/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Suppose that the length of stay ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the length of stay (in hours) at a hospital emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4 .\) Determine the following in parts (a) and (b): a. Mean and variance b. \(P(X<8)\) c. Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.

Short Answer

Expert verified
a. Mean: \( e^{1.58} \), Variance: \( (e^{0.16} - 1)e^{3.16} \); b. Use the z-score: \( Z = \frac{\log(8) - 1.5}{0.4} \); c. \( P(X<0)=0 \) for lognormal, non-zero for normal.

Step by step solution

01

Understanding Lognormal Distribution

A lognormal random variable is defined as a random variable whose logarithm is normally distributed. It is characterized by a scale parameter \(\theta\) and a shape parameter \(\omega\), where \(\theta\) describes the location of the distribution and \(\omega\) represents the spread.
02

Calculating the Mean

The mean of a lognormal distribution is given by the formula: \( E(X) = e^{\theta + \frac{\omega^2}{2}} \). Substitute the given values: \( \theta = 1.5 \) and \( \omega = 0.4 \), compute: \[E(X) = e^{1.5 + \frac{0.4^2}{2}} = e^{1.5 + 0.08} = e^{1.58}.\]
03

Calculating the Variance

The variance of a lognormal distribution is given by: \( \text{Var}(X) = (e^{\omega^2} - 1)e^{2\theta + \omega^2} \). Substitute the values: \( \theta = 1.5 \) and \( \omega = 0.4 \), compute: \[\text{Var}(X) = (e^{0.16} - 1)e^{3 + 0.16} = (e^{0.16} - 1)e^{3.16}.\]
04

Calculating Probability \( P(X

To find \( P(X<8) \), first find the z-score of 8 in terms of a normal distribution of \( \log(X) \): \[Z = \frac{\log(8) - 1.5}{0.4}.\] Use the standard normal distribution table or a calculator to find \( P(Z < \text{calculated value})\).
05

Interpretation of Probability Results

Since a lognormal distribution cannot take negative values (because it is the exponential of a normal distribution), \( P(X<0) \) for a lognormal distribution is 0. In contrast, for a normal distribution with the same mean and variance there will be a non-zero probability \( P(X<0) \) because normal distributions include all real numbers.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Random Variable
The concept of a random variable is pivotal in understanding probability distributions such as the lognormal distribution. A random variable is essentially a variable whose values depend on the outcomes of a random phenomenon. In simpler terms, it is a way to map outcomes of a probability event to numbers.
In the context of the exercise, the length of stay at a hospital emergency department is represented by a random variable, denoted as \(X\). This variable follows a specific probability distribution, in our case, a lognormal distribution.
This means that while the natural logarithm of this variable is normally distributed, the variable itself exhibits a lognormal distribution. This distinction is crucial as it affects how we compute probabilities and characteristics such as mean and variance.
A good way to grasp random variables is to consider them as a blend of randomness and structure, where each outcome of a random event maps to a numerical result, often represented by a function.
Mean and Variance
In any probability distribution, understanding the mean and variance helps in grasping the overall behavior of the random variable.
For a lognormal distribution, the mean is calculated using the formula \(E(X) = e^{\theta + \frac{\omega^2}{2}}\), where \(\theta\) and \(\omega\) are the location and scale parameters, respectively. In our exercise, these values are given as \(\theta = 1.5\) and \(\omega = 0.4\). By substituting these into the formula, we compute the mean as \(e^{1.58}\). This mean represents the central tendency or the expected value of the hospital stay.
Variance, on the other hand, offers insight into the spread and variability of the data. The variance formula for a lognormal distribution is \(\text{Var}(X) = (e^{\omega^2} - 1)e^{2\theta + \omega^2}\). By calculating this using the same parameter values, we gain understanding into how much the durations might vary from the mean.
Both mean and variance are essential as they help in predicting average outcomes and understanding the degree of variation or uncertainty in those outcomes.
Probability Calculations
Calculating probabilities associated with a random variable gives us the likelihood of specific events occurring.
In the case of a lognormal distribution like ours, probabilities are determined considering the normal distribution of the logarithm of the random variable. For instance, to find the probability \(P(X < 8)\), we transform the problem by relating \(\log(X)\) to a normal distribution: \[Z = \frac{\log(8) - \theta}{\omega}\] where \(\theta = 1.5\) and \(\omega = 0.4\). This approach essentially standardizes \(\log(X)\) and allows us to employ the standard normal distribution to find \(P(Z < \text{calculated value})\).
It's also important to consider the behavior of a lognormal distribution at boundaries. Since it is derived from an exponential transformation, \(P(X < 0)\) is always zero, unlike its normal counterpart which may have non-zero probabilities even for negative numbers due its continuous range. Understanding these nuances enhances comprehension of how probabilities under a lognormal distribution are determined.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Given the probability density function \(f(x)=\) \(0.01^{3} x^{2} e^{-0.01 x} / \Gamma(3),\) determine the mean and variance of the distribution.

The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes. a. What is the probability that there are no calls within a 30-minute interval? b. What is the probability that at least one call arrives within a 10 -minute interval? c. What is the probability that the first call arrives within 5 to 10 minutes after opening? d. Determine the length of an interval of time such that the probability of at least one call in the interval is \(0.90 .\)

A show is scheduled to start at 9: 00 A.M., 9: 30 A.M., and 10: 00 A.M. Once the show starts, the gate will be closed. A visitor will arrive at the gate at a time uniformly distributed between 8: 30 A.M. and 10: 00 A.M. Determine the following: a. Cumulative distribution function of the time (in minutes) between arrival and 8: 30 A.M. b. Mean and variance of the distribution in the previous part c. Probability that a visitor waits less than 10 minutes for a show d. Probability that a visitor waits more than 20 minutes for a show

Suppose \(X\) has a continuous uniform distribution over the interval \([-1,1] .\) Determine the following: a. Mean, variance, and standard deviation of \(X\) b. Value for \(x\) such that \(P(-x

Suppose that the probability density function of the length of computer cables is f(x) = 0.1 from 1200 to 1210 millimeters. a. Determine the mean and standard deviation of the cable length. b. If the length specifications are 1195 < x < 1205 millimeters, what proportion of cables is within specifications?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.