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Probability is a way of quantifying the likelihood of an event occurring. It ranges from 0 to 1, where 0 means the event is impossible and 1 means it is certain. Consider flipping a fair coin: since there are two equally likely outcomes, the probability of getting heads is 0.5 (or 50%). In the exercise, we're dealing with the probability of a muffler being replaced, which is 20%, meaning out of every set of five mufflers, on average, one would be replaced.

The binomial distribution becomes useful when we want to find out probabilities for a certain number of successes (replacements in this case) in a series of independent trials. However, when the sample size is large, like 400 purchases, calculating each possible outcome becomes impractical. Instead, we use a normal approximation to simplify the calculation, which leads us to the next concept.

Normal approximation is a technique used to estimate the probabilities of a binomial distribution when the number of trials, represented by 'n', is large. The binomial distribution of 'success' probabilities can be approximated to a normal distribution if the sample size and the probabilities meet certain conditions (np and n(1-p) are both greater than 5). Through this method, complex binomial probability calculations can be streamlined and approximated with a good degree of accuracy.

In our muffler scenario, instead of working out the probability for each number of replacements from 75 to 100, we use normal approximation. This simplification requires us to calculate the mean (average outcome) and standard deviation (measure of dispersion) for the distribution, and then use the z-score to find probabilities. The steps outlined in the solution highlight this process, converting a binomial problem into a normal distribution scenario for easier computation.

A z-score, also known as a standard score, measures the number of standard deviations an element is from the mean of the distribution. It's a way to compare results from different sets of data. A positive z-score indicates higher than average, while a negative z-score shows lower.

In the context of our textbook exercise, z-scores are used to find probabilities corresponding to specific numbers of mufflers replaced. We take the number of replacements, subtract the mean (80 mufflers), and divide by the standard deviation (8). This calculation places our observed value within a standardized normal distribution where we can easily find its probability. It's a powerful tool when dealing with problems like these, as it turns the abstract concept of 'how likely is this?' into a concrete number we can look up on the z-table or calculate using software. The solution process illustrates that extreme z-scores (like -3.75 for fewer than 50 replacements) often correspond to very unlikely events, thus raising red flags about the assumed probabilities.