Problem 42 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

PUBLIC HEALTH As part of a campaign to combat a new strain of influenza, public health authorities are planning to inoculate 1 million people. It is estimated that the probability of an individual having a bad reaction to the vaccine is \(0.0005\). Suppose the number of people inoculated who have bad reactions to the vaccine is modeled by a random variable with a Poisson distribution. a. What is \(\lambda\) for the distribution? b. What is the probability that of the 1 million people inoculated, exactly five will have a bad reaction? c. What is the probability that of the 1 million people inoculated more than 10 will have a bad reaction?

Step-by-Step Solution

Answer

(a) \( \lambda = 500 \). (b) The probability of exactly 5 bad reactions is approximately \( 1.74 \times 10^{-210} \). (c) The probability of more than 10 bad reactions is nearly 1.

Step 1: Understanding the Problem

In this exercise, public health authorities are inoculating 1 million people with a vaccine. The probability of a single person having a bad reaction is 0.0005. We need to use the Poisson distribution to find the probabilities for different scenarios.

Step 2: Defining \( \lambda \)

The parameter \( \lambda \) for a Poisson distribution is the expected number of occurrences. This is calculated by multiplying the probability of a single event by the number of trials. Here, \( \lambda = 1{,}000{,}000 \times 0.0005 = 500 \).

Step 3: Probability of Exactly 5 Bad Reactions

The probability of exactly k occurrences in a Poisson distribution with parameter \( \lambda \) is given by \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]. For \( k = 5 \) and \( \lambda = 500 \), \ P(X = 5) = \frac{500^5 e^{-500}}{5!}. \To compute this: \ P(X = 5) = \frac{500^5 e^{-500}}{120} \approx 1.74 \times 10^{-210} \.

Step 4: Probability of More than 10 Bad Reactions

To find the probability of more than 10 bad reactions, we need to sum the probabilities from 11 to infinity: \ P(X > 10) = 1 - P(X \leq 10).\We sum the probabilities from \( P(X = 0) \) to \( P(X = 10) \). Each probability is calculated using the formula \[ P(X = k) = \frac{500^k e^{-500}}{k!}. \]Finally, \[ P(X > 10) = 1 - \sum_{k=0}^{10} \frac{500^k e^{-500}}{k!}. \]Calculating the sum, \ P(X > 10) \ is practically equal to 1 due to the significantly small probabilities for \ k \ from 0 to 10.

Key Concepts

Poisson distribution

The Poisson distribution is a probability distribution that describes the number of events occurring within a fixed interval of time or space. In the context of public health, it's useful for modeling rare events, such as adverse reactions to a vaccine.

To use the Poisson distribution effectively, you need to define the parameter, \( \lambda \). This parameter represents the average number of occurrences in the given interval. It's calculated by multiplying the probability of a single event by the number of trials. In our public health example, with 1 million people inoculated and a reaction probability of 0.0005, our \( \lambda \) is \( 1,000,000 \times 0.0005 = 500 \).

The Poisson distribution formula for exactly \( k \) occurrences is given by:
\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]
This formula helps us compute probabilities for specific numbers of bad reactions to the vaccine.

Probability

Probability is a measure of the likelihood of an event occurring. In public health, understanding probability helps predict outcomes and make informed decisions. For instance, the probability of a single person having a bad reaction to a vaccine is given as 0.0005.

When dealing with large populations and rare events, like adverse vaccine reactions, the Poisson distribution is often used because it approximates the probability well.

In our exercise, we find the probability of exactly 5 bad reactions using:
\[ P(X = 5) = \frac{500^5 e^{-500}}{5!} \approx 1.74 \times 10^{-210} \]
This shows the event is extremely rare. Similarly, for more than 10 bad reactions, we use:
\[ P(X > 10) = 1 - \sum_{k=0}^{10} \frac{500^k e^{-500}}{k!}. \]
The calculation reveals that having more than 10 adverse reactions is nearly certain, highlighting the importance of understanding and managing risk in public health.

Expected value

The expected value in probability and statistics is the average outcome if an experiment is repeated many times. It provides a sense of the 'center' of a probability distribution. For the Poisson distribution, the expected value is simply the parameter \( \lambda \).

In our specific case, with 1 million inoculated individuals and a 0.0005 probability of a bad reaction, the expected number of bad reactions—our \( \lambda \)—is 500. This value is crucial as it summarizes the entire distribution and informs public health authorities about what to anticipate. Knowing the expected value helps in resource planning and risk assessment.

Public health

Public health focuses on protecting and improving the health of populations. Accurate modeling and prediction using probability distributions, like the Poisson distribution, are vital tools.

When introducing a new vaccine, understanding the likelihood and expected number of adverse reactions aids in planning and preventing panic. For instance, by knowing that out of 1 million vaccinations, we expect 500 adverse reactions, authorities can prepare adequate medical support.

The use of statistical models ensures informed decision-making and effective resource allocation. This helps build public trust and improves response strategies to health challenges.