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Aerosolized vaccine for measles. An aerosolized vaccine for measles was developed in Mexico and has been used on more than 4 million childiren since 1980. Aerosolized vaccines have the advantages of being able to be administered by people without clinical training and do not cause injectionassociated infections. The percentage of children developing an immune response to measles after receiving the subcutaneous injection of the vaccine is \(95 \%\), and for those receiving the aerosolized vaccine, it is \(85 \%\). 9 There are 20 children to be vaccinated for measles using the aerosolized vaccine in a small rural village in India. We are going to count the number who developed an immune response to measles after vaccination. (a) Explain why this is a binomial setting. (b) What is the probability that at least one child does not develop an immune response to measles after receiving the aerosolized vaccine? What would be the probability that at least one child does not develop an immune response to measles if all children were vaccinated using the subcutaneous injection of the vaccine?

Short Answer

Expert verified
This is a binomial setting. The probability for aerosolized is 0.9641 and for subcutaneous is 0.6407.

Step by step solution

01

Identifying Binomial Setting

In probability theory, a binomial setting is defined when there are more than one independent trials, each trial has only two possible outcomes (success or failure), and the probability of success is the same for each trial. Here, we are vaccinating 20 children with the aerosolized vaccine, each trial (child) has two outcomes (immune response/no immune response), and each child has an 85% chance of developing an immune response. Thus, this is a binomial setting.
02

Calculating Probability for Aerosolized Vaccine

First, we need to determine the probability that at least one child does not develop an immune response. This can be calculated by first finding the probability that all children develop an immune response, and then subtracting this from 1. The probability of all 20 children developing an immune response is given by \((0.85)^{20}\). Then subtract from 1 to find the probability that at least one child does not develop an immune response: \(1 - (0.85)^{20}\).
03

Calculating Probability for Subcutaneous Vaccine

For the subcutaneous vaccine, apply the same approach. The probability of developing an immune response for all 20 children is \((0.95)^{20}\). So, the probability that at least one child does not develop an immune response is \(1 - (0.95)^{20}\).

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Key Concepts

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

Probability Theory
Probability theory is a mathematical framework used to predict the likelihood of events. It is essential for statistical analysis and everyday decision-making. At its core, it involves understanding and calculating chances, which are expressed as probabilities.
In this exercise, probability theory is used to analyze the outcomes of a vaccination study. Each child's response to the vaccine represents a separate probability trial. The concept we are dealing with is a **binomial distribution**. This means that the situation involves several trials (vaccinations), each with two possible outcomes, like success (develop immune response) or failure (no immune response). Here, success is defined by the child developing an immune response.
  • Independent Trials: Each child's immune response is independent of the others.
  • Two Possible Outcomes: Success (immune response) or failure (no immune response).
  • Constant Probability: The probability of success is constant for each trial (85% for aerosolized, 95% for subcutaneous).
This binomial setting allows us to calculate the likelihood of different outcomes using the binomial probability formula. This is crucial in understanding how many children will develop immunity.
Vaccination Study
In this scenario, the vaccination study focuses on the use of an aerosolized versus a subcutaneous vaccination for measles. These vaccines aim to trigger an immune response in children, protecting them from the disease.
Aerosolized vaccines offer several benefits:
  • Easy Administration: They can be given by individuals without clinical training.
  • Safer Delivery: They avoid risks like needle-related infections.
However, the study highlights that only 85% of children develop an immune response compared to the 95% success rate of the subcutaneous vaccine.
This difference poses significant implications for public health strategies, especially in remote areas. By analyzing the vaccination study through probability, we can better understand the trade-offs between ease of administration and effectiveness, allowing for data-driven decisions regarding vaccination programs.
Statistical Setting
The statistical setting of the problem is essential to understanding its analysis. Here, we use statistical methods to examine the vaccination's impact on the children's immune response.
The study employs a binomial distribution, which is a powerful statistical tool for analyzing experiments where outcomes fall into two categories. With 20 trials (children), each representing a chance of success (immune response) or failure (no immune response), we calculate probabilities for two settings:
  • Aerosolized Vaccine: Each child has an 85% chance of showing an immune response.
  • Subcutaneous Vaccine: Each child has a 95% chance of showing an immune response.
In both cases, we aim to find the likelihood that at least one child does not develop an immune response. This is essential in evaluating the vaccine's effectiveness and reliability.
The statistical analysis helps compare the performance of the two vaccination methods, guiding decision-makers on choosing the best approach based on their situation.
Immune Response Analysis
Analyzing the immune response is crucial in evaluating the vaccine's effectiveness. In this study, we focus on how many children develop immunity after vaccination.
Calculating the probability that at least one child does not develop an immune response involves subtracting the probability that all children develop the response from 1. Using the binomial probability formula for each method:
  • Aerosolized Vaccine: The probability that all 20 children develop a response is \((0.85)^{20}\), therefore, the probability that at least one does not is \1 - (0.85)^{20}\.
  • Subcutaneous Vaccine: The probability that all 20 children develop a response is \((0.95)^{20}\), thus, the probability that at least one does not is \1 - (0.95)^{20}\.
These calculations help illuminate how likely it is for the vaccines to fail in producing immunity. For public health, knowing this likelihood ensures better planning and management of vaccination programs to maximize immunity and minimize measles outbreaks.

