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When working with binomial distributions, if the sample size, denoted by \(n\), is large enough, the binomial distribution can be approximated by a normal distribution. This concept is known as the "Normal Approximation to the Binomial Distribution". It is particularly useful because normal distributions are continuous and therefore easier to work with in terms of finding probabilities.
This method is valid when the sample size \(n\) is large, and the probability \(p\) does not equal zero or one. More concretely, a good rule of thumb for applying this approximation is if both \(np\) and \(n(1-p)\) are greater than 5. In our case, \(n = 1000\) and \(p = 0.5\), which satisfies this condition.
The mean (\(\mu\)) and variance (\(\sigma^2\)) of the approximating normal distribution are calculated using the formulas \(\mu = np\) and \(\sigma^2 = np(1-p)\). For this exercise, we have \(\mu = 500\) and \(\sigma = \sqrt{250} \approx 15.81\). This allows us to use the properties of the normal distribution to easily find probability values.

The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. In practical terms, the z-score tells us how far and in what direction our statistic is from the expected result under the normal approximation.
To calculate the z-score for a particular value \(x\), use the formula: \[ z = \frac{x - \mu}{\sigma} \] where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the distribution. For this exercise, we are calculating the z-score for \(x = 750\). Given \(\mu = 500\) and \(\sigma = 15.81\), we find that: \[ z = \frac{750 - 500}{15.81} \approx 15.8 \] This indicates that the value 750 is 15.8 standard deviations away from the mean. Such a high z-score signifies a very rare event in a normal distribution.
Z-scores are highly useful as they allow us to determine probabilities efficiently using standard normal distribution tables or calculators.

Probability calculation using the normal approximation involves transforming our binomial problem into a question about the normal distribution. When the z-score is as high as 15.8, we can determine its probability using a standard normal distribution table or calculator.
The question is to find \(P(Z \geq 15.8)\), which represents the probability that a randomly selected value from the distribution is greater than 15.8 standard deviations from the mean. This probability is practically zero, indicating an outlier extreme. In such cases, software or z-tables can confirm this probability is close to zero for a normal distribution.
This essentially means that observing 750 or more sets being more dispersed than the opteron is extremely unlikely if the true probability \(p = 0.5\). Such exceptionally low probability values lead us to conclude that there is strong evidence against assuming a 0.5 probability for the dispersion, thus questioning the initial hypothesis significantly.