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When working with a binomial distribution, calculating probabilities directly can be challenging, especially when the number of trials is large. This is where the normal approximation becomes useful. The idea is to approximate the binomial distribution with a normal distribution because it simplifies calculations and allows us to use techniques developed for normal distributions.

For a binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success), we can use a normal distribution to approximate it if \( np \) and \( n(1-p) \) are both greater than 5.

• The mean (\( \mu \)) of the normal approximation is the same as the binomial mean: \( np \).
• The standard deviation (\( \sigma \)) is calculated by \( \sqrt{np(1-p)} \).

This allows the binomial distribution \( B(n, p) \) to be approximated by the normal distribution \( N(np, \sqrt{np(1-p)}) \).

In our exercise, since \( np = 5 \) and \( n(1-p) = 995 \), we meet this condition, which makes the normal approximation valid.

In the context of the normal approximation to the binomial distribution, a continuity correction is used to improve the accuracy of calculation. The continuity correction accounts for the fact that a binomial distribution is discrete (individual, separate pieces) while a normal distribution is continuous (smooth and flowing everywhere).

To apply the continuity correction, we adjust our range of interest by 0.5 units. This ensures that the approximated area under the curve is more aligned with the discrete nature of the original binomial distribution.

• If calculating a probability like \( P(a \leq X \leq b) \), you adjust it to \( P(a - 0.5 \leq X \leq b + 0.5) \).
• For probabilities such as \( P(X \geq k) \), adjust to \( P(X \geq k - 0.5) \).
• For \( P(X < k) \), adjust to \( P(X < k + 0.5) \).

So in exercise part (a), \( P(5 \leq X \leq 8) \) adjusts to \( P(4.5 \leq X \leq 8.5) \), and for part (b), \( P(X \geq 8) \) adjusts to \( P(X \geq 7.5) \), making our approximation more precise.

The cumulative distribution function (CDF) helps determine the probability that a random variable is less than or equal to a specific value. For a continuous probability distribution, the CDF is expressed as \( F(x) = P(X \leq x) \). This function is extremely helpful in finding probabilities in our exercise when dealing with normal approximations.

Once we apply the continuity correction and convert our problem to a normal distribution setup, we utilize the CDF to find these probabilities.

• For part (a) of the exercise, we calculate \( P(4.5 < X < 8.5) \), which involves finding \( F(8.5) - F(4.5) \).
• For part (b), \( P(X \geq 7.5) = 1 - P(X < 7.5) \), which uses the CDF to find \( F(7.5) \) and subtracting it from 1.

Computing these CDF values for a normal distribution often involves looking up standard normal distribution tables or using computational tools to find the required probabilities.

A random variable is a concept that helps describe numerical outcomes of random processes. In probability and statistics, it can be thought of as a function that assigns a real number to each outcome of a statistical experiment. In our gene-carrying example, the random variable \( X \) represents the number of people in a sample of 1000 who carry a specific defective gene.

Understanding the nature of a random variable is crucial because it sets the stage for identifying the distribution type and computing probabilities based on that distribution.

• Discrete random variables take on specific, isolated values, like our variable \( X \) which counts people.
• Continuous random variables can assume any value within a given range and are typically associated with measurements.

Therefore, by defining \( X \) in the context of this exercise as the number of gene carriers among 1000 individuals, it becomes possible to apply the binomial distribution, and subsequently, approximate it using a normal distribution for probability calculations.