Problem 167 - Applied Statistics and Probability for Engineers

Question

An article in Proceeding of the \(33 \mathrm{rd}\) International ACM SIGIR Conference on Research and Development in Information Retrieval ["Understanding Web Browsing Behaviors Through Weibull Analysis of Dwell Time" \((2010,\) p. \(3791-\) 386 ) ] proposed that a Weibull distribution can be used to model Web page dwell time (the length of time a Web visitor spends on a Web page). For a specific Web page, the shape and scale parameters are 1 and 300 seconds, respectively. Determine the following: (a) Mean and variance of dwell time (b) Probability that a Web user spends more than four minutes on this Web page (c) Dwell time exceeded with probability 0.25

Step-by-Step Solution

Answer

(a) Mean: 300 seconds; Variance: 90000 seconds². (b) Probability: 0.449. (c) Time: approximately 415.89 seconds.

Step 1: Understanding the Weibull Parameters

The Weibull distribution is specified by two parameters: the shape parameter (\( k \)) and the scale parameter (\( \lambda \)). Given in the question, the shape parameter is 1 and the scale parameter is 300 seconds. When the shape parameter is 1, it characterizes the Exponential distribution.

Step 2: Calculating Mean of the Weibull Distribution

The mean of a Weibull distribution is given by the formula: \( \lambda \Gamma(1 + 1/k) \), where \( \Gamma \) is the Gamma function. Since the shape parameter \( k = 1 \), the distribution is exponential, and \( \Gamma(2) = 1! = 1 \). Therefore, the mean is \( 300 \times 1 = 300 \) seconds.

Step 3: Calculating Variance of the Weibull Distribution

The variance of a Weibull distribution is given by the formula: \( \lambda^2 \left[ \Gamma(1 + 2/k) - (\Gamma(1 + 1/k))^2 \right] \). For the exponential case where \( k = 1 \), \( \Gamma(3) = 2 \) and \( \Gamma(2)^2 = 1^2 = 1 \). Thus, the variance is \( 300^2 \times (2 - 1) = 90000 \) seconds^2.

Step 4: Converting Time to Match Distribution

For part (b), the task is to calculate the probability of spending more than four minutes on the web page. First, convert four minutes to seconds: \( 4 \times 60 = 240 \) seconds.

Step 5: Calculating Probability for More Than 240 Seconds

The survival function for the Weibull distribution (probability of a time being greater than \( t \)) is given by: \( S(t) = e^{-(t/\lambda)^k} \). With \( k = 1 \) and \( \lambda = 300 \), the formula becomes \( S(240) = e^{-(240/300)^1} = e^{-0.8} \). This gives a probability of approximately 0.449.

Step 6: Finding Time with Specific Probability

To find the time \( t \) exceeded with probability 0.25, we use the formula for the survival function: \( S(t) = e^{-(t/\lambda)^k} = 0.25 \). Solving for \( t \) with \( k = 1 \) and \( \lambda = 300 \), we have \( e^{-t/300} = 0.25 \). Taking the natural logarithm, \( -t/300 = \ln(0.25) \). Solving for \( t \), we find \( t = -300 \ln(0.25) \approx 415.89 \) seconds.

Key Concepts

Probability Calculations

Probability calculations are essential in understanding behaviors that follow a Weibull distribution. In general, probabilities help us determine the likelihood of certain events occurring. For our specific case, we want to calculate the probability that a web user spends more than a certain amount of time, such as four minutes, on a webpage.

This is done using the survival function, which represents the probability of exceeding a certain time, denoted as \( S(t) = e^{-(t/\lambda)^k} \). Here, \( t \) stands for the time in seconds, \( \lambda \) is the scale parameter, and \( k \) is the shape parameter. In our exercise with a Weibull distribution having \( \lambda = 300 \) seconds and \( k = 1 \), the distribution is exponential, simplifying calculations.

For our example, to find the probability of spending more than 240 seconds on a webpage (since 4 minutes equals 240 seconds), we substitute these values into the survival function: \( S(240) = e^{-(240/300)^1} = e^{-0.8} \). This results in a probability of approximately 0.449, indicating a 44.9% chance of a web session exceeding four minutes.

Mean and Variance of Distributions

The mean and variance of a distribution provide vital information about its central behavior and variability. For the Weibull distribution, these metrics help us understand the average dwell time on a page and its variation.

The mean of a Weibull distribution is calculated using the formula \( \lambda \Gamma(1 + 1/k) \). With our shape parameter \( k = 1 \) and scale parameter \( \lambda = 300 \), the equation simplifies to \( 300 \times 1 = 300 \) seconds, using the fact that \( \Gamma(2) = 1 \). This means, on average, a user spends 300 seconds on this particular web page.

The variance formula for the Weibull distribution is \( \lambda^2 [ \Gamma(1 + 2/k) - (\Gamma(1 + 1/k))^2 ] \). Again, for \( k = 1 \), it becomes \( 300^2 \times (2 - 1) = 90000 \) seconds squared. Variance tells us about the spread of the dwell times around the mean of 300 seconds, indicating the expected fluctuations.

Survival Analysis

Survival analysis is a statistical method used to analyze the expected duration until one or more events happen, like a user exiting a webpage. This analysis is crucial for understanding web user retention and session lengths. The Weibull distribution is highly effective in modeling these scenarios due to its flexibility in handling different shapes of survival curves.

In our exercise, we used survival analysis to calculate the time exceeded with a probability of 0.25. This involves finding a "cut-off" time, \( t \), such that there is a 25% chance that a user's session will last longer than \( t \).

We set up the survival function: \( S(t) = e^{-(t/300)} = 0.25 \). By solving for \( t \), we get \( -t/300 = \ln(0.25) \). Therefore, \( t = -300 \ln(0.25) \approx 415.89 \) seconds. This analysis tells us that about 25% of users will stay on the page for longer than approximately 416 seconds, helpful for assessing user engagement and optimizing web content.