Chapter 4: Problem 138
Suppose that \(X\) has a lognormal distribution with parameters \(\theta=5\) and \(\omega^{2}=9\). Determine the following: (a) \(P(X<13,300)\) (b) The value for \(x\) such that \(P(X \leq x)=0.95\) (c) The mean and variance of \(X\)
Short Answer
Expert verified
(a) 0.9332; (b) 20,634; (c) Mean is 13,399.93, Variance is 1.446 x 10^12.
Step by step solution
01
Identify distribution properties
Since we have a lognormal distribution with parameters \(\theta = 5\) and \(\omega^2 = 9\), this means the natural logarithm of \(X\), say \(Y = \ln X\), follows a normal distribution with mean \(\theta = 5\) and variance \(\omega^2 = 9\). Therefore, \(Y \sim N(5, 3)\) because standard deviation \(\omega = \sqrt{9} = 3\).
02
Convert to normal distribution for part (a)
To find \(P(X < 13,300)\), we translate this into the normal distribution: \(P(\ln X < \ln 13,300)\). Calculate \(\ln 13,300\) and standardize: \(Z = \frac{\ln 13,300 - 5}{3}\).
03
Calculating \(Z\) for part (a)
Calculate \(\ln 13,300 \approx 9.4972\). Thus, \(Z = \frac{9.4972 - 5}{3} = 1.4991\). Use the standard normal distribution table to find \(P(Z < 1.4991)\).
04
Standard Normal Table Lookup
From the standard normal distribution table, \(P(Z < 1.4991) \approx 0.9332\). Thus, \(P(X < 13,300) \approx 0.9332\).
05
Find quantile for part (b)
To find \(x\) such that \(P(X \leq x) = 0.95\), first find z-value for \(P(Z \leq z) = 0.95\) which is about 1.645.
06
Back transformation to find \(x\)
Convert \(z = 1.645\) back to the \(Y\) scale: \(y = \theta + \omega \times z = 5 + 3 \times 1.645 = 9.935\). Exponentiate to return to \(X\) scale: \(x = e^{9.935} \approx 20,634\).
07
Mean of lognormal distribution
For a lognormal distribution, the mean \(\mu_X = e^{\theta + \frac{\omega^2}{2}} = e^{5 + \frac{9}{2}} = e^{9.5}\). Calculate \(\mu_X \approx 13,399.93\).
08
Variance of lognormal distribution
The variance \(\sigma_X^2 = (e^{\omega^2} - 1)e^{2\theta + \omega^2} = (e^9 - 1)e^{19}\). Estimate \(e^9 \approx 8103.0839\) and \(e^{19} \approx 1.78482 \times 10^8\). Therefore, \(\sigma_X^2 \approx (8102.0839) (1.78482 \times 10^8) \approx 1.446 \times 10^{12}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Distribution
A normal distribution is one of the most common and well-known statistical distributions. It is described by a bell-shaped curve, often referred to as the Gaussian distribution. The main characteristics of a normal distribution include:
- Symmetrical shape around its mean.
- Total area under the curve equals 1, representing total probability.
- Densely clustered values around the mean, with fewer occurrences further away.
Probability Calculation
Calculating probabilities under a distribution curve is key to understanding statistical patterns. With a normal distribution, we often calculate probabilities by "standardizing" a variable. This involves transforming it into a standard normal distribution with mean 0 and variance 1 using a z-score.
To determine \(P(X < 13,300)\) for our lognormal distribution, we first compute \(P(\ln X < \ln 13,300)\). The value \(\ln 13,300\) serves as the threshold when transformed to the normal scale. This produces a z-score: \[Z = \frac{\ln 13,300 - 5}{3}\].After standardizing, the probability is easily obtained using a standard normal distribution table or technology tools, which yields \(P(Z < 1.4991) \approx 0.9332\). This result highlights that there's about a 93.32% chance that the lognormal variable X is less than 13,300.
To determine \(P(X < 13,300)\) for our lognormal distribution, we first compute \(P(\ln X < \ln 13,300)\). The value \(\ln 13,300\) serves as the threshold when transformed to the normal scale. This produces a z-score: \[Z = \frac{\ln 13,300 - 5}{3}\].After standardizing, the probability is easily obtained using a standard normal distribution table or technology tools, which yields \(P(Z < 1.4991) \approx 0.9332\). This result highlights that there's about a 93.32% chance that the lognormal variable X is less than 13,300.
Statistical Distribution Properties
Understanding the properties of distributions aids in interpreting the behavior of random variables. A lognormal distribution like the one in this problem is asymmetrical, differing from the normal distribution's symmetry. Its shape is commonly right-skewed due to the exponential nature of its development from a normally distributed variable.
- The parameter \(\theta\) affects the location, akin to the mean in the transformed normal distribution.
- The parameter \(\omega^2\) determines the spread or variation.
- Unlike a linear transformation, the back-transformation to linear scale increases skewness.
Mean and Variance Calculation
Determining the mean and variance of a lognormal distribution requires specific transformations due to its exponential nature.
For this exercise, the mean of X, denoted as \(\mu_X\), is calculated using the formula \(\mu_X = e^{\theta + \frac{\omega^2}{2}}\). Plugging in \(\theta = 5\) and \(\omega^2 = 9\), we obtain \(\mu_X = e^{9.5}\), providing an approximate mean of 13,399.93.
The variance, on the other hand, is derived with the formula \(\sigma_X^2 = (e^{\omega^2} - 1)e^{2\theta + \omega^2}\). This combines both parameters to determine spread on a logarithmic scale. Calculation methods reveal a substantial variance: roughly 1.446 x 10^{12}. These values highlight how unique parameters substantially adjust the mean and variance of lognormal compared to normal distributions.
For this exercise, the mean of X, denoted as \(\mu_X\), is calculated using the formula \(\mu_X = e^{\theta + \frac{\omega^2}{2}}\). Plugging in \(\theta = 5\) and \(\omega^2 = 9\), we obtain \(\mu_X = e^{9.5}\), providing an approximate mean of 13,399.93.
The variance, on the other hand, is derived with the formula \(\sigma_X^2 = (e^{\omega^2} - 1)e^{2\theta + \omega^2}\). This combines both parameters to determine spread on a logarithmic scale. Calculation methods reveal a substantial variance: roughly 1.446 x 10^{12}. These values highlight how unique parameters substantially adjust the mean and variance of lognormal compared to normal distributions.