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Normal approximation allows us to use the normal distribution to make probability calculations for sample proportions. When you have a large enough sample size, the sampling distribution of the sample proportion - which tells us how these proportions vary with different samples - can be approximated as normal. This is because, according to the Central Limit Theorem, if the sample size is large, the distribution of the sample proportion should be approximately normal even if the population distribution is not.

In the exercise, we want to find the probability of getting less than 50% Democrats in a sample of 100 voters. Although the original data is of proportions, the normal approximation applies because of the adequate sample size (n = 100), allowing us to use normal distribution properties to calculate probabilities which would be harder with other distributions.

The standard error (SE) measures how much the sample proportion is expected to vary from the true population proportion. It serves as a measure of the sampling distribution's spread. To compute the SE, you use the formula: \[ \text{SE} = \sqrt{ \frac{p(1-p)}{n} } \]where \( p \) is the population proportion and \( n \) is the size of the sample. In our situation, with 55% of voters being Democrats, \( p = 0.55 \) and \( n = 100 \). By calculating this:

- \( p(1-p) \) = 0.55 x 0.45 = 0.2475- Dividing by 100: 0.2475/100 = 0.002475- Taking the square root gives us the SE: about 0.0497

The standard error tells us that different samples could have sample proportions varying by around 4.97% from the true proportion.

A Z-score indicates how many standard deviations an element is from the mean of a distribution. In the context of sample proportions, it helps to standardize the sample proportion observations to find probabilities using the standard normal distribution.

To find a Z-score, use the formula: \[ Z = \frac{\text{Sample Proportion} - \text{Mean of Sampling Distribution}}{\text{Standard Error}}.\]We look at our desired sample proportion, 0.50, and compare it to the mean, which is 0.55 in this case. The calculated SE is 0.0497. Substituting these values, the Z-score is: \[ Z = \frac{0.50 - 0.55}{0.0497} \]Calculating gives a Z-score of about -1.005, which tells us that 0.50 is 1.005 standard deviations below the mean of 0.55. The Z-score allows us to use standard normal distribution tables to calculate the probability of observing such a sample proportion or less.

The concept of a proportion is vital to understanding the original problem. In statistical terms, a proportion is an expression that tells us the quantity of a part with respect to the whole.

In this context, 55% of registered voters are Democrats. The proportion is: - \( p = 0.55 \) which indicates that 55 out of every 100 voters are Democrats. We seek the probability of a sample proportion being lower than 0.50, or less than 50 Democrats in a sample of 100.

Proportions are essential because they summarize data in a meaningful way, showing relationships and variations in categorical data sets. They also facilitate comparison between different samples and populations, particularly when it comes to approximating the sampling distribution of the sample proportion using methods like the normal approximation. Recognizing and knowing how to manipulate proportions allow for easier probability calculations and greater insight into data trends.