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The **standard deviation** of a sampling distribution is a measure that tells us how much the sample mean is expected to vary from the true population mean. Imagine it as a way to understand the "spread" or "dispersion" of the sample outcomes around the expected value. The smaller the standard deviation, the closer our sample estimate is likely to be to the actual population parameter.

In statistical terms, for a sample proportion like in our problem, the formula for standard deviation is:

• \( sd = \sqrt{\frac{p(1-p)}{n}} \)

Here, \(p\) is the proportion of the population, \(1-p\) represents the complement of that proportion, and \(n\) is the number of observations or sample size.

For our specific scenario, where \(p = 0.48\) and \(n = 500\), the calculation would involve substituting these values into the formula to find the precise measure of variability. Understanding the standard deviation helps us appreciate how typical or atypical certain sample results are, given the known characteristics of the population.

A **Z-score** is a statistical measure that describes a value's position relative to the mean of a group of values. It indicates how many standard deviations an element is from the mean. Z-scores are especially useful because they let us easily determine the probability of a score occurring within a normal distribution, as well as comparing two scores from different normal distributions.

To compute a Z-score, we use the formula:

• \( Z = \frac{X - \text{mean}}{\text{sd}} \)

Here, \(X\) is the value of interest, the mean is the average of the distribution, and \(sd\) is the standard deviation of the distribution. Using Z-scores transforms the data to a standard normal distribution where the mean is 0 and the standard deviation is 1.

In our exercise, calculating the Z-score helps determine how much the sample proportion \(\hat{p}\) deviates from the population proportion \(p\). Specifically, by changing the threshold \(0.5\) into a Z-score, we can use statistical tables or software to find out the probability of obtaining a sample proportion greater than this threshold.

The **normal distribution** is a continuous probability distribution characterized by its symmetric bell-shaped curve. It is fundamental in statistics because it models many natural phenomena. The key features of a normal distribution include:

• The mean, median, and mode of the distribution are equal.
• The curve is symmetrical around the mean.
• About 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations, following the empirical rule.

In the context of the problem, the sample proportion \(\hat{p}\) is approximately normal due to the large sample size (thanks to the Central Limit Theorem). Therefore, even though the original proportion \(p\) does not have to be normally distributed, the distribution of \(\hat{p}\) will approach normality, especially since \(n = 500\) is a sizeable number.

This normal distribution enables us to utilize Z-scores and probability tables to find the likelihood of observing sample outcomes, which is the crux of answering the exercise's question. Understanding the properties of normal distribution makes it easier to predict how sample data behaves relative to population parameters.