Question

Insulin-dependent diabetes (IDD) among children occurs most frequently in persons of northern European descent. The incidence ranges from a low of \(1-2\) cases per 100,000 per year to a high of more than 40 per 100,000 in parts of Finland. \(^{9}\) Let us assume that an area in Europe has an incidence of 5 cases per 100,000 per year. a. Can the distribution of the number of cases of IDD in this area be approximated by a Poisson distribution? If so, what is the mean? b. What is the probability that the number of cases of IDD in this area is less than or equal to 3 per \(100,000 ?\) c. What is the probability that the number of cases is greater than or equal to 3 but less than or equal to 7 per \(100,000 ?\) d. Would you expect to observe 10 or more cases of IDD per 100,000 in this area in a given year? Why or why not?

Short answer

Based on the provided data and calculations using the Poisson distribution, we determined that the probability of observing 3 or fewer IDD cases per 100,000 children in a given year is approximately 0.265, and the probability of observing between 3 and 7 cases is approximately 0.616. The probability of observing 10 or more cases is quite low at approximately 0.0318, making it unlikely to observe such a high number of cases per 100,000 children in a single year.

Step-by-step solution

Determine if Poisson distribution can be used and find the mean

The given data represents the number of occurrences of an event (IDD) within a fixed space (population) over a fixed time period (one year). As such, the Poisson distribution can be used as an approximation. In this case, the mean (lambda) is given as 5 cases per 100,000 per year.

Calculate the probability for the number of cases being less than or equal to 3 per 100,000

We need to find the probability P(X <= 3) using the Poisson distribution formula, where lambda = 5 cases per 100,000 per year. The Poisson probability formula is given by:\[P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\]Applying this formula, we have:\[P(X \le 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\] Using the formula for each value and substituting lambda = 5, we get: \[P(X \le 3) = \frac{e^{-5}5^0}{0!} + \frac{e^{-5}5^1}{1!} + \frac{e^{-5}5^2}{2!} + \frac{e^{-5}5^3}{3!}\] Calculating it, we get: \[P(X \le 3) \approx 0.265\]

Calculate the probability for the number of cases being greater than or equal to 3 but less than or equal to 7 per 100,000

Now, we need to find the probability P(3 <= X <= 7) using the Poisson distribution formula. We can calculate this by finding the sum of probabilities for X = 3, 4, 5, 6, and 7. \[P(3 \le X \le 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X =7)\] Applying the formula and substituting lambda = 5, we get: \[P(3 \lesssim X \lesssim 7) = \frac{e^{-5}5^3}{3!} + \frac{e^{-5}5^4}{4!} + \frac{e^{-5}5^5}{5!} + \frac{e^{-5}5^6}{6!} + \frac{e^{-5}5^7}{7!}\] Calculating, we get: \[P(3 \lesssim X \lesssim 7) \approx 0.616\]

Determine if we should expect 10 or more cases per 100,000 in a given year

To determine if we should expect 10 or more cases per 100,000 in a given year, we need to find P(X >= 10). One way to calculate this probability is to use the complement rule: P(X >= 10) = 1 - P(X < 10). We can find P(X < 10) by summing up the probabilities for X = 0, 1, ... 9 and then calculate the complement. It can be simplified as; \[P(X \ge 10) = 1 - \sum_{k=0}^{9} \frac{e^{-5}5^k}{k!}\] Calculating this, we get: \[P(X \ge 10) \approx 0.0318\] This probability is very low. Therefore, it's unlikely to observe 10 or more cases of IDD per 100,000 in this area in a given year.

Key concepts

Probability Calculation

Probability calculation is a crucial concept in statistics used to determine the likelihood of certain events happening. In the context of Poisson distribution, which deals with events happening in a fixed interval of time or space, the probability is calculated using a specific formula. This formula is:

• \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \)

This allows us to find the likelihood of different numbers of events occurring within the specified interval. The term \( \lambda \) represents the average number of events in that interval, while \( e \) is the base of the natural logarithm, and \( k \) is the actual number of events you are calculating the probability for. Understanding this formula and how to use it is key for solving problems involving Poisson distribution.

Insulin-Dependent Diabetes

Insulin-dependent diabetes (IDD) is a chronic condition that primarily affects children and occurs more commonly among individuals of northern European descent. Understanding the distribution and occurrence of IDD is important for public health and medical research. In statistical terms, IDD incidents can serve as a case study for Poisson distribution, where the focus is on the number of certain events (in this case, IDD cases) occurring in a fixed population over a set period.
The example of IDD in this exercise assumes a specific location in Europe with an incidence rate of 5 cases per 100,000 per year. Such real-world applications underscore the relevance of statistical methods like Poisson distribution to estimate the number of occurrences in health-related studies.

Mean Value Estimation

Mean value estimation is the process of finding the average number of occurrences within a given dataset or sample. In a Poisson distribution, the mean value \( \lambda \) is fundamental, as it serves as the average rate of occurrence for an event within a specified time or space.

• The mean is crucial because it directly affects the shape and properties of the distribution.
• In our exercise, \( \lambda \) is given as 5 cases per 100,000 per year, representing the estimated average number of IDD cases.
• This mean value allows us to use the Poisson formula to predict different probabilities for IDD case counts that help in planning and crafting public health responses.

The precision of the mean value affects the accuracy of probability predictions derived from the Poisson model.

Complement Rule

The complement rule is a useful probability principle, which states that the probability of an event happening is equal to one minus the probability of it not happening. Formally, this can be represented as:

• \( P(A^c) = 1 - P(A) \)

where \( P(A^c) \) is the probability of the event not happening. In the context of this exercise, the complement rule helps in determining the probability of observing 10 or more cases of IDD. Instead of calculating every possible outcome above 9, which can be cumbersome, we calculate the probability of fewer than 10 cases occurring and subtract this value from one.
This simplification significantly eases the calculation process and highlights the utility of the complement rule in practical applications.

Probability Distribution Approximation

Probability distribution approximation involves using known statistical models to estimate the distribution of events. The Poisson distribution is a common approximation method used when events are rare and occur independently over a fixed period or area. It is particularly useful for modeling rare diseases, like IDD, as it provides a way to approximate the actual distribution of observed case counts.

• The assumption of randomness and independence is key to applying this approach.
• In this exercise, we approximate the number of IDD incidents in a specific location using a Poisson distribution with a mean value.
• This allows for calculating probabilities of various counts of IDD cases and aids in effective decision-making in healthcare and policy.

Probability distribution approximation through the Poisson model offers a simplified view of potential outcomes which is essential for informed predictions and strategies.