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A bone density test is a medical procedure that measures the strength and density of bones. It is often used to diagnose osteoporosis and assess the risk of fractures. During this test, a machine measures the amount of calcium and other minerals in a segment of bone, typically the spine, hip, and forearm.

Bone density tests produce scores that indicate the bone's density compared to a healthy young adult's average density. These scores are usually presented as z-scores, which we'll explain later. The results can help doctors make decisions about the necessity of medication or lifestyle changes to strengthen bones.

In our exercise, the bone density test scores are given with a mean of 0 and a standard deviation of 1, which allows us to use the standard normal distribution to find probabilities related to different scores.

The standard normal distribution is a special type of normal distribution. It has a mean (\( \mu \))) of 0 and a standard deviation (\( \sigma \))) of 1. A normal distribution is symmetrical, meaning it forms a perfect bell curve when graphed.

This distribution allows us to use z-scores to find probabilities and percentages for different values within the distribution. In other words, if we know a particular z-score, we can determine how far that score is from the mean and what percentage of the distribution lies below that score.

For our bone density test example, since the scores are normally distributed with a mean of 0 and standard deviation of 1, we can directly use the standard normal distribution to find the probability of a given score. Specifically, to find the probability of a score less than 2.56, we use the cumulative distribution function (CDF) for the standard normal distribution.

A z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. It is calculated using the formula:
\[ z = \frac{x - \mu}{\sigma} \]
where \( x \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the context of the bone density test, we use the z-score to find out how far a particular test score is from the mean (which is 0) in terms of standard deviations. By doing this, we can use the standard normal distribution to find the probability associated with that z-score.

For example, a z-score of 2.56 means the bone density score is 2.56 standard deviations above the mean. Using the standard normal distribution (or z-table), we can find the cumulative probability for this z-score, which in this case is 0.9948. This means there is a 99.48% probability that a randomly selected bone density test score will be less than 2.56.