Problem 173 - Applied Statistics and Probability for Engineers

Question

The length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters \(\theta=0.5\) and \(\omega^{2}=1\). (a) What is the probability that a page is viewed for more than 10 seconds? (b) By what length of time have \(50 \%\) of the users moved to another page? (c) What are the mean and standard deviation of the time until a user moves from the page?

Step-by-Step Solution

Answer

(a) P(T > 10) ≈ 0.0357; (b) Median is 1.6487 seconds; (c) Mean = 2.7183, Std Dev ≈ 3.313.

Step 1: Understanding the Lognormal Distribution

A lognormal random variable is of the form \( T = e^X \), where \( X \) is normally distributed. Given: \( \theta = 0.5 \) and \( \omega^2 = 1 \). Here, we will use the properties of the lognormal distribution, where \( X \sim N(\theta, \omega^2) \).

Step 2: Transformations and Parameter Identification

For a lognormal distribution, if \( T = e^X \), where \( X \sim N(\mu, \sigma^2) \), then the parameters of \( X \) are \( \mu = \theta = 0.5 \) and \( \sigma^2 = \omega^2 = 1 \).

Step 3: Solving Part A Using a Z-score

The question asks for \( P(T > 10) \). First we convert to the normal variable: if \( T > 10 \), \( \ln(T) > \ln(10) \). We standardize this to find \( P\left( \frac{\ln(10) - \mu}{\sigma} > Z \right) \), where \( Z \) is the standard normal variable. Here, \( \ln(10) = 2.302 \), \( \mu = 0.5 \), \( \sigma = 1 \), so we find the Z-score as \( Z > \frac{2.302 - 0.5}{1} = 1.802 \). Use the standard normal distribution table or calculator to find \( P(Z > 1.802) \).

Step 4: Calculating Cumulative Probability for Part B

For part (b), we need the median of the lognormal distribution, which is given by the transformation of the median of \( X \), \( e^{\mu} = e^{0.5} \). So, the time by which 50% of users have moved is \( \approx 1.6487 \) seconds (from \( e^{0.5} \) calculation).

Step 5: Calculating Mean and Standard Deviation for Part C

The mean of a lognormal distribution is \( e^{\mu + \frac{\sigma^2}{2}} = e^{0.5 + 0.5} = e^1 = 2.7183 \). The standard deviation is \( \sqrt{(e^{\sigma^2} - 1) \cdot e^{2\mu + \sigma^2}} = \sqrt{(e^1 - 1) \cdot e^{1.5}} = \sqrt{1.7183 \cdot 4.4817} \approx 3.313 \).

Key Concepts

Probability Calculation

The probability calculation in the context of a lognormal distribution, such as the one given for the webpage viewing time problem, involves transforming our lognormal variable to a normal one. Since the lognormal distribution is defined as the exponentiation of a normal distribution (i.e., if \( T \) is lognormally distributed, \( T = e^X \), where \( X \) is normally distributed), we work with the natural logarithm to perform our calculations.

To find the probability that the page is viewed for more than 10 seconds, we calculate \( P(T > 10) \). First, we transform this into \( X \) by applying the logarithm: \( \ln(T) > \ln(10) \). This gives us \( \ln(10) = 2.302 \).

Next, we standardize this to use normal distribution tables or a calculator. The Z-score is computed as \( Z = \frac{\ln(10) - \mu}{\sigma} \), where \( \mu = 0.5 \) and \( \sigma = 1 \). Plug in the values to get \( Z > \frac{2.302 - 0.5}{1} = 1.802 \). Use standard normal distribution tables to find \( P(Z > 1.802) \). This yields the probability that a page is viewed for more than 10 seconds.

Median of Lognormal Distribution

The median of a lognormal distribution is a straightforward calculation because it comes directly from the median of the normally distributed variable \( X \). The relationship is:\[\text{Median of lognormal } = e^{\mu}.\]

Because the median of \( X \sim N(\mu, \sigma^2) \) is \( \mu \), exponentiating \( \mu \) gives us the median of \( T \). In this exercise, \( \mu = 0.5 \).

Therefore, the median of the lognormal distribution, the point by which 50% of users move to another page, is calculated as \( e^{0.5} \). This results in approximately 1.6487 seconds. This means that half of the users spend this amount of time or less on the page.

Mean and Standard Deviation Calculation

The mean and standard deviation for a lognormal distribution require a bit more computation, using the parameters \( \mu \) and \( \sigma^2 \). For the mean of a lognormal distribution, the formula is:
\[\text{Mean } = e^{\mu + \frac{\sigma^2}{2}}.\]
For the given values, we have \( \mu = 0.5 \) and \( \sigma^2 = 1 \). Substitute these in:
\( e^{0.5 + 0.5} = e^1 \approx 2.7183 \).

The standard deviation requires more detailed computation with the formula:
\[\text{Standard Deviation} = \sqrt{(e^{\sigma^2} - 1) \cdot e^{2\mu + \sigma^2}}.\]

For the given parameters (i.e., \( \sigma^2 = 1 \) and \( \mu = 0.5 \)):
\( \sqrt{(e^1 - 1) \cdot e^{1.5}} \approx \sqrt{1.7183 \times 4.4817} \approx 3.313 \).

Here, you first compute \( e^{\sigma^2} - 1 = e^1 - 1 = 1.7183 \) and \( e^{1.5} = 4.4817 \), then use them to calculate the standard deviation. This information helps describe how spread out the viewing times are around the mean, within this lognormal distribution.