/*! 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 24 Assume the number of episodes pe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume the number of episodes per year of otitis media, a common disease of the middle ear in early childhood, follows a Poisson distribution with parameter \(\lambda=1.6\) episodes per year. Find the probability of getting 3 or more episodes of otitis media in the first 2 years of life.

Short Answer

Expert verified
The probability of having 3 or more episodes is approximately 0.6199.

Step by step solution

01

Understand the Poisson Distribution

The Poisson distribution is used to model the number of times an event occurs within a given interval of time or space, with a known average rate \( \lambda \) and independence between occurrences. Here, \( \lambda = 1.6 \) episodes per year for otitis media.
02

Adjust Parameters for a 2-Year Period

Since the problem involves the first 2 years of life, we need to calculate the parameter \( \lambda \) over this period: \( \lambda_{2\text{ years}} = 1.6 \times 2 = 3.2 \).
03

Define Probability of 3 or More Episodes

We need to find the probability of having 3 or more episodes, which is the complement of having fewer than 3 episodes. Mathematically, this can be expressed as \( P(X \geq 3) = 1 - P(X < 3) \).
04

Calculate Probability of Fewer Than 3 Episodes

We compute \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \) using the Poisson probability mass function: \( P(X = k) = \frac{{e^{-\lambda} \lambda^k}}{k!} \).
05

Compute Individual Probabilities

Using \( \lambda = 3.2 \), compute each component:\( P(X = 0) = \frac{e^{-3.2} 3.2^0}{0!} \approx 0.0408 \)\( P(X = 1) = \frac{e^{-3.2} 3.2^1}{1!} \approx 0.1305 \)\( P(X = 2) = \frac{e^{-3.2} 3.2^2}{2!} \approx 0.2088 \).
06

Sum Probabilities for 0, 1, and 2 Episodes

Add the probabilities: \( P(X < 3) = 0.0408 + 0.1305 + 0.2088 = 0.3801 \).
07

Compute Complement Probability for 3 or More Episodes

Finally, compute \( P(X \geq 3) = 1 - 0.3801 = 0.6199 \).

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.

Probability Calculation
Probability calculation is a mathematical technique used to determine the likelihood of an event occurring. In the context of the Poisson distribution, we use probability to evaluate how many times an event, such as an episode of otitis media, occurs within a fixed period.

The probability of an event using the Poisson distribution is calculated using the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \], where \( \lambda \) represents the average number of occurrences, and \( k \) is the number of occurrences we are interested in. For instance, if you want to calculate the probability of a child having 0, 1, or 2 episodes of otitis media, you would plug these values into the formula to find each probability.

By summing these individual probabilities, you can determine the cumulative probability for having fewer episodes, and subsequently find the complement to get the likelihood of having more episodes.
Otitis Media
Otitis media is a common infection of the middle ear, particularly in young children. It is characterized by pain, fever, and sometimes fluid in the ear, which can lead to hearing problems if untreated. Understanding the likelihood of occurrence, like through modeling with a Poisson distribution, can help parents and healthcare providers anticipate and manage the condition effectively.

Since otitis media episodes can happen multiple times within a given year, it serves as a good example for using the Poisson distribution in probabilistic modeling. By using statistical techniques, we can understand the expected number of episodes and better plan for medical interventions or parental guidance.

This type of prediction can assist in preparing resources in healthcare settings and helping with early diagnosis and treatment.
Parameter Adjustment
Parameter adjustment is necessary when the time period in question changes, and it involves altering the parameter \( \lambda \) to reflect this new interval. For a Poisson distribution problem that originally presents \( \lambda = 1.6 \) episodes per year, and now considers a 2-year span, you would adjust \( \lambda \) by multiplying it by the number of years in interest.

Thus, for a 2-year period: \[ \lambda_{2\text{ years}} = 1.6 \times 2 = 3.2 \].

This adjusted parameter reflects the mean number of otitis media episodes expected over the longer period. Properly adjusting the parameter ensures that the probability calculations remain accurate and relevant for the scenario being examined.
Cumulative Probability
Cumulative probability helps in determining the probability of an event occurring at most up to a certain number of times. Using the Poisson distribution, one can sum the probabilities for all occurrences up to a specific maximum to obtain a cumulative probability.

