91Ó°ÊÓ

A binomial distribution is used to model scenarios where there are a fixed number of independent trials, each with two possible outcomes. In this exercise, the outcomes are a child having the gene or not having it.

• The probability of a child having the gene is represented by \( p = 0.04 \).
• The probability of not having the gene is \( 1 - p = 0.96 \).
• The number of trials, \( n \), is 732 as each child tested is a trial.

The goal is to determine how likely it is to find at least 20 children with the gene. Since we're interested in the number of successful outcomes (children with the gene), this is ideal for a binomial distribution represented by \( X \sim \text{Binomial}(732, 0.04) \). The binomial distribution formula helps calculate the probability of a given number of successes in these trials.

The normal approximation can simplify calculations in binomial distributions, especially when dealing with a large number of trials. Since we are working with 732 children, using the binomial formula directly can be cumbersome. Hence, a normal distribution approximation is used here.To apply normal approximation on a binomial distribution, conditions \( np \) and \( n(1-p) \) should both be greater than 5, which is clearly satisfied in this problem:

• \( np = 732 \times 0.04 = 29.28 \)
• \( n(1-p) = 732 \times 0.96 = 702.72 \)

Thus, we can use a normal distribution with mean \( \mu = np = 29.28 \) and variance \( \sigma^2 = np(1-p) \approx 28.1088 \). This approximation enables easier probability calculations using Z-scores.

Cumulative probability is a measure of the probability that a random variable takes on a value less than or equal to a certain number. In this context, we want to compute the probability that fewer than 20 children have the gene and use it to find the probability of finding at least 20 children.The task is to determine \( P(X \geq 20) \), which equates to \( 1 - P(X < 20) \). This involves calculating \( P(X < 20) \) using the cumulative property of the normal distribution. It leverages the symmetry and properties of the normal distribution to compute probabilities without manually calculating every specific outcome. This makes the process efficient and practical for large datasets.

Z-scores convert data points from a normal distribution into a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This conversion makes it possible to use standard normal distribution tables to find cumulative probabilities.In this exercise, once the normal approximation is applied, the Z-score is calculated as follows:

• \( Z = \frac{X - \mu}{\sigma} \)

For the specific adjustment using continuity correction (explained below),\( X \) becomes 19.5. Hence, \( Z = \frac{19.5 - 29.28}{5.3} = -1.846 \).Determining \( P(Z < -1.846) \) from the Z-table gives us the cumulative probability needed to find \( P(X \geq 20) \). The conversion to a Z-score simplifies interaction with standard tables.

Continuity correction is a technique used when approximating a discrete distribution, like the binomial distribution, with a continuous one, such as the normal distribution. This adjustment accounts for the fact that the normal distribution is continuous, whereas the binomial is not.In the scenario of this exercise:

• We initially desire \( P(X < 20) \).
• To make use of the continuous normal curve, we consider \( P(X \leq 19.5) \) rather than \( P(X < 20) \).
• This adjusts for the discrete nature of the original data and gives a more accurate approximation.

The continuity correction helps bridge the gap between continuous and discrete distributions, ensuring that our probability approximations are as accurate as possible.