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Standard deviation is a measure of the amount of variation or dispersion in a set of values. It's an essential concept in statistics, giving insight into the spread of data points around the mean. In a Normal distribution, standard deviation determines the width of the bell curve. The smaller the standard deviation, the narrower and steeper the curve. Conversely, a larger standard deviation indicates a wider and flatter curve.

In the context of the exercise, the mean score is 70, and the standard deviation is 10. This means that most students' scores lie within 10 points of the mean. Normal distributions follow a specific rule called the Empirical Rule or 68-95-99.7 rule. This rule states:

• About 68% of data points fall within one standard deviation (from 60 to 80 in this case)
• 95% lie within two standard deviations (50 to 90)
• 99.7% are within three standard deviations (40 to 100)

The standard deviation helps identify how scores deviate from the mean and can hint towards outliers or unusually high or low performances.

The Z-score, also known as the standard score, indicates how many standard deviations an element is from the mean. It is a way to standardize the data point within the distribution, allowing comparison between different datasets.

The Z-score is calculated with the formula: \[ Z = \frac{X - \mu}{\sigma} \]Where:

• \(Z\) is the Z-score
• \(X\) is the value of the data point
• \(\mu\) is the mean of the distribution
• \(\sigma\) is the standard deviation

In our exercise, the score of 80 yielded a Z-score of 1. This means 80 is one standard deviation above the mean score of the exam. Z-scores are valuable for understanding how extreme a particular score is in comparison to the average. They are also necessary for calculating probabilities associated with different parts of the distribution.

Probability is the measure of the likelihood that an event will occur. Under a Normal distribution, different scores have different probabilities depending on their position in the distribution. Probability calculations help determine the area under the curve for specific ranges, which represents the likelihood of obtaining those scores.

In the given problem, finding the probability that a score is greater than 80 requires the use of Z-scores and a standard Normal distribution table. After computing the Z-score for 80, which is 1, you can reference the table to find the probability corresponding to a Z-score of 1. This table provides the cumulative probability from the far left to any given Z-score.

For Z = 1, the cumulative probability is 0.8413, meaning 84.13% of students score below 80. To find the probability of scores above 80, you subtract this from 1: \[ 1 - 0.8413 = 0.1587 \]The result, 0.1587, indicates there is a 15.87% chance of randomly selecting a student who scored above 80. Understanding these probabilities can assist in gauging the commonality or rarity of specific scores in a dataset.