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Chapter 11: Problem 28

In a group of 100 twentyyear-old Americans, an average of 2.15 will live to be 100 years old ("centenarians"). Use the Poisson distribution to find the probability that in such a group of \(100,\) at least one will be a centenarian. [Hint: Find the probability that none will be a centenarian, and subtract it from \(1 .]\) Source: National Vital Statistics Report

Short Answer

Expert verified
The probability is approximately 0.884.

Step by step solution

01

Define the Poisson Distribution

The Poisson distribution is used to model the probability of a given number of events happening in a fixed interval of time or space, given the average number of times the event occurs over that interval. Here, the average number, or the expected value, for the number of centenarians out of 100 individuals is given as \( \lambda = 2.15 \). This \( \lambda \) represents the average rate (mean) of people who will become centenarians.
02

Calculate Probability of Zero Centenarians

In a Poisson distribution, the probability of observing exactly \( k \) events (centenarians, in this case) is given by the formula: \[ P(k) = \frac{{e^{-\lambda} \lambda^k}}{{k!}} \] To find the probability that none of the 100 individuals will be a centenarian, set \( k = 0 \): \[ P(0) = \frac{{e^{-2.15} \times 2.15^0}}{{0!}} = e^{-2.15} \times 1\] Inserting \( \lambda = 2.15 \) into the formula: \[ P(0) = e^{-2.15} \approx 0.116 \]
03

Calculate the Probability of At Least One Centenarian

The problem requires finding the probability of at least one individual being a centenarian. We can find this by subtracting the probability of having zero centenarians from 1: \[ P(\text{at least one}) = 1 - P(0) = 1 - 0.116 \approx 0.884 \]
04

Conclusion

The probability that in a group of 100 individuals at least one person will live to be 100 years old is approximately 0.884. This calculation is derived using the complement rule in probability for Poisson distribution.

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Key Concepts

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

Probability
The concept of probability is foundational in understanding events that have uncertain outcomes. Probability quantifies the likelihood that a particular event will occur. It is expressed as a number between 0 and 1, where:
  • 0 indicates impossibility
  • 1 indicates certainty
In the context of the Poisson distribution exercise, we're tasked with finding the probability that at least one out of 100 individuals will become a centenarian. Knowing how to calculate the probability helps to assess risks and make informed predictions about future outcomes.
To find probabilities using a Poisson distribution, you use the provided formula, which depends on the average expected value (in this case, 2.15) and the specific number of events (like zero centenarians). The calculated number allows us to deduce more complex probabilities, such as the likelihood of at least one event happening.
Expected Value
Expected value, often denoted by \( \lambda \) in the Poisson distribution, is a measure of the center of a probability distribution. It is essentially the average or mean outcome that you can expect if an action is repeated many times. For our exercise:
  • The expected value \( \lambda = 2.15 \).
  • This represents the average number of centenarians among 100 individuals.
Understanding the expected value allows us to anticipate the outcome over numerous trials even though each individual event remains random.
In our context, knowing that, on average, 2.15 individuals out of 100 could be centenarians helps in preparing for the range of outcomes observed in different sample groups. It serves as a benchmark to gauge the number of occurrences that fit within the scope of normal expectation as described by the Poisson process.
Complement Rule
The complement rule is a fundamental concept in probability used to simplify calculations, particularly when it is easier to calculate the probability of an event not occurring than the event itself. Mathematically, it can be expressed as:
  • \( P(A) = 1 - P(A^c) \)
Where \( A \) is the event of interest, and \( A^c \) is the complement of \( A \) (i.e., the event of \( A \) not happening).
In this exercise, to find the probability of at least one centenarian, we calculated the probability of having zero centenarians \( P(0) \) using the Poisson formula and then subtracted it from 1. This way, we efficiently determined the likelihood of the complement event (at least one centenarian), reflecting how useful the complement rule is in simplifying complex probability assessments.

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Most popular questions from this chapter

For a random variable \(X\) with mean \(\mu\) that takes values \(x_{1}, x_{2}, \ldots, x_{n}\) with probabilities \(p_{1}, p_{2}, \ldots, p_{n}\) prove the following "alternative" formula for the variance $$ \operatorname{Var}(X)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i}-\mu^{2} $$ by giving a justification for each numbered equality in the following derivation. $$ \begin{aligned} \operatorname{Var}(X) & \stackrel{1}{=} \sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2} p_{i} \stackrel{2}{=} \sum_{i=1}^{n}\left(x_{i}^{2}-2 x_{i} \mu+\mu^{2}\right) p_{i} \\ & \stackrel{3}{=} \sum_{i=1}^{n} x_{i}^{2} p_{i}-\sum_{i=1}^{n} 2 x_{i} \mu p_{i}+\sum_{i=1}^{n} \mu^{2} p_{i} \\ & \stackrel{4}{=} \sum_{i=1}^{n} x_{i}^{2} p_{i}-2 \mu \sum_{i=1}^{n} x_{i} p_{i}+\mu^{2} \sum_{i=1}^{n} p_{i} \\ & \stackrel{5}{=} \sum_{i=1}^{n} x_{i}^{2} p_{i}-2 \mu \mu+\mu^{2} 1 \stackrel{6}{=} \sum_{i=1}^{n} x_{i}^{2} p_{i}-\mu^{2} \end{aligned} $$

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