Z-Scores, Standardization, and the Standard Normal Distribution (5.3)

This section introduces the standard normal distribution, which has a mean of 0 and a standard deviation of 1. Z-scores indicate how many standard deviations an observation is from the mean. A Z-score table helps find the exact area (proportion) associated with a specific Z-score.

The Z-score table shows the total area to the left of any Z-value. To find the area to the right, subtract the Z-score's left-side area from 1. The table can also be used for reverse look-ups to find the Z-score corresponding to a given area.

Standardization is the process of transforming any normal distribution into a standard normal distribution (mean of 0, standard deviation of 1). This allows the use of the Z-score table to calculate exact areas for any normally distributed population using the formula: Z = (X - μ) / σ.

An example demonstrates standardizing chemistry exam scores with a mean of 60 and standard deviation of 10. By converting X-values to Z-scores, the distribution aligns with the standard normal distribution, where the mean always becomes 0 and standard deviation 1.

Another example calculates the proportion of students with heights between 5.81 and 6.3 feet from a normally distributed population with a mean of 5.5 feet and standard deviation of 0.5 feet. This involves standardizing both height values, finding their respective areas, and subtracting them to get the proportion of the shaded region.

This part illustrates how to find the proportion of students scoring less than 49 on the exam by standardizing 49 to a Z-score of -1.1 and then using the Z-score table to find the corresponding area.