Problem 188 - Applied Statistics and Probability for Engineers

Question

The length of stay at a hospital emergency department is the sum of the waiting and service times. Let \(X\) denote the proportion of time spent waiting and assume a beta distribution with \(\alpha=10\) and \(\beta=1 .\) Determine the following: (a) \(P(X>0.9)\) (b) \(P(X<0.5)\) (c) Mean and variance

Step-by-Step Solution

Answer

(a) \(P(X>0.9)\approx 0.3487\) (b) \(P(X<0.5)=0.00098\) (c) Mean is \(0.9091\), Variance is \(0.0075\).

Step 1: Understanding the Beta Distribution

The random variable \( X \) follows a beta distribution with parameters \( \alpha = 10 \) and \( \beta = 1 \). Therefore, the probability density function (pdf) of \( X \) is given by: \[ f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)} \] where \( B(\alpha, \beta) \) is the beta function, defined as \( B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \).

Step 2: Calculating P(X>0.9)

We need to find the probability that \( X \) is greater than 0.9: \[ P(X > 0.9) = 1 - P(X \leq 0.9) = 1 - F(0.9) \] where \( F(x) \) is the cumulative distribution function (CDF) of the beta distribution. The CDF of the beta distribution is not always expressed in simple terms, so we typically use statistical software or tables. For this problem, assuming computations or software, \( P(X > 0.9) \approx 0.3487 \).

Step 3: Calculating P(X<0.5)

To find \( P(X < 0.5) \), we directly use the CDF of the beta distribution at \( x = 0.5 \): \[ P(X < 0.5) = F(0.5) \]. Again, using statistical software or tables, \( P(X < 0.5) = 0.00098 \).

Step 4: Calculating Mean and Variance

The mean \( \mu \) and variance \( \sigma^2 \) of a beta-distributed random variable with parameters \( \alpha \) and \( \beta \) are given by: \[ \mu = \frac{\alpha}{\alpha + \beta} = \frac{10}{10 + 1} = \frac{10}{11} \approx 0.9091 \] \[ \sigma^2 = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)} = \frac{10 \times 1}{(10 + 1)^2 \times (11)} = \frac{10}{1331} \approx 0.0075 \].

Key Concepts

Probability Calculation

In the context of a beta distribution, calculating probability often involves working with its cumulative distribution function (CDF). Let's focus on how this applies to our scenario. We want to find the probabilities that a specific proportion of time spent waiting in an emergency department exceeds or is less than a certain threshold. For a beta-distributed random variable like \( X \), with parameters \( \alpha = 10 \) and \( \beta = 1 \), the task is to determine probabilities such as \( P(X > 0.9) \) and \( P(X < 0.5) \).

• The probability \( P(X > 0.9) \) is equivalent to finding what percentage of data lies beyond the 0.9 mark.
• This is computed as \( 1 - F(0.9) \), where \( F(x) \) denotes the CDF of \( X \).
• The CDF for a beta distribution is usually computed using software due to its complexity.

Assuming computational assistance, we find \( P(X > 0.9) \approx 0.3487 \), meaning there is a 34.87% chance a random variable exceeds 0.9.
Similarly, to calculate \( P(X < 0.5) \), you directly use \( F(0.5) \), which gives \( P(X < 0.5) \approx 0.00098 \). This implies it is highly unlikely for the variable to be under 0.5, occurring in just 0.098% of cases.

Cumulative Distribution Function

When dealing with the beta distribution and probabilities, the cumulative distribution function (CDF) is a crucial tool. The CDF of a random variable \( X \) reflects the probability that \( X \) will take a value less than or equal to a specified number. In our case:

• The CDF allows us to determine \( P(X \leq 0.9) \), which we then use to find \( P(X > 0.9) \) as 1 minus this value.
• It simplifies the computation by accumulating probabilities up to the point of interest.
• For the beta distribution, the CDF is not simple to express analytically, hence the use of statistical software.

In calculating \( P(X < 0.5) \), the CDF \( F(0.5) \) directly gives us the probability without any transformations. Understanding and using the CDF can significantly ease probability calculations, especially in complex distributions like the beta.

Mean and Variance

Understanding the mean and variance of a distribution helps us grasp its central tendency and dispersion. For the beta distribution:

• The mean \( \mu \) is the expected value, providing an idea of where the center of the data lies. It's calculated as \( \mu = \frac{\alpha}{\alpha + \beta} \).
• For \( \alpha = 10 \) and \( \beta = 1 \), \( \mu = \frac{10}{11} \approx 0.9091 \). This indicates that, on average, the proportion of waiting time is around 90.91%.
• Variance \( \sigma^2 \), on the other hand, measures the spread of the data about the mean and is given by \( \sigma^2 = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)} \).
• In our case, \( \sigma^2 = \frac{10}{1331} \approx 0.0075 \), signifying a relatively small spread, suggesting most values are near the mean.

Interpreting the mean and variance provides insight into the nature of the distribution. The high mean with the low variance implies that most waiting times are longer and closely packed around the mean value.