/*! 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 50 Peanut Allergies A mong Children... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Peanut Allergies A mong Children. About \(2 \%\) of children in the United States are allergic to peanuts. \(\underline{23}\) Choose three children at random and let the random variable \(X\) be the number in this sample who are allergic to peanuts. The possible values \(X\) can take are \(0,1,2\), and 3. Make a three-stage tree diagram of the outcomes (allergic or not allergic) for the three individuals and use it to find the probability distribution of \(X\).

Short Answer

Expert verified
The probability distribution is: \(P(X=0) = 0.9412\), \(P(X=1) = 0.0576\), \(P(X=2) = 0.0012\), \(P(X=3) = 0.000008\).

Step by step solution

01

Identify Outcomes for One Child

Each child has two possible outcomes: allergic to peanuts or not allergic. We will denote allergic by 'A' and not allergic by 'N'.
02

Construct a Tree Diagram - First Level

Start with one child and create two branches from a point: one for 'A' (allergic) and another for 'N' (not allergic).
03

Construct a Tree Diagram - Second Level

From each outcome of the first child ('A' and 'N'), draw two branches for the second child, again noting 'A' or 'N' for each branch.
04

Construct a Tree Diagram - Third Level

From each of the four paths created by the second child, draw two more branches representing 'A' or 'N' for the third child, resulting in eight total final paths.
05

Label End Outcomes

Each path of the tree ends in a combination of 'A' and 'N'. The eight outcomes are: 'AAA', 'AAN', 'ANA', 'ANN', 'NAA', 'NAN', 'NNA', 'NNN'.
06

Assign Probabilities to Each Path

The probability a child is allergic is 0.02, and not allergic is 0.98. Calculate the probability for each path using these probabilities. For example, the probability of 'AAA' is \(0.02 \times 0.02 \times 0.02 = 0.000008\).
07

Count Number of Allergic Individuals in Each Outcome

For each path, count how many 'A's are present to determine the value of the random variable \(X\).
08

Calculate Probability Distribution of X

Sum the probabilities of paths that correspond to each possible value of \(X\). For \(X = 0\), use 'NNN'; for \(X = 1\), use 'ANN', 'NAN', and 'NNA'; for \(X = 2\), use 'AAN', 'ANA', and 'NAA'; for \(X = 3\), use 'AAA'. Calculate each sum.
09

Present the Probability Distribution

Summarize the probability distribution as follows: \(P(X=0) = (0.98)^3 = 0.941192\), \(P(X=1) = \binom{3}{1} (0.02) (0.98)^2 = 0.057624\), \(P(X=2) = \binom{3}{2} (0.02)^2 (0.98) = 0.001176\), \(P(X=3) = (0.02)^3 = 0.000008\).

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.

Tree Diagram
A tree diagram is a visual representation used to map out all possible outcomes of a probability scenario. It is extremely useful in simplifying complex probability problems and ensuring all potential outcomes are considered. In our peanut allergy scenario, we begin by identifying the two possibilities for each child: allergic (A) and not allergic (N). We create branches from a starting point for a single child, which splits into 'A' and 'N'. For three children, this process repeats three times, expanding the tree to eight unique paths: 'AAA', 'AAN', 'ANA', 'ANN', 'NAA', 'NAN', 'NNA', 'NNN'.

Each path represents a sequence of outcomes for the three children. From this, we can easily visualize and calculate the different combinations of allergic and non-allergic outcomes. A tree diagram, like this, provides a detailed view of how probabilities develop step by step. It gives a clear structure to understand complex probability problems by breaking them down into smaller, more manageable parts.
Binomial Probability
Binomial probability is a statistical method used to determine the likelihood of a given number of successful outcomes in a fixed number of trials, with each trial having two possible outcomes: success or failure. It's essential when dealing with cases like our scenario about peanut allergies—where each child is either allergic or not allergic. Here, the probability of success (becoming allergic) is given as 0.02 for an individual child, whereas the probability of failure (not allergic) is 0.98.

