Problem 36 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

MEDICINE Suppose that the number of children who die each year from leukemia follows a Poisson distribution and that on average, \(7.3\) children per 100,000 die from leukemia. For a city with 100,000 children, find the probability of each of the following events: a. Exactly seven children in the city die from leukemia each year. b. Fewer than two children in the city die from leukemia each year. c. More than five children in the city die from leukemia each year.

Step-by-Step Solution

Answer

a. 0.2203, b. 0.0051, c. 0.7335

Step 1: Understand the Poisson Distribution

The Poisson distribution can be used to model the number of events that occur in a fixed interval of time or space. The parameter \( \lambda \) represents the average number of occurrences in the interval. In this problem, \( \lambda = 7.3 \), which means on average, 7.3 children per 100,000 die from leukemia each year.

Step 2: Poisson Probability Formula

The probability of observing exactly k events in a Poisson distribution is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( e \) is the base of the natural logarithm, \( \lambda = 7.3 \) and \( k \) is the number of events.

Step 3: Calculate Probability for Exactly Seven Children (Part a)

For exactly 7 children: \[ P(X = 7) = \frac{e^{-7.3} \times 7.3^7}{7!} \] Compute the values: \[ P(X = 7) = \frac{0.00061525 \times 1808.564}{5040} \approx 0.2203 \] So, the probability of exactly seven children dying is approximately 0.2203.

Step 4: Calculate Probability for Fewer Than Two Children (Part b)

Sum the probabilities for 0 and 1 child: \[ P(X < 2) = P(X = 0) + P(X = 1) \] First, for 0 children: \[ P(X = 0) = \frac{e^{-7.3} \times 7.3^0}{0!} = e^{-7.3} \approx 0.00061525 \] Then, for 1 child: \[ P(X = 1) = \frac{e^{-7.3} \times 7.3^1}{1!} = 7.3 \times 0.00061525 \approx 0.00449 \] Add them: \[ P(X < 2) = 0.00061525 + 0.00449 \approx 0.0051 \] Therefore, the probability is approximately 0.0051.

Step 5: Calculate Probability for More Than Five Children (Part c)

Calculate the probability for 0 to 5 children and subtract from 1: \[ P(X > 5) = 1 - P(X \leq 5) \] Compute for k = 0 to 5 and sum: \[ P(X = 0) = 0.00061525 \] \[ P(X = 1) = 0.00449 \] \[ P(X = 2) = \frac{e^{-7.3} \times 7.3^2}{2!} = 26.729 \times 0.00061525 \approx 0.0138 \] \[ P(X = 3) = \frac{e^{-7.3} \times 7.3^3}{3!} = 194.12 \times 0.00061525 \approx 0.0452 \] \[ P(X = 4) = \frac{e^{-7.3} \times 7.3^4}{4!} = 1417.076 \times 0.00061525 \approx 0.082 \] \[ P(X = 5) = \frac{e^{-7.3} \times 7.3^5}{5!} = 10344.66 \times 0.00061525 \approx 0.1204 \] Sum these probabilities: \[ P(X \leq 5) = 0.00061525 + 0.00449 + 0.0138 + 0.0452 + 0.082 + 0.1204 \approx 0.2665 \] Finally: \[ P(X > 5) = 1 - 0.2665 \approx 0.7335 \] Therefore, the probability of more than five children dying is approximately 0.7335.

Key Concepts

Poisson distribution

The Poisson distribution is a powerful statistical tool used to model the number of events that occur within a fixed interval of time or space. This distribution is best suited for predictably rare events. In this context, the parameter \(\lambda\) (lambda) stands for the average rate of occurrence within the interval. For this exercise, the average number of child leukemia deaths per year per 100,000 children is given as \( \lambda = 7.3 \). This means that over a large number of years, we expect there to be an average of 7.3 child leukemia deaths each year in such a city.

Probability calculation

To find the probability of different events in a Poisson distribution, we use the Poisson probability formula. The formula to calculate the probability of exactly \( k \) events occurring is:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here:

• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
• \( \lambda \) is the average number of occurrences (7.3 in this problem).
• \( k \) represents the number of occurrences we want to find the probability for.

Using this formula, we can calculate the probability of various scenarios:
1. Exactly \( 7 \) children dying: \[ P(X = 7) = \frac{e^{-7.3} \times 7.3^7}{7!} \approx 0.2203 \]
2. Fewer than \( 2 \) children dying involves summing the probabilities of 0 and 1 child:
\[P(X < 2) = P(X = 0) + P(X = 1) \] \[ P(X = 0) = e^{-7.3} \approx 0.00061525 \] \[ P(X = 1) = 7.3 \times 0.00061525 \approx 0.00449 \] Combining them: \[ P(X < 2) \approx 0.0051 \]
3. More than \( 5 \) children dying requires calculating probabilities for 0 to 5 children and subtracting from 1:
\[ P(X > 5) = 1 - P(X \leq 5) \] \[ P(X \leq 5) \approx 0.2665 \] \[ P(X > 5) \approx 0.7335 \]

Average occurrences

Understanding \( \lambda \), the average rate of occurrence, is crucial in the Poisson distribution. It tells us how often an event is expected to happen within a given interval. For example, in this exercise, the parameter \( \lambda = 7.3 \) indicates that, on average, 7.3 children die from leukemia per year per 100,000 children. The idea is that, over a long period, this average rate will hold true, although individual years may vary.

Usually, the average rate \( \lambda \) is derived from extensive historical data or observations. This average can then be used in the Poisson probability formula to predict the likelihood of various outcomes in similar scenarios. By using the Poisson distribution, we can efficiently determine the probabilities for specified numbers of occurrences over time. This makes it a valuable tool for planning and decision-making in areas like public health, insurance, and risk management.