In our example, we find the cumulative probability for having less than 3 episodes by summing the probabilities of having 0, 1, and 2 episodes: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \].

Once these probabilities are calculated and added together, the result is used to calculate the probability of the complementary event––which in this instance is having 3 or more episodes, ensuring a comprehensive understanding of all possible outcomes.
Independent Occurrences
The concept of independent occurrences is central to understanding the Poisson distribution. Events are independent if the occurrence of one does not affect the probability of another occurring. This is true for many processes modeled by the Poisson distribution, such as the random occurrences of otitis media episodes.

This independence means each episode arises without influencing or being influenced by others, making the model appropriate for predicting episodes over time.

Such independence simplifies the calculation of probabilities because each event is considered separately, with its occurrence solely determined by the average rate \( \lambda \), rather than any previous or future events.

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

The two-stage model of carcinogenesis is based on the premise that for a cancer to develop, a normal cell must first undergo a "first hit" and mutate to become a susceptible or intermediate cell. An intermediate cell then must undergo a "second hit" and mutate to become a malignant cell. A cancer develops if at least one cell becomes a malignant cell. This model has been applied to the development of breast cancer in females (Moolgavkar et al. \([15])\) Suppose there are \(10^{8}\) normal breast cells and 0 intermediate or malignant breast cells among 20 -year-old females. The probability that a normal breast cell will mutate to become an intermediate cell is \(10^{-7}\) per year. What is the probability that there will be at least 5 intermediate cells by age 21 ? (Hint: Use the Poisson distribution.)

One interesting phenomenon of bacteriuria is that there is a "turnover"; that is, if bacteriuria is measured on the same woman at two different points in time, the results are not necessarily the same. Assume that \(20 \%\) of all women who are bacteriuric at time 0 are again bacteriuric at time 1 (1 year later), whereas only 4.2\% of women who were not bacteriuric at time 0 are bacteriuric at time 1 . Let \(X\) be the random variable representing the number of bacteriuric events over the two time periods for 1 woman and still assume that the probability that a woman will be positive for bacteriuria at any one exam is \(5 \%\) What is the mean of \(X ?\)

A study was conducted among 234 people who had expressed a desire to stop smoking but who had not yet stopped. On the day they quit smoking, their carbonmonoxide level (CO) was measured and the time was noted from the time they smoked their last cigarette to the time of the CO measurement. The CO level provides an "objective" indicator of the number of cigarettes smoked per day during the time immediately before the quit attempt. However, it is known to also be influenced by the time since the last cigarette was smoked. Thus, this time is provided as well as a "corrected CO level," which is adjusted for the time since the last cigarette was smoked. Information is also provided on the age and sex of the participants as well as each participant's self-report of the number of cigarettes smoked per day. The participants were followed for 1 year for the purpose of determining the number of days they remained abstinent. Number of days abstinent ranged from 0 for those who quit for less than 1 day to 365 for those who were abstinent for the full year. Assume all people were followed for the entire year. (TABLE CAN NOT COPY) Develop life tables for subsets of the data based on age, gender, number of cigarettes per day, and CO level (one variable at a time). Given these data, do you feel age, gender, number of cigarettes per day, and/or CO level are related to success in quitting? (Methods of analysis for lifetable data are discussed in more detail in Chapter \(14 .\) )

An experiment is designed to test the potency of a drug on 20 rats. Previous animal studies have shown that a \(10-\mathrm{mg}\) dose of the drug is lethal \(5 \%\) of the time within the first 4 hours; of the animals alive at 4 hours, \(10 \%\) will die in the next 4 hours. What is the probability that 3 or more rats will die in the first 4 hours?

Suppose 6 of 15 students in a grade-school class develop influenza, whereas \(20 \%\) of grade-school students nationwide develop influenza. Is there evidence of an excessive number of cases in the class? That is, what is the probability of obtaining at least 6 cases in this class if the nationwide rate holds true?

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.