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The binomial distribution is a probability distribution that is used to model the number of successes in a fixed number of trials. Each trial must be independent, and there are only two outcomes: success or failure. Let's imagine flipping a coin. Each flip (trial) can result in either heads or tails, just like in the gene-carrying problem, where each individual either carries the gene or doesn't.
For the exercise about colon cancer genes, we are asked to find how many individuals out of a sample of 1000 people carry a defective gene. Here, the carrying of the gene is a success. We can define the number of trials as 1000, with a small probability of success of 0.005, representing the chance that any one individual carries the gene. Therefore, the problem initially uses a binomial distribution:

• Number of trials (\( n \)): 1000
• Probability of success (\( p \)): 0.005

Understanding the binomial distribution helps in setting the stage to approximate the distribution needed for further calculations in this type of scenario.

When dealing with large sample sizes in a binomial distribution, calculations can become complex. However, if the sample size is large enough, the binomial distribution can be approximated by a normal distribution due to the Central Limit Theorem. This is helpful because the normal distribution has simpler mathematical properties, making calculations more straightforward.
For our gene problem, since we are analyzing 1000 people, we can use the normal approximation. We first need to calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the binomial distribution. The mean is given by \( \mu = np = 1000 \times 0.005 = 5 \), and the standard deviation is \( \sigma = \sqrt{np(1-p)} = \sqrt{1000 \times 0.005 \times 0.995} \approx 2.23 \).
The normally distributed approximation would thus be expressed as \( N(5, 2.23) \), which significantly eases the calculations required in finding the probabilities we seek.

The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. Often, problems cannot be directly solved using the raw normal distribution, so we convert to standard normal distribution, making them easier to analyze using Z-scores.
In the task of finding the probability that between 5 and 8 individuals carry the gene, using normal approximation, we need to convert our normal distribution values to standard normal values using the formula:\[ Z = \frac{X - \mu}{\sigma} \]Continuity correction is applied by adjusting the values to 4.5 and 8.5 before calculating their respective Z-scores. For example, when \( X = 4.5 \), the Z-score would be \(\frac{4.5 - 5}{2.23} \approx -0.22 \). Similarly, converting 8.5 gives us \(\frac{8.5 - 5}{2.23} \approx 1.57 \).
With Z-scores, we can use standard normal distribution tables or calculators to easily find the probabilities such as \( P(-0.22 < Z < 1.57) \), helping to determine the likelihood of specific outcomes.