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Most popular questions from this chapter

Larry reads that one out of four eggs contains salmonella bacteria. So he never uses more than three eggs in cooking. If eggs do or don't contain salmonella independent of each other, the number of contaminated eggs when Larry uses three chosen at random has the distribution (a) binomial with \(n=4\) and \(p=1 / 4\). (b) binomial with \(n=3\) and \(p=1 / 4\). (c) binomial with \(n=3\) and \(p=1 / 3\).

Binomial setting? A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. (a) A National Football League kicker has made \(80 \%\) of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and 50 on. (b) A National Basketball Association player has made \(80 \%\) of his free- throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away from the basket with no interference from other players.

Hiking the High Sierra loop in Yosemite. Yosemite National Park has five High Sierra camps arranged in a 49-mile loop. The camps are separated by 8-10 miles, which allows you to hike from one camp to the next in a day. Each camp provides tent cabins with four to six beds per cabin and serves family-style meals. Because of the popularity of the loop, you must enter a lottery in the fall for the following summer, and in a given year, approximately \(25 \%\) of the groups are selected at random to hike the loop." You plan to enter the lottery each year for the next five years. Let \(X\) be the number of years in which you are selected to hike the loop. (a) \(X\) has a binomial distribution. What are \(n\) and \(p\) ? (b) What are the possible values that \(X\) can take? (c) Find the probability of each value of \(X\). Draw a probability histogram for the distribution of \(\boldsymbol{X}\). (See Figure 14.2, page 339, for an example of a probability histogram-) (d) What are the mean and standard deviation of this distribution? Mark the location of the mean on your histogram.

Binomial setting? In each of the following situations, is it reasonable to use a bìnomial distribution for the random variable \(X\) ? Give reasons for your answer in each case. (a) An auto manufacturer chooses one car from each hour's production for a detailed quality inspection. One variable recorded is the count \(X\) of finish defects (dimples, ripples, etc.) in the car's paint. (b) The pool of potential jurors for a murder case contains 100 persons chosen at random from the adult residents of a large city. Each person in the pool is asked whether he or she opposes the death penalty; \(X\) is the number who say "Yes." (c) Joe buys a ticket in his state's Pick 3 lottery game every week; \(X\) is the number of times in a year that he wins a prize.

Is this coin balanced? While he was a prisoner of war during World War II, 14.36 John Kerrich tossed a coin 10,000 times. He got 5067 heads. If the coin is perfectly balanced, the probability of a head is \(0.5\). Is there reason to think that Kerrich's coin was not balanced? To answer this question, find the probability that tossing a balanced coin 10,000 times would give a count of heads at least this far from 5000 (that is, at least 5067 heads or no more than 4933 heads).

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