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The normal approximation is a technique used when dealing with binomial distributions, which are inherently discrete. To apply the normal approximation, we convert the binomial distribution into a continuous one, specifically a normal distribution. This is particularly useful because normal distributions are often easier to work with due to their well-understood properties.
For the normal approximation to be valid, two key conditions need to be met:

• The product of the sample size and the probability of success, denoted as \(np\), should be at least 10.
• The product of the sample size and the probability of failure, denoted as \(n(1-p)\), should also be at least 10.

In the given exercise, with \(n = 25\) and \(p = 0.6\), both these conditions hold true as \(np = 15\) and \(n(1-p) = 10\). Therefore, the normal approximation can be reliably used in this scenario.

Understanding the mean and standard deviation is crucial when working with any probability distribution. These metrics give insight into the data's central tendency and variability.
For a binomial distribution, the mean \(\mu\) and standard deviation \(\sigma\) can be calculated using the following formulas:

• Mean \(\mu\) is given by \(np\), which denotes the expected value or average outcome.
• Standard deviation \(\sigma\) is given by \(\sqrt{np(1-p)}\), representing the degree of variation from the mean.

In our exercise, with \(n = 25\) and \(p = 0.6\), the mean is calculated as \(\mu = 25 \times 0.6 = 15\). The standard deviation is found by \(\sigma = \sqrt{25 \times 0.6 \times 0.4} = \sqrt{6} \approx 2.45\). Thus, these calculations help in approximating probabilities when using a normal distribution.

When transitioning from a discrete binomial distribution to a continuous normal distribution, a continuity correction is applied. This is essential to account for the difference between discrete and continuous representations of data.
In simpler terms, a continuity correction adjusts our point of interest by 0.5 units. This helps improve the approximation accuracy when assessing probabilities. For example, to approximate \(P(x > 9)\), instead of using 9 directly, we adjust to 9 - 0.5 = 8.5.
This adjustment ensures that the discrete nature of the original probability distribution is respected when calculating the corresponding z-score:\[ z = \frac{x - \mu}{\sigma} \]Applying the continuity correction, the z-score calculation becomes \(z = \frac{8.5 - 15}{2.45} \approx -2.65\). Using the standard normal distribution table or calculator, we find \(P(z \approx -2.65) \approx 0.004\). Consequently, the probability \(P(x > 9)\) is approximately 1 - 0.004, resulting in 0.996, thus demonstrating the effect and necessity of continuity correction.