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The population proportion (denoted as \(p\)) represents the fraction of the entire population that possesses a particular characteristic. In our exercise, the population proportion, \(p\), is 0.42. This means that 42% of the population has the specified characteristic.
This proportion serves as a key reference point for many statistical calculations.
For any population, understanding the population proportion helps in predicting and generalizing the findings from a sample back to the whole population.
In this case, knowing that 42% of a population of 1,500,000 people has the characteristic helps in developing expectations about smaller samples from this population.

Sample size (denoted as \(n\)) is the number of observations or individuals in the sample. In our given problem, the sample size is \(n=1460\).
Sample size is crucial in determining the reliability of our estimations. Larger sample sizes generally provide more accurate estimates of the population parameters.
In our example, the large sample size of 1,460 helps to make our sampling distribution of \(\hat{p}\)\ more stable and closer to a normal distribution. Mathematically, increasing the sample size decreases the standard error of the sample proportion.

The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. The formula for the z-score in our context is:
\[z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \]
First, we calculate the sample proportion \( \hat{p} \) for a given \(x\). For instance, for \(x=657\), \( \hat{p}= \frac{657}{1460} = 0.45\).
Next, we find the z-score:
\[z = \frac{0.45 - 0.42}{0.0129} \approx 2.33\]
A positive z-score means that the value is above the mean, while a negative z-score means it is below the mean. The z-score tells us how far from the mean a particular \( \hat{p} \) is in terms of standard errors.

Probabilities provide insight into the likelihood of certain outcomes occurring. Using z-scores, we can find probabilities from standard normal distribution tables (z-tables).
For our exercise, we wanted to find two probabilities:

• The probability of obtaining \(x=657\) or more individuals with the characteristic:
• With z=2.33, the probability is approximately 0.0099 or 0.99%.

• The probability of obtaining \(x=584\) or fewer individuals with the characteristic:
• With z=-1.55, the probability is approximately 0.0606 or 6.06%.

These probabilities help us understand how likely or unlikely certain sample outcomes are, based on our knowledge of the population proportion.