Problem 36 - Calculus for Biology and Medicine

Question

Use the following facts: Cystic fibrosis is an inherited disorder that causes abnormally thick body secretions. About 1 in 2500 white babies in the United States has this disorder. About 3 in 100 children with cystic fibrosis develop diabetes mellitus, and about 1 in 5 females with cystic fibrosis is infertile. Find the probability that, in a group of 1000 children with cystic fibrosis, at least 25 will develop diabetes mellitus.

Step-by-Step Solution

Answer

The probability is approximately 0.8485.

Step 1: Determine the Probability of Diabetes in Cystic Fibrosis

We are given that 3 out of 100 children with cystic fibrosis develop diabetes mellitus. Therefore, the probability that a child with cystic fibrosis develops diabetes mellitus is \( p = \frac{3}{100} = 0.03 \).

Step 2: Understand the Binomial Distribution

Since we are dealing with a fixed number of children (1000) and each child has the same probability of developing diabetes, the situation follows a binomial distribution. The binomial model is appropriate here with parameters \( n = 1000 \) and \( p = 0.03 \).

Step 3: Define the Binomial Random Variable

Define a random variable \( X \) that represents the number of children with cystic fibrosis developing diabetes mellitus out of 1000 children. \( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n = 1000, p = 0.03) \).

Step 4: Calculate the Probability Using Normal Approximation

Since \( n \) is large, we can use the normal approximation to the binomial distribution. The mean \( \mu = np = 1000 \times 0.03 = 30 \) and the standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{1000 \times 0.03 \times 0.97} \approx 5.36 \).

Step 5: Find the Probability of 25 or More

Using the normal approximation, we want to find \( P(X \geq 25) \). For continuity correction, calculate \( P(X \geq 24.5) \) first. Convert to a Z-score: \( Z = \frac{24.5 - 30}{5.36} \approx -1.03 \).

Step 6: Look Up the Z-score in a Standard Normal Table

Using the Z-score table, find \( P(Z \geq -1.03) \). This is approximately 0.8485. Thus, the probability that at least 25 children out of 1000 with cystic fibrosis will develop diabetes mellitus is about \( 0.8485 \).

Key Concepts

Cystic Fibrosis

Cystic fibrosis (CF) is a genetic disorder that primarily affects the lungs and digestive system, leading to thick and sticky mucus production. This mucus can obstruct airways and glands, causing severe respiratory and digestive issues. The condition is autosomal recessive, meaning a child needs to inherit one defective gene from each parent to develop CF. In the United States, cystic fibrosis is most common among white individuals, with an occurrence rate of about 1 in 2,500 live births.

Due to advances in medical care and treatment, the life expectancy and quality of life for individuals with CF have improved remarkably. Nonetheless, CF patients often face complications such as lung infections, respiratory failure, and digestive problems. Additionally, cystic fibrosis can lead to other health concerns like diabetes mellitus and infertility, especially in females.

The complexity of CF requires comprehensive management from a multidisciplinary healthcare team. Treatment strategies focus on preventing infections, maintaining lung function, and addressing nutritional needs to manage the disease effectively.

Probability

Probability is a branch of mathematics that deals with the likelihood of an event occurring. It provides a numerical measure, ranging from 0 to 1, where 0 indicates impossibility, and 1 indicates certainty. Probability helps to model and predict the outcomes of random events.

When examining the situation with cystic fibrosis and diabetes, the probability that a child with CF develops diabetes mellitus is given as 3%. This can be represented as a probability value of 0.03, derived from 3 out of every 100 children. In applied contexts, such as the problem of interest, calculating probabilities involves determining the likelihood of a specified outcome (like developing diabetes among 1,000 children with CF) using methods such as the binomial distribution.

Understanding probability is crucial in various fields, including statistics, finance, and science, as it allows individuals to make informed decisions based on risk and uncertainty.

Normal Approximation

Normal approximation is a statistical technique used to approximate the binomial distribution with a normal distribution when the sample size is large. This method leverages the Central Limit Theorem, which states that given a sufficiently large sample size, the distribution of the sample mean will be approximately normal, regardless of the distribution of the underlying population.

In the context of the problem involving cystic fibrosis, we look at a situation where there are 1,000 trials (children), each with the same chance of success (developing diabetes). The parameters of the binomial distribution are given as:

• Number of trials: \( n = 1000 \)
• Probability of success: \( p = 0.03 \)

The mean or expected value is calculated as \( \mu = np = 30 \), and the standard deviation is \( \sigma = \sqrt{np(1-p)} \approx 5.36 \). Using these parameters, the situation can be approximated using the normal distribution to find probabilities, such as the probability that at least 25 out of 1,000 children develop diabetes. By applying continuity correction, the calculation adjusts to \( X \geq 24.5 \), converted to a Z-score for lookup in a Z-table.

The normal approximation simplifies calculations and is a powerful tool for handling large sample size problems, as it allows us to estimate binomial distribution probabilities using the more manageable normal distribution.