Problem 179 - Applied Statistics and Probability for Engineers

Question

Suppose that the length of stay (in hours) at a hospital emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4\). Determine the following in parts (a) and (b): (a) Mean and variance (b) \(P(X<8)\) (c) Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.

Step-by-Step Solution

Answer

(a) Mean is 4.85, variance is 3.77. (b) Probability is 0.926. (c) Lognormal distribution has no negative probabilities; normal does.

Step 1: Mean of Lognormal Distribution

For a lognormal random variable, the mean is given by the formula:\[ \mu = e^{\theta + \frac{\omega^2}{2}} \]Substitute \( \theta = 1.5 \) and \( \omega = 0.4 \):\[ \mu = e^{1.5 + \frac{0.4^2}{2}} = e^{1.5 + 0.08} = e^{1.58} \approx 4.85 \]So, the mean is approximately 4.85 hours.

Step 2: Variance of Lognormal Distribution

The variance of a lognormal distribution is given by:\[ \sigma^2 = (e^{\omega^2} - 1) e^{2\theta + \omega^2} \]Substitute \( \theta = 1.5 \) and \( \omega = 0.4 \):\[ \sigma^2 = (e^{0.4^2} - 1) e^{2(1.5) + 0.4^2} = (e^{0.16} - 1) e^{3.08} \approx 0.1732 \times 21.75 \approx 3.77 \]The variance is approximately 3.77 hours.

Step 3: Probability Calculation for P(X<8)

To find \( P(X<8) \), we first convert the problem to the standard normal distribution. We use the property that if \( X \) is lognormal, then \( \ln(X) \) is normal with mean \( \theta \) and variance \( \omega^2 \).\[ P(X < 8) = P(\ln(X) < \ln(8)) \]\[ \ln(8) \approx 2.079 \]The standardized version is \( Z = \frac{\ln(X) - \theta}{\omega} \), hence:\[ P\left( Z < \frac{2.079 - 1.5}{0.4} \right) = P(Z < 1.4475) \]Check standard normal distribution tables, \( P(Z < 1.4475) \approx 0.926 \).

Step 4: Discussion of P(X<0) for Lognormal and Normal

In a lognormal distribution, \( X \) cannot be less than 0 because it models positive values only (since \( e^Y > 0 \) where \( Y \) is normal). Therefore, \( P(X<0) = 0 \) for the lognormal distribution. Conversely, for a normal distribution, there's a non-zero probability for values less than 0 because a normal distribution covers the entire real line. For the same mean and variance, the lognormal distribution more realistically models scenarios where the outcome cannot be negative.

Key Concepts

Mean and Variance of a Lognormal Distribution

The mean and variance of a lognormal distribution are different from those of a normal distribution due to the transformation involved. In a lognormal distribution, the variable is the exponential of a normally distributed variable, resulting in positive skewness.
Let's dive into each one separately:**Mean**: - The mean is calculated using the formula: \[ \mu = e^{\theta + \frac{\omega^2}{2}} \]- Here, \( \theta \) and \( \omega \) represent the parameters of the underlying normal distribution.- For our example, substituting \( \theta = 1.5 \) and \( \omega = 0.4 \), we get: \[ \mu = e^{1.5 + \frac{0.4^2}{2}} = e^{1.58} \approx 4.85 \]- This tells us that the average length of stay is approximately 4.85 hours.**Variance**:- The variance is found using: \[ \sigma^2 = (e^{\omega^2} - 1) e^{2\theta + \omega^2} \]- For the given parameters: \[ \sigma^2 = (e^{0.16} - 1) e^{3.08} \approx 0.1732 \times 21.75 \approx 3.77 \]- This means there is a spread of about 3.77 hours squared in the length of stay.
The mean and variance here indicate how data is spread in a lognormal scenario, taking into account the exponential nature of the distribution.

Probability Calculation in a Lognormal Distribution

Understanding probability calculations in a lognormal distribution involves transforming the problem to a normal context. Since the natural logarithm of a lognormal variable is normally distributed, we can leverage this property to simplify calculations.To compute the probability, \( P(X<8) \):- We start with the inequality \( X < 8 \), which is transformed using the natural logarithm to \( \ln(X) < \ln(8) \).- Solving \( \ln(8) \) gives approximately 2.079.Next, adjust to a standard normal variable using the formula for \( Z \):- \[ Z = \frac{\ln(X) - \theta}{\omega} \]- Plugging in our values gives: \[ P\left( Z < \frac{2.079 - 1.5}{0.4} \right) = P(Z < 1.4475) \]- Looking this up in the standard normal distribution table gives approximately 0.926.This means there's a 92.6% chance that the length of stay is less than 8 hours. This transformation and simplification are standard procedures when dealing with lognormal distributions.

Comparing Lognormal and Normal Distributions

When we compare lognormal and normal distributions, it is essential to recognize their capabilities and limitations. The two distributions are inherently different due to their structural properties.In a **lognormal distribution**:- All values of \( X \) must be positive because it is derived by exponentiating a normal variable.- Thus, \( P(X<0) \) is exactly 0.
This property makes the lognormal distribution ideal for modeling naturally positive data.In contrast, the **normal distribution**:- Extends over the entire real line, meaning it includes both negative and positive values.- Hence, even if the mean and variance are the same, \( P(X<0) \) is non-zero since negative values are possible.This technical difference crucially affects applications. For situations like modeling stays in a hospital, which can't be negative, the lognormal is the better fit, capturing the natural constraint of positivity. This serves as a reminder of choosing the correct model for real-world situations based on their theoretical foundation.