Problem 213 - Applied Statistics and Probability for Engineers

Question

The waiting time for service at a hospital emergency department follows an exponential distribution with a mean of three hours. Determine the following: (a) Waiting time is greater than four hours (b) Waiting time is greater than six hours given that you have already waited two hours (c) Value \(x\) (in hours) exceeded with probability 0.25

Step-by-Step Solution

Answer

(a) 0.2636; (b) 0.2636; (c) 4.16 hours.

Step 1: Understand the Exponential Distribution

The problem states that the waiting time follows an exponential distribution with a mean of three hours. In an exponential distribution, the probability density function (PDF) is given by \( f(x) = \lambda e^{-\lambda x} \), where \( \lambda = \frac{1}{\text{mean}} \). For this problem, \( \lambda = \frac{1}{3} \). The cumulative distribution function (CDF) is \( F(x) = 1 - e^{-\lambda x} \).

Step 2: Calculate Probability for Waiting Time Greater Than Four Hours

To find the probability that waiting time is greater than 4 hours, calculate \( P(X > 4) = 1 - F(4) \), where \( F(4) \) is the CDF at \( x = 4 \):\[ F(4) = 1 - e^{-\frac{1}{3} \times 4} = 1 - e^{-\frac{4}{3}} \] Therefore, \( P(X > 4) = e^{-\frac{4}{3}} \).

Step 3: Calculate Conditional Probability for Waiting Time Greater Than Six Hours

The conditional probability that the waiting time exceeds 6 hours given that it has already exceeded 2 hours is given by:\[ P(X > 6 \mid X > 2) = \frac{P(X > 6)}{P(X > 2)} \]We previously calculated \(P(X > 2) = e^{-\frac{2}{3}} \) and \( P(X > 6) = e^{-2} \). Substitute these into the formula:\[ P(X > 6 \mid X > 2) = \frac{e^{-2}}{e^{-\frac{2}{3}}} = e^{-\frac{4}{3}} \]

Step 4: Determine Value x Exceeded with Probability 0.25

To find the value of \( x \) that is exceeded with probability 0.25, we solve for \( x \) in the equation:\[ P(X > x) = 0.25 \Rightarrow 1 - F(x) = 0.25 \Rightarrow F(x) = 0.75 \]Substitute the expression for \( F(x) \):\[ 1 - e^{-\frac{x}{3}} = 0.75 \Rightarrow e^{-\frac{x}{3}} = 0.25 \]Taking the natural logarithm, solve for \( x \):\[ -\frac{x}{3} = \ln(0.25) \Rightarrow x = -3 \ln(0.25) \]

Step 5: Calculate Numerical Values

Now, compute the values numerically:- \( e^{-\frac{4}{3}} \approx 0.2636 \), so the probability that the waiting time is greater than 4 hours is approximately 0.2636.- \( e^{-\frac{2}{3}} \approx 0.5134 \) and \( e^{-2} \approx 0.1353 \), thus \( P(X > 6 \mid X > 2) \approx 0.2636 \).- \( x = -3 \ln(0.25) \approx 4.1589 \), so the value exceeded with probability 0.25 is approximately 4.16 hours.

Key Concepts

Probability Density Function

In an exponential distribution, the concept of a probability density function (PDF) is essential to understanding how it models random events that happen continuously and independently at a constant average rate. The formula for the PDF of an exponential distribution is:

• \( f(x) = \lambda e^{-\lambda x} \)

In this equation, \( \lambda \) is a rate parameter, calculated as the reciprocal of the mean of the distribution. For example, if an event like waiting time has an average of 3 hours, \( \lambda \) would be \( \frac{1}{3} \) per hour.

This rate parameter influences how fast or slow an event is likely to happen. The PDF shows the likelihood of a different specific outcome within the sample space. It is important for determining probabilities of continuous random variables, giving us the probability per unit on the horizontal axis.

Understanding the PDF provides insights into the characteristic "memorylessness" property of exponential distributions. This means that the probability of an event occurring in the upcoming time phase is the same, regardless of the past.

Cumulative Distribution Function

The cumulative distribution function (CDF) complements the PDF by representing the probability that a random variable is less than or equal to a certain value, \( x \). For an exponential distribution, the CDF can be expressed as:

• \( F(x) = 1 - e^{-\lambda x} \)

Here, \( F(x) \) accumulates the probabilities from the start (zero) up to some point \( x \). So, if you have a mean waiting time of 3 hours, \( \lambda = \frac{1}{3} \), then plugging in specific values for \( x \) tells us the cumulative probability up till that time.

For example, using the CDF, you can find the probability that the waiting time is less than or equal to four hours. Calculating \( F(4) \) gives you an idea of how long you might expect to wait on average at a hospital emergency department, up to a specific cutoff.

The CDF is crucial for calculating probabilities over intervals, providing a more complete picture by accumulating probabilities across the range of interest.

Conditional Probability

Conditional probability is a concept that allows us to find the probability of an event happening, given that another event has already occurred. Under exponential distributions, this is straightforward due to their "memorylessness" feature. The conditional probability for the exponential distribution is given by:

• \( P(X > a \mid X > b) = \frac{P(X > a)}{P(X > b)} \)

This formula illustrates how we can update our probability calculations under new conditions. In the context of waiting times, you might calculate the probability of waiting more than 6 hours, given that you've already waited 2 hours.

Memorylessness simplifies calculations, as past events don't influence future probabilities beyond the initial conditions. That means the wait time distribution beyond those 2 hours behaves the same as if you had just started waiting, making this approach efficient and helpful in predicting ongoing processes.

In practice, this type of probability is valuable for decisions in real-time scenarios, where understanding and updating probabilities quickly is crucial.

Exceedance Probability

The exceedance probability is the likelihood that a random variable is greater than a specific threshold value. In an exponential distribution, finding this probability can help answer questions like, "What value will be exceeded 25% of the time?" This is calculated by solving for \( x \) in:

• \( P(X > x) = 0.25 \)

Using the CDF, which connects directly to exceedance probabilities, we convert it into:

• \( 1 - F(x) = 0.25 \)

From here, algebraically solve for \( x \):
\[ e^{-\frac{x}{3}} = 0.25 \]
Taking logarithms helps determine \( x \), indicating this is where 25% of the waiting scenarios surpass.

Exceedance probability answers are significant in planning and preparation. Knowing such thresholds ensures that systems can be managed more efficiently by preparing for outliers. This kind of insight assists stakeholders in making data-driven decisions, anticipating events occurring less frequently, and ensuring resources are allocated effectively.