# 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.