# Z-Distributions

## Introduction

Up to this point we have been concerned with how scores are distributed and how best to describe the distribution. There are several different measures, but the mean will be the measure that we use to describe the center of the distribution and the standard deviation will be the measure we use to describe the spread (dispersion) of the distribution. Knowing these two facts gives us ample information to make statements about the probability of observing a certain value within that distribution. If I know, for example, that the average I.Q. score is 100 with a standard deviation of 20, then I know that someone with an I.Q. of 140 is very smart. I know this because 140 deviates from the mean by twice the average amount as the rest of the scores in the distribution. Thus, it is unlikely to see a score as extreme as 140 because most the I.Q. scores are clustered around 100 and on average only deviate 20 points from the mean. The following example demonstrates how knowing the mean and standard deviation enables us to make probability statements. In this population, Nine (N=9) I.Q. scores are normally distributed. For this population what is the probability of observing a value that is 2 standard deviations away from the mean? ## Probability and Normal Distributions

Anytime a set of values is distributed normally the mean and standard deviation are good markers for the position of most of the scores in the distribution. We can count on the fact that a certain percentage of scores will fall within a certain range of the mean. For example, when N is fairly large and the population has a normal distribution the scores will be distributed about the mean as follows: Thus, we can make statements about the probability of observing any value in a distribution based solely on deviation of the value from the mean of the distribution.  For example, what is the probability that if I pick someone at random from the above population that they will be somewhere between 1 standard deviation below the mean and one standard deviation above the mean?  Since 68.26% lie within one standard deviation of the mean, then there is the same chance a random pick will yield a value in that range.

What is the probability that if I choose someone at random that they will be between the mean and one standard deviation above the mean?  Notice that this is exactly half the area in the last example.  This time we are only interested in the area between the mean and 1 standard deviation above the mean, so, 34.13% will be in that area.

Instead of relying on this limited information, we will use a table of probabilities (areas) associated with different z-scores.