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To comprehend the idea of a z-score, imagine you have a set of data that follows a common pattern, known as a normal distribution. A z-score, in simple terms, is a measure of how many standard deviations a particular data point is from the mean of the distribution.

To compute a z-score, the formula used is \( z = \frac{(X - \mu)}{\sigma} \), where \( X \) represents the value you're investigating, \( \mu \) is the mean average, and \( \sigma \) is the standard deviation, which is a measure of how spread out the numbers in the data set are.

In the context of the problem, if you want to find out the likelihood of there being at least 650 vehicles departing at the turnpike, a z-score tells you how far and in what direction that number deviates from the average, which is 500. In this case, a z-score of 2 implies the event is 2 standard deviations above the mean – a rare occurrence in a normal distribution, thereby indicating a lower probability.

A standard normal table, also known as a z-table, provides the probability of a z-score occurring in a normal distribution. It's a crucial tool for statisticians and students alike, converting z-scores to probabilities. Standard normal tables are structured such that they illustrate the area (probability) under the curve from the far left up to the z-score value.

After calculating a z-score, you then use the table to find the corresponding probability. This is where step-by-step problem-solving comes in hand, as shown in the exercise. The higher the absolute value of the z-score, the further it is from the mean, and thus the smaller the probability of occurrence. For the given z-scores in the problem, matching them to the table and understanding the area they represent translates to finding out the probability for various numbers of cars exiting the turnpike.

Probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. In the given problem, we are dealing with a normal distribution, which is a probability distribution that is symmetrically centered around its mean, representing the expected value.

In practice, various scenarios might ask for different probabilities such as at least, less than, or between certain values. The principles of probability distribution guide us to compute these by integrating the normal curve, or simply by using the standard normal table for simplicity. For question (b) and (c) from the exercise, the normal probability distribution is used to determine the chances of the number of cars being strictly between or inclusively between the numbers 400 and 550, respectively. We capture these probabilities using the areas under the curve, delineated by the calculated z-scores.