Problem 37 - 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 250 women with cystic fibrosis, no more than 60 are infertile.

Step-by-Step Solution

Answer

The probability that no more than 60 out of 250 women are infertile is approximately 0.9429.

Step 1: Determine Infertility Probability

First, identify the probability that any given female with cystic fibrosis is infertile. From the problem, we know that 1 in 5 females are infertile. Therefore, the probability \( p \) of infertility is \( \frac{1}{5} = 0.20 \).

Step 2: Calculate Expected Number of Infertile Women

We are dealing with a group of 250 women with cystic fibrosis. Calculate the expected number of infertile women by multiplying the probability of infertility by the total number of women: \( \text{Expected number} = 250 \times 0.20 = 50 \).

Step 3: Understand the Problem Scope

The problem asks for the probability that no more than 60 out of 250 women with cystic fibrosis are infertile. This situation can be modeled with a binomial distribution where \( n = 250 \), \( p = 0.20 \), and we are looking for \( P(X \leq 60) \).

Step 4: Use Normal Approximation

Since we are dealing with a large sample size, we can use the normal approximation for the binomial distribution. For this, calculate the mean \( \mu = n \times p = 50 \) and variance \( \sigma^2 = n \times p \times (1-p) = 250 \times 0.20 \times 0.80 = 40 \). Hence, \( \sigma = \sqrt{40} \approx 6.32 \).

Step 5: Find Z-score for X = 60

Convert \( X = 60 \) to a Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \). Therefore, \( Z = \frac{60 - 50}{6.32} \approx 1.58 \).

Step 6: Calculate the Cumulative Probability

Using a standard normal distribution table or calculator, find \( P(Z \leq 1.58) \). This gives approximately 0.9429. Thus, the probability that no more than 60 women are infertile is about 0.9429.

Key Concepts

Probability

Probability is a fundamental concept in statistics that measures the likelihood of a particular event occurring. It is expressed as a number between 0 and 1, where 0 indicates an impossible event and 1 signifies a certain event.
In the context of our exercise, probability helps us determine how likely it is for women with cystic fibrosis to be infertile. We use the given data that "1 in 5 females with cystic fibrosis is infertile," meaning there is a probability of 0.20 (or 20%) that any given female with cystic fibrosis is infertile.

It's crucial to understand probabilities when working with binomial distributions, as they help us calculate expected results and assess the variability of outcomes in different scenarios.

• The probability of infertility (p) is 0.20.
• We want to find the probability of at most 60 infertile women in 250, i.e., the probability of the event where 0 to 60 women could be infertile.

This understanding of probability sets the foundation for applying further statistical techniques, such as the normal approximation, in our problem.

Normal Approximation

Normal approximation is a technique used to simplify the calculations involved in a binomial distribution when dealing with large sample sizes.
When a binomial distribution has a large number of trials (n) and a probability (p) that leads to both np and n(1-p) being sufficiently large, it resembles a normal distribution. This allows us to use the normal distribution to approximate the binomial distribution.

In this problem, we are dealing with 250 trials (women), which makes it suitable for normal approximation. The parameters for our normal distribution are:

• Mean (\(\mu\)) of 50, calculated as \(\mu = np = 250 \times 0.20\).
• Standard deviation (\(\sigma\)) of approximately 6.32, calculated using \(\sqrt{np(1-p)}\) or \(\sqrt{250 \times 0.20 \times 0.80 }\).

When using the normal approximation, we can convert the number of infertile women (like 60) into a Z-score and find the cumulative probability from a standard normal distribution table for the calculated Z-score.

Statistics

Statistics is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data.
In this exercise, statistics involves using statistical methods to find the probability of an event within a defined population. This involves understanding and applying concepts such as mean, variance, probability distributions, and Z-scores.

Key statistical approaches used in our exercise include:

• Calculating the expected number of event occurrences, which involves multiplying the probability of the event by the number of trials (in our case, the expected number of infertile women).
• Using the binomial distribution to model the outcomes of a process in which each trial has two possible outcomes.
• Applying normal approximation to convert the binomial probability calculations into a problem that involves the standard normal distribution.

Statistics allows us to quantify and manage uncertainty, making it applicable across various fields, such as healthcare, which includes understanding diseases like cystic fibrosis and their characteristics.

Cystic Fibrosis

Cystic fibrosis is a genetic disorder that affects various organs in the body, primarily the lungs and pancreas, due to the production of thick, sticky mucus. This mucus can cause severe respiratory and digestive problems.
Cystic fibrosis is inherited, which means that the disease is passed down from parents to children through genes. In the U.S., about 1 in every 2,500 white children is born with this condition.

This disorder also has various secondary effects, such as increased prevalence of diabetes mellitus and infertility among females. For example, in this exercise, we're focused on infertility, which affects about 1 in 5 females with the disorder.

• Infertility in women with cystic fibrosis results from thicker cervical mucus that can impede sperm passage.
• Understanding these probabilities and their implications in diseases such as cystic fibrosis helps in shaping healthcare strategies and personal health decisions.

Statistics and probability are essential for research in cystic fibrosis, allowing healthcare professionals and researchers to predict outcomes and manage long-term care more effectively.