilities for specific time intervals.

Understanding the PDF provides insights into event likelihood, but to find probabilities over an interval—such as between specific minutes—you'll typically use the cumulative distribution function, which we'll explore next.

Cumulative Distribution Function

The cumulative distribution function (CDF) is all about accumulation. It helps us determine the total probability that a random variable is less than or equal to a certain value. In mathematical terms, for the exponential distribution, it's defined as:\[ F(t) = 1 - e^{-\lambda t} \]This formula accumulates the probability from zero up to each point \( t \).

For example, to find the probability of arrivals being between 20 and 30 minutes apart, you calculate \( F(30) - F(20) \). When \( t = 30 \), \( F(30) = 1 - e^{-6} \), and for \( t = 20 \), \( F(20) = 1 - e^{-4} \). These results allow you to see the probability of the time falling within that specific range.

The beauty of the CDF lies in its efficiency. Instead of considering each tiny interval separately, you use a single mathematical expression to cover wide ranges, giving a broad overview of probabilities over time.

Complementary Probability

Complementary probability is a neat trick to quickly find the probability of the opposite event occurring. In our example, we want the probability of more than 15 minutes elapsing between arrivals. Instead of directly calculating it, you use the complementary probability of \( F(15) \), which is the probability of 15 minutes or less.

This concept hinges on the simple idea: the sum of probabilities for all possible outcomes is 1. Therefore, the probability of anything not occurring is 1 minus the probability of it occurring. For example:\[ P(t > 15) = 1 - F(15) = e^{-3} \]Here, \( e^{-3} \) gives the probability of more than 15 minutes, elegantly utilizing the exponential decay characteristic of this type of distribution.

Utilizing complementary probabilities simplifies solving exponential distribution problems, making it easy to find elusive outcomes by focusing on their opposites.