Problem 132 - Applied Statistics and Probability for Engineers

Question

The length of stay at a specific emergency department in a hospital in Phoenix, Arizona had a mean of 4.6 hours. Assume that the length of stay is exponentially distributed. (a) What is the standard deviation of the length of stay? (b) What is the probability of a length of stay of more than 10 hours? (c) What length of stay is exceeded by \(25 \%\) of the visits?

Step-by-Step Solution

Answer

(a) The standard deviation is 4.6 hours. (b) The probability is approximately 11.34%. (c) 25% of visits exceed 6.38 hours.

Step 1: Understand Exponential Distribution

The exponential distribution is characterized by the parameter \( \lambda \), which is the rate parameter. The mean \( \mu \) of an exponential distribution is given by \( \mu = \frac{1}{\lambda} \). In this exercise, the mean is 4.6 hours.

Step 2: Calculate the Rate Parameter \( \lambda \)

Given the mean \( \mu = 4.6 \), we can find the rate parameter \( \lambda \) using \( \lambda = \frac{1}{\mu} \). Thus, \( \lambda = \frac{1}{4.6} \approx 0.2174 \text{ per hour}.\)

Step 3: Determine the Standard Deviation

The standard deviation for an exponential distribution is given by \( \sigma = \frac{1}{\lambda} \), which is equal to the mean. Therefore, the standard deviation is also 4.6 hours.

Step 4: Calculate Probability for Stay Longer Than 10 Hours

The probability that the length of stay \( X \) is greater than 10 hours is given by \( P(X > 10) = 1 - P(X \leq 10) \). According to the exponential distribution, \( P(X \leq 10) = 1 - e^{-\lambda \times 10} \). Compute \( 1 - e^{-0.2174 \times 10} \).

Step 5: Compute Probability Value

Calculate \( P(X > 10) = e^{-0.2174 \times 10} \approx e^{-2.174} \approx 0.1134 \). Thus, the probability is approximately 0.1134 or 11.34%.

Step 6: Find the 25th Percentile of Stay Length

To find the length of stay exceeded by \( 25\% \) of the visits, we need to find \( x \) such that \( P(X > x) = 0.25 \) or equivalently \( P(X \leq x) = 0.75 \). Use the exponential CDF: \( P(X \leq x) = 1 - e^{-\lambda x} = 0.75 \).

Step 7: Solve for \( x \) in the CDF Equation

Rearrange the CDF equation to \( e^{-\lambda x} = 0.25 \). Taking natural logs gives \( -\lambda x = \ln(0.25) \). Solving for \( x \), we get \( x = -\frac{\ln(0.25)}{\lambda} \approx \frac{1.386}{0.2174} \approx 6.38 \text{ hours}.\)

Key Concepts

Mean and Standard Deviation

In an exponential distribution, the **mean** and the **standard deviation** are directly linked. This makes it unique compared to other distributions. The mean is denoted by \( \mu \), and it represents the average value we expect from our dataset. For our exercise, it is given as 4.6 hours for the length of stay in the emergency department.

The intriguing part of an exponential distribution is that its standard deviation \( \sigma \) is the same as the mean. Therefore, once you have the mean, you automatically know the standard deviation as well.

Understanding this can simplify many calculations and make it easier to work with exponential data. For instance, knowing the mean of 4.6 hours immediately tells us that the standard deviation is also 4.6 hours.

Probability Calculation

Calculating probabilities with an exponential distribution involves understanding the nature of continuous probability. When we want to find the probability of a data point, like a length of stay, exceeding a certain value, we use the cumulative distribution function (CDF).

For this exercise, determining the probability of a stay longer than 10 hours involves calculating \( P(X > 10) \) in an exponential distribution. This is done using:

• The rate parameter \( \lambda \), which we found earlier to be approximately 0.2174 per hour.
• The formula \( P(X > x) = e^{-\lambda x} \).

Substitute into the formula: \( P(X > 10) = e^{-0.2174 \times 10} \approx e^{-2.174} \approx 0.1134 \).

This tells us there is about an 11.34% probability that a patient will stay longer than 10 hours. Understanding this calculation helps in predicting outcomes in real-world situations where events like these follow an exponential pattern.

Percentile Calculation

Percentile calculations in exponential distributions allow us to understand which data points represent specific slices of our data. A percentile is a value below which a certain percentage of data falls. For this problem, we are interested in the value which is exceeded by \( 25\% \) of the visits.

This means we seek the 75th percentile (since \( 0.75 = 1 - 0.25 \)), calculated using:

• The exponential CDF: \( P(X \leq x) = 1 - e^{-\lambda x} \).
• Solving for \( x \) when \( P(X \leq x) = 0.75 \).

Substituting and rearranging gives \( e^{-\lambda x} = 0.25 \).
Taking logs and solving for \( x \), we get \( x = -\frac{\ln(0.25)}{\lambda} \).
Applying our values leads to \( x \approx \frac{1.386}{0.2174} \approx 6.38 \) hours.

This calculation shows that approximately 6.38 hours is the cut-off point where 25% of the visit lengths will be shorter, providing essential insights for hospital staffing and planning.