91Ó°ÊÓ

When dealing with normal distribution in statistics, two key parameters are fundamental: the mean and the standard deviation. The mean, often symbolized by \( \mu \), represents the average value in the dataset. In our problem, the average body temperature for the healthy men studied is \(98.1^\circ \mathrm{F}\).
This average gives us a central point around which other values are dispersed. Now, speaking of dispersion, the standard deviation \( \sigma \) measures how spread out the numbers are in relation to the mean. In our exercise, it is \(0.70^\circ \mathrm{F}\), indicating the temperatures don't vary wildly. The smaller the standard deviation, the closer the data points tend to be to the mean.

The z-score is a powerful statistic that tells us how many standard deviations an individual data point is from the mean. To calculate the z-score for a given value, we use the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value we're comparing to the mean \( \mu \), and \( \sigma \) is the standard deviation.
In the body temperature study, we're interested in how a temperature of \(98.6^\circ \mathrm{F}\) compares to the average based on our standard deviation. Substituting the figures into our formula yields a z-score around 0.71. This calculation shifts our interpretation of data from the original values to a standardized scale, making it easier to understand where a particular value like \(98.6^\circ \mathrm{F}\) stands in relation to the group average.

The percentile rank is a relative score that tells us the percentage of data points in a distribution that fall below a particular value. This concept is incredibly useful when trying to position an individual score within a larger context.
To find the temperature corresponding with a particular percentile, we turn to the z-score. For example, to determine the 76th percentile of our body temperature study, we looked up the z-score that marks the point below which 76% of the values lie. The closer we get to 100th percentile, the fewer instances we would find above this value.
By locating a z-score of approximately 0.72, which aligns with our 76th percentile, and applying the reverse z-score formula, we were able to find the corresponding body temperature, concluding it is roughly \(98.6^\circ \mathrm{F}\). This concept allows us to map individual instances onto the broader landscape of our dataset.