In our exercise, we apply binomial probability to calculate the probability distribution for the random variable, X, which factors all the different ways the outcomes can occur for three children. This is summed up by using the formula for binomial probability: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
  • \( n \) is the number of trials (children),
  • \( k \) is the number of successes (allergic children),
  • \( p \) is the probability of success (0.02),
  • \( 1-p \) is the probability of failure (0.98).
By using this formula, we systematically assess the probability of encountering 0, 1, 2, or 3 allergic children in our group of three.
Random Variable
A random variable is a key concept in statistics used to quantify outcomes of a random phenomenon. In our scenario, the random variable, denoted as X, represents the number of children out of three who are allergic to peanuts. This random variable can take on the discrete values of 0, 1, 2, or 3, each representing the possible number of allergic children in the sample.

Once the tree diagram is constructed, and probabilities are calculated for each outcome, these can be applied to define the probability distribution of the random variable X. This probability distribution captures how likely it is for different values of X to occur, effectively characterizing the "randomness" in the outcome of selecting three children at random from the population. Understanding random variables is essential for interpreting the results in probability and statistics, as they provide a bridge to linking probability with real-world data.
Statistics Education
Statistics education focuses on teaching and learning the principles of data analysis, probability, and statistical reasoning. It emphasizes the development of critical thinking skills to interpret data and draw meaningful conclusions. In exercises like our peanut allergy problem, students learn to apply their knowledge of probability distributions and statistical tools such as tree diagrams and binomial probability formulas.

By crafting a clear and methodical approach to solving statistical problems, such as calculating probability distributions, learners gain a foundation for making data-driven decisions. This educational approach not only helps students solve textbook exercises but also prepares them to handle complex data-related challenges in various fields. Effective statistics education empowers students with the essential skills to analyze real-world problems, understand uncertainty, and communicate their interpretations effectively.

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

White Cats and Deafness. Although cats generally possess an acute sense of hearing, due to an anomaly in their genetic makeup, deafness among white cats with blue eyes is quite common. Approximately \(95 \%\) of the general cat population are non-white cats (i.e., not pure white), and congenital deafness is extremely rare in non-white cats. However, among white cats, approximately \(75 \%\) with two blue eyes are deaf, \(40 \%\) with one blue eye are deaf, and only \(19 \%\) with eyes of other colors are deaf. In addition, among white cats, approximately \(23 \%\) have two blue eyes, \(4 \%\) have one blue eye, and the remainder have eyes of other colors. 10 a. Draw a tree diagram for selecting a white cat (outcomes: one blue eye, two blue eyes, or eyes of other colors) and deafness (outcomes: deaf or not deaf). b. What is the probability that a randomly chosen white cat is deaf?

Tendon Surgery (continued). You have torn a tendon and are facing surgery to repair it. The surgeon explains the risks to you: infection occurs in \(3 \%\) of such operations, the repair fails in \(14 \%\), and both infection and failure occur together in 1\%. What is the probability of infection, given that the repair is successful? Follow the four-step process in your answer.

Suppose the locus D21S11 has two alleles, called 29 and 31. The proportion of the Caucasian population with allele 29 is \(0.181\) and with allele 31 is \(0.071\). What proportion of the Caucasian population has the combination \((29,31)\) at the locus D21S11? What proportion has the combination \((29,29) ?\)

Lost Internet References. Internet sites often vanish or move so that references to them can't be followed. In fact, \(47 \%\) of Internet sites referenced in major medical journals are lost. 5 If a paper contains seven Internet references, what is the probability that all seven are still good? What specific assumptions did you make to calculate this probability?

Playing the Lottery. New York State's "Quick Draw" lottery moves right along. Players choose between 1 and 10 numbers from the range 1 to \(80 ; 20\) winning numbers are displayed on a screen every four minutes. If you choose just one number, your probability of winning is \(20 / 80\), or \(0.25\). Lester plays one number eight times as he sits in a bar. What is the probability that all eight bets lose?

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.