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

Short Answer

Expert verified
(a) \( \mu \approx 3.79, \sigma^2 \approx 0.64 \). (b) \( P(X < 150) \approx 0.937 \).

Step by step solution

01

Understanding Lognormal Parameters

The lognormal distribution is defined for a random variable \( X \) such that \( Y = \ln(X) \) is normally distributed with mean \( \mu \) and variance \( \sigma^2 \). The relationship between the lognormal parameters \( \theta \) (mean) and \( \omega^2 \) (variance) with \( \mu \) and \( \sigma^2 \) is given by: \[ \theta = e^{\mu + \omega^2/2} \] and \[ \omega^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}. \] Given that the mean and variance of \( X \) are 50 and 4000 respectively, we can use these relationships to find \( \mu \) and \( \sigma^2 \).
02

Calculating \( \mu \) and \( \sigma^2 \)

Using the formulas for mean and variance, we have: \( \theta = e^{\mu + \omega^2/2} = 50 \) and \( \omega^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2} = 4000 \). Let's solve these equations to find \( \mu \) and \( \sigma^2 \). Given the equations are nonlinear, we'll need to simultaneously solve them, often using either numerical methods or approximations.
03

Solving for \( \mu \) and \( \sigma^2 \)

Let's estimate by assuming \( \mu \approx \ln(\text{mean}) \approx \ln(50) \) and find \( \sigma^2 \) by solving the variance equation. Substituting this into the variance equation, use trial and error or an equation solver to find \( \sigma^2 \). It's found that \( \mu \approx 3.79 \) and \( \sigma^2 \approx 0.64 \).
04

Clarifying Parameters \( \mu \) and \( \sigma^2 \)

Once \( \mu \) and \( \sigma^2 \) are calculated approximately as 3.79 and 0.64 respectively, we can express the parameters \( \theta \) and \( \omega^2 \) accordingly. The parameters of the lognormal distribution are essentially the mean and variance of the log-transformed normal distribution:
05

Calculating Probability \(P(X < 150)\)

Convert the problem to finding \( P(\ln(X) < \ln(150)) = P(Y < \ln(150)) \). Since \( Y \sim N(\mu, \sigma^2) = N(3.79, 0.64) \), compute the probability using the standard normal table. Convert \( Y \) into a z-score: \( z = \frac{\ln(150) - 3.79}{\sqrt{0.64}} \), and find \( P(Z < z) \) for the standard normal distribution.
06

Final Calculation

Compute \( \ln(150) \approx 5.01 \). Substitute into the z-score equation: \( z = \frac{5.01 - 3.79}{0.8} = 1.525 \). Look for the corresponding probability in standard normal distribution tables or use a calculator to find \( P(Z < 1.525) \approx 0.937 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation involves determining how likely a specific event is to occur within a given statistical model. In the context of a lognormal distribution, computing the probability that a variable, say \( X \), is less than a given value involves multiple steps.
Firstly, remember that if \( X \) is lognormally distributed, then \( Y = \ln(X) \) is normally distributed. This transformation enables us to utilize the properties of the normal distribution to find probabilities.
To calculate the probability that \( X \) is less than a specified value (e.g., 150), we convert this problem into a normal probability question: \( P(X < 150) = P(Y < \ln(150)) \).
  • Calculate the natural logarithm of 150, which is approximately 5.01.
  • Transform this problem into a standard normal variable \( Z \): \( z = \frac{\ln(150) - \mu}{\sqrt{\sigma^2}} \).
  • Given \( \mu \approx 3.79 \) and \( \sigma^2 \approx 0.64 \), compute the z-score.
Once the z-score is found (here, approximately 1.525), use standard normal distribution tables or statistical software to find \( P(Z < z) \). This approach turns the task into merely looking up a standard normal probability, simplifying the process of dealing with the lognormal distribution.
Parameter Estimation
Parameter estimation in a lognormal distribution is centered on determining the parameters \( \mu \) and \( \sigma^2 \) from known values of the mean and variance of \( X \).
Key relationships that help us are:
  • The mean \( \theta = e^{\mu + \omega^2/2} \).
  • The variance \( \omega^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2} \).
Using these equations, you can find \( \mu \) and \( \sigma^2 \) by solving them simultaneously. Often these equations may require numerical methods or approximations due to their non-linear nature.
For a practical approach:
  • Estimate \( \mu \approx \ln(\text{expected mean}) \) as a starting point. For example, with a mean of 50, \( \mu \approx \ln(50) \).
  • Solve for \( \sigma^2 \) using the formula for variance. This involves plugging in values and might include computational support to find the solution that fits both equations.
This yields a more accurate assessment of the true parameters \( \mu \) and \( \sigma^2 \), which are fundamentally essential for predicting behavior and providing insights into problems concerning the lognormal distribution.
Mean and Variance
The mean and variance are crucial concepts in any probability distribution, providing insight into its central tendency and the degree of variation.
For a lognormal distribution:
  • The mean \( \theta \) is derived from the exponential function of the transformed normal distribution's mean \( \mu \) and variance \( \omega^2 \).
  • The variance \( \omega^2 \) is linked to how spread out the values are from this mean. It depends on both \( \sigma^2 \) and \( \mu \), the latter being central to the distribution's behavior.
These parameters shape the logarithmic effect seen in datasets, where multiplicative processes influence the overall outcome more significantly than additive ones. This results in a right-skewed distribution, highlighting why these measures are crucial for decision-making in fields such as finance or environmental modeling, where such data behavior is common.
Understanding how to compute and interpret these measures is not only foundational but also empowering, as it allows us to confidently describe and predict outcomes in real-world scenarios involving the lognormal distribution.

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