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Understanding the z-score is essential for handling problems involving normal distributions in statistics. A z-score, also known as a standard score, represents the number of standard deviations a data point is from the mean of a set of data. In simpler terms, it measures how far a specific point is from the norm, or average, and in which direction.

Calculating a z-score involves the formula:
\( z = \frac{{x - \mu}}{{\sigma}} \)
where \( x \) is the data point, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation. A positive z-score indicates that the data point is above the mean, while a negative score indicates it is below the mean.

To illustrate, in the given problem, we calculated the z-score using the provided percentage and determined it to be approximately -1.88. It implies that the weight of 15 lb is 1.88 standard deviations below the mean weight of Mr. Smith's ripe watermelons.

The standard deviation is a measure of how spread out numbers are in a data set. It is a key concept in statistics that lets us quantify the amount of variation or dispersion there is from the average (mean).

A low standard deviation means that most of the numbers are close to the mean. A high standard deviation indicates a greater range of values and more spread out numbers. The standard deviation is the square root of the variance, which averages the squared distances from the mean.

In our exercise, the standard deviation given is 2.8 lb, which means that the weights of the watermelons typically vary by 2.8 lb from the average weight of the melons. This value is pivotal to understanding the distribution of weights and is used directly in calculating the mean when we have a z-score, as demonstrated in the solution.

The mean, often referred to as the average, is a central aspect of any normal distribution. In a perfectly shaped bell curve, it is both the midpoint and the peak. All individual data points contribute to the calculation of the mean, which determines the center of the distribution's bell curve.

When dealing with a normal distribution and you need to find the mean given a certain percentile (like 3% in our problem), you can rearrange the z-score formula to solve for the mean:
\( \mu = x - z \times \sigma \).
After calculating the z-score related to the specific percentile and knowing the standard deviation, this equation allows us to compute the mean. In our case, the mean was found to be 20.264 lb, indicating that on average, Mr. Smith's ripe watermelons weigh just over 20 lb, with 3% of them weighing less than 15 lb.