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An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

Short Answer

Expert verified
P(5 ≤ X ≤ 8) ≈ 0.580, P(X ≥ 8) ≈ 0.1314.

Step by step solution

01

Understand the Problem

We need to find the distribution of individuals in a sample of 1000 who carry a defective gene prevalent in 1 out of every 200 people. We will then calculate the probability of certain numbers of individuals carrying this gene within the population sample.
02

Determine the Appropriate Probability Distribution

Since the probability of an individual carrying the gene is 1/200, and we have a large sample size of 1000, we will model this situation using a binomial distribution. The number of trials, denoted as \( n \), is 1000 and the probability of success, \( p \), is \( \frac{1}{200} = 0.005 \).
03

Calculate the Binomial Mean and Variance

The mean \( \mu \) of the binomial distribution is given by \( n \times p = 1000 \times 0.005 = 5 \). The variance \( \sigma^2 \) is \( n \times p \times (1-p) = 1000 \times 0.005 \times 0.995 = 4.975 \).
04

Approximate Binomial Distribution with Normal Distribution

Since \( n \) is large and \( p \) is small, the binomial distribution can be approximated by a normal distribution with mean \( \mu = 5 \) and variance \( \sigma^2 = 4.975 \), so \( \sigma = \sqrt{4.975} \approx 2.23 \).
05

Calculate Probability for Between 5 and 8 (Inclusive)

Using the normal approximation, find the probability that the number of individuals carrying the gene is between 5 and 8. Calculate the Z-scores for 4.5 and 8.5 (continuity correction) and use the standard normal distribution table.\[ P(5 \leq X \leq 8) = P\left(\frac{4.5-5}{2.23} \leq Z \leq \frac{8.5-5}{2.23}\right) \]\[ P(5 \leq X \leq 8) = P(-0.22 \leq Z \leq 1.57) \approx 0.580 \]
06

Calculate Probability for At Least 8

For finding the probability that at least 8 individuals carry the gene, use the Z-score for 7.5 (continuity correction).\[ P(X \geq 8) = P(Z \geq \frac{7.5-5}{2.23}) \]\[ P(X \geq 8) = P(Z \geq 1.12) \approx 1 - 0.8686 = 0.1314 \]
07

Conclusion

The probabilities calculated reveal how these health-related gene distributions might appear in large samples, which has implications for genetic screening outcomes and decisions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
Binomial distribution is a probability distribution that applies to experiments where there are exactly two possible outcomes: success or failure. In the context of genetic screening, the outcomes could be: carrying a defective gene or not carrying it.

This distribution is useful when you want to know the probability of a specific number of successes in a certain number of trials. For our exercise, we looked at 1000 individuals and checked how many carry a defective gene with a probability of 1 in 200 (or 0.005 as a decimal).

It's essential to remember: - **n** is the number of trials (people in the sample), which is 1000 in this case. - **p** is the probability of success in each trial, here being 0.005. The binomial distribution helps calculate the likelihood of different numbers of successes (carriers of the gene) within the sample.
Normal Approximation
When dealing with a large sample size, the binomial distribution can be difficult to calculate manually. That's where normal approximation comes in handy. It simplifies things significantly.

Given that our sample size is large and the probability is quite small, the distribution of the defective gene carriers follows a bell curve. The binomial distribution can be approximated to a normal distribution for this case. Here:- The mean (\(\mu\)) is calculated as \( n imes p \), which for us is 5.- The variance (\(\sigma^2\)) is \( n imes p imes (1-p) \), which comes out to be approximately 4.975.Once you have the mean and variance, you can derive the standard deviation (\(\sigma\)) as \(\sqrt{4.975} \approx 2.23\), making it easier to calculate probabilities.
Genetic Screening
Genetic screening is a critical tool for identifying individuals who might be at risk for certain inherited conditions, like colon cancer from a defective gene.

In our case, this involves testing a sample of individuals to estimate how many might carry the gene that could lead to such health issues. With a probability model, you can predict how often you might find gene carriers in different sample sizes, helping in understanding broader health impacts.

The calculations give us insight into how often we might expect to observe this genetic trait in populations, contributing to planning public health measures and informing individuals about their genetic risks.
Sample Size and Probability
Sample size plays a significant role when working with probabilities. A larger sample size generally provides a more reliable estimate of the underlying probability distribution.

In our example, the sample size is 1000. This large size allows us to use the normal approximation effectively when calculating the distribution of defective gene carriers.

The key points to consider when dealing with sample size and probability are: - The probability (p) of an event (carrying the gene) remains constant. - Larger samples give a more accurate picture of how this probability plays out in real-world scenarios. Accurately understanding how sample size interacts with probability helps in designing studies and experiments that yield meaningful information.

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Most popular questions from this chapter

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