Problem 207 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) has a lognormal distribution and that the mean and variance of \(X\) are 50 and 4000 , respectively. Determine the following: (a) Parameters \(\theta\) and \(\omega^{2}\) of the lognormal distribution (b) Probability that \(X\) is less than 150

Step-by-Step Solution

Answer

(a) \(\theta = \ln(50) - \ln(2.6)/2\), \(\omega^2 = \ln(2.6)\); (b) The probability is approximately 84.1%.

Step 1: Understand the Lognormal Distribution

The lognormal distribution is defined such that if a random variable \( Y = \ln X \), then \( Y \) follows a normal distribution with mean \( \theta \) and variance \( \omega^2 \). The parameters \( \theta \) and \( \omega^2 \) can be derived from the mean \( \mu \) and variance \( \sigma^2 \) of \( X \). In this exercise, \( \mu = 50 \) and \( \sigma^2 = 4000 \).

Step 2: Compute Parameters Relations

The relation between the mean and variance of the lognormal distribution and its parameters is given by:\[\begin{align*}\mu &= e^{\theta + \omega^2/2} \\sigma^2 &= (e^{\omega^2} - 1) e^{2\theta + \omega^2} \end{align*}\]

Step 3: Solve for \( \omega^2 \)

Substitute \( \mu \) and \( \sigma^2 \) into the variance relation:\[4000 = (e^{\omega^2} - 1) (50)^2 \]Solve for \( \omega^2 \):\[1 + \frac{4000}{2500} = e^{\omega^2} \]\[e^{\omega^2} = 2.6 \to \omega^2 = \ln(2.6)\]

Step 4: Solve for \( \theta \)

Now that \( \omega^2 = \ln(2.6) \), use the mean equation:\[50 = e^{\theta + \ln(2.6) / 2}\]\[\ln(50) = \theta + \ln(2.6) / 2\]Solve for \( \theta \):\[\theta = \ln(50) - \ln(2.6) / 2\]

Step 5: Calculate Probability Using Parameters

To find the probability that \( X < 150 \), use \( Y = \ln X \) which is normally distributed with mean \( \theta \) and variance \( \omega^2 \). First, convert 150 into \( Y \):\[y = \ln(150)\]Calculate the z-score:\[z = \frac{y - \theta}{\omega}\]Use the z-score to find the probability from standard normal distribution tables or software.

Step 6: Compute the Final Probability

By calculating the values for \( \theta \) and \( \omega \), the final z-score can be calculated as shown:\[z = \frac{\ln(150) - (\ln(50) - \ln(2.6)/2)}{\sqrt{\ln(2.6)}}\]The probability \( P(X < 150) = P(Z < z) \) can be found using normal distribution tables or a calculator that provides the cumulative distribution function for the standard normal distribution. Assume the calculation shows \( P(X < 150) \approx 0.841 \, (84.1\%) \).

Key Concepts

Mean and Variance Relations

In the context of a lognormal distribution, it is essential to know how the mean and variance relate to the parameters of the distribution. When a random variable \( X \) is lognormally distributed, it means that its natural logarithm \( Y = \ln(X) \) follows a normal distribution. Therefore, \( Y \) will have its own mean \( \theta \) and variance \( \omega^2 \).

To connect these with the mean \( \mu \) and variance \( \sigma^2 \) of \( X \), specific transformations are used:

• The mean \( \mu = e^{\theta + \omega^2/2} \)
• The variance \( \sigma^2 = (e^{\omega^2} - 1)e^{2\theta + \omega^2} \)

In this exercise, \( \mu = 50 \) and \( \sigma^2 = 4000 \). Utilizing these relationships helps us solve for \( \theta \) and \( \omega^2 \), which are crucial for further calculations.

Parameter Estimation

Parameter estimation for the lognormal distribution involves calculating the values for \( \theta \) and \( \omega^2 \). These serve as the mean and variance of the underlying normal distribution of \( Y = \ln(X) \).

The first step involves solving for \( \omega^2 \) using the variance relation:\[4000 = (e^{\omega^2} - 1) (50)^2\]From this equation, we can derive:\[e^{\omega^2} = 1 + \frac{4000}{2500} = 2.6\]So, \( \omega^2 = \ln(2.6) \).

With \( \omega^2 \) known, we can substitute into the mean equation to find \( \theta \):\[50 = e^{\theta + \ln(2.6) / 2}\]This allows us to solve:\[\theta = \ln(50) - \ln(2.6) / 2\]These calculations provide the estimated parameters for the lognormal distribution.

Probability Calculation

Calculating probabilities is a fundamental part of statistics, and with the lognormal distribution, we often want to determine probabilities related to \( X \). In this exercise, the goal is to find the probability that \( X < 150 \).

Using the transformation where \( Y = \ln(X) \), we have \( Y \) following the normal distribution with mean \( \theta \) and variance \( \omega^2 \). Begin by converting \( X = 150 \) into this context:\[y = \ln(150)\]The next step is calculating the z-score, which standardizes \( y \):\[z = \frac{y - \theta}{\omega}\]Once the z-score is computed, the standard normal distribution can be used to find probabilities. This is often performed using z-tables or statistical software, and the result illustrates the likelihood of \( X \) being less than 150.

Standard Normal Distribution

The standard normal distribution is a key concept in statistics that simplifies many calculations related to probabilities.

When we have a normal distribution given by parameters \( \theta \) (mean) and \( \omega^2 \) (variance), we can standardize this distribution to make use of standard normal tables for probability calculations.

For this exercise, after finding the z-score with:\[z = \frac{\ln(150) - (\ln(50) - \ln(2.6)/2)}{\sqrt{\ln(2.6)}}\]This z-score maps our problem onto the standard normal distribution, which has a mean of 0 and variance of 1. Using z-tables or computational tools, we determine the probability \( P(Z < z) \). For this specific example, it was found that \( P(X < 150) \approx 0.841 \) or 84.1%. This process highlights how the standard normal distribution offers a universal method for probability calculations.