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Probability is a way to quantify the likelihood of an event happening. It ranges from 0 to 1, where 0 means the event is impossible, and 1 means it is certain to happen.

For example, flipping a fair coin gives us two equally likely outcomes: heads or tails. Since these are the only two outcomes and both are equally likely, the probability of getting heads is \( \frac{1}{2} \) or 0.5, as is the probability of getting tails.

Now, let’s relate this to our exercise about the restaurant in Portland. The event we're interested in is 'vegans attending the opening night.' We know from the survey that about 4% of the population in Portland are vegans. The probability (\(p\)) of one randomly selected individual being a vegan is therefore 0.04. In a group of 300 patrons, we would expect \( 300 \times 0.04 = 12 \) vegans.

Here, we're concerned with whether or not there will be at least 15 vegans, which is more than what the chef prepared for. Calculating this involves understanding the distribution of probabilities for different numbers of vegans arriving and is where our binomial distribution and normal approximation come into play.

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. Mathematically, a binomial distribution can be quite complex to calculate, especially with a higher number of trials.

The normal approximation simplifies this calculation under certain conditions. When we have a large enough sample size and the probability of success isn’t too close to 0 or 1 (common thumb rule is \(np \) and \(n(1-p) \) both greater than 5), the binomial distribution can be approximated by the normal distribution thanks to the Central Limit Theorem.

In our case with the restaurant, given \(n = 300\) and \(p = 0.04\), it is appropriate to use the normal approximation because \(np = 12\) and \(n(1-p) = 288\), which are both greater than 5. With normal approximation, we can easily calculate probabilities using the Z-score method.

A Z-score measures how many standard deviations an element is from the mean. It's used to find the probability of a data point occurring within a normal distribution.

To calculate a Z-score in our context, we subtract the expected number of vegans (mean, \(\mu\)) from the number we are investigating (just above 14 in this case for 'more than 14', hence 14.5), and then divide by the standard deviation (\(\sigma\)). This tells us how far from the expected value our query is in terms of standard deviations.

After finding the Z-score, we use it to determine the probability we're looking for. This is done by subtracting the Z-score's cumulative probability from 1, as we're interested in the likelihood of an outcome greater than our Z-score. In this restaurant scenario, the calculated Z-score is approximately 0.456, and using the Z-table, we find the probability of having less than 15 vegans, then subtract this from 1 to get the final answer.

Remember, this is an approximation because we used the normal distribution instead of the true binomial distribution, which is valid due to the large sample size and other favorable conditions.