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The Z-score helps us understand how far and in what direction a value is from the mean of the data. In this exercise, we use it to find out how unusual a specific service time is compared to the average service time of 2 minutes. The Z-score formula is given by:

• \( Z = \frac{X - \mu}{\sigma} \)
• \( X \) is the value we're interested in, for example, 1 minute.
• \( \mu \) is the average, here it's 2 minutes.
• \( \sigma \) is the standard deviation, which is 0.5 minutes.

The Z-score tells us how many standard deviations our value is from the mean. For instance, if the Z-score is \(-2\), it means the value is 2 standard deviations below the average. By converting data into Z-scores, we can use common Z-tables or software to find corresponding probabilities. This transforms different values into a common language of standard deviations, allowing for universal application of probability assessment.

Probability is the measure of the likelihood that an event will occur. In part a of the exercise, we're asked to find the probability that two customers are served in less than 1 minute each. Here’s how it's done:

• First, determine the probability for one customer to be served in less than 1 minute using its Z-score.
• Next, because both events are independent, multiply that probability by itself to find out the chance of both events happening simultaneously.

To tackle part b, where it's about the probability that at least one of four customers will need more than 2.25 minutes, we first find the likelihood a single customer takes more than 2.25 minutes using the Z-score calculation. Then considering the complementary approach:

• Calculate the probability that all customers are served in 2.25 or less minutes.
• Subtract this value from 1 to get the probability that at least one customer takes more than 2.25 minutes.

This exercise highlights different tools and methods used to find probabilities in various scenarios.

Independent events refer to those occurrences where the outcome of one event does not influence the other. In our exercise, when we check if two customers are served in under 1 minute, it's assumed that the service time for one customer has no bearing on the service time for another. This means:

• We can multiply their probabilities to find the joint probability. For example, if the probability for each to take less than a minute is \(0.16\), then the combined probability is \(0.16 \times 0.16 = 0.0256\) or 2.56%.

Similarly, when calculating the probability of at least one customer out of four taking more than 2.25 minutes, we're considering each visit as separate and independent. This independence greatly simplifies the calculations and is a key assumption in the use of many probabilistic models.

The standard normal distribution is a fundamental concept in statistics that represents a normal distribution with a mean of 0 and a standard deviation of 1. To use this in practice, we convert data to this scale using Z-scores, making it possible to use standard normal tables or software for probability calculations. For example, in part a of the exercise, converting a 1-minute service time into a Z-score allows us to apply the standard normal distribution table to find the probability of such an event. The transformation involves:

• Standardizing data points through Z-scores, aligning them with the standard normal curve.
• Utilizing the properties of the standard normal distribution, which includes the symmetry about the mean and known probabilities for different ranges of Z.

The standard normal distribution is a key tool because it's universal, allowing comparisons across different datasets and contexts once data is standardized. It effectively unlocks probability insights through its well-documented curve characteristics.