Calculating and Interpreting Z-Scores

Date: 12/13/2005 at 14:04:49
From: Catherine
Subject: Z-Scores

Please help me by providing a step-by-step process for calculating the
z-score, or standard score.  My class textbook is too confusing.

There seem to be many ways of calculating z-scores, but no standard 
method that I can apply.  I know how to do standard deviation, but I
fall apart when it comes to the z-scores.  Can you help?

Date: 12/13/2005 at 17:28:48
From: Doctor Minter
Subject: Re: Z-Scores

Hi Catherine!

A z-score is used in statistics to model any normal distribution as a 
standard normal distribution.  You can consider it "rescaling" the 
original distribution to fit the properties of a normal distribution.  
Recall that a normal distribution has a mean of 0 and a standard 
deviation of 1.

If you understand percentages, the concept is very much the same.  If 
you know what proportion of people in a room have brown hair, say 
three out of four, you know that 75% of the people in the room have 
brown hair.

Does that mean that there are exactly 100 people in the room and that 
exactly 75 of them have brown hair?  That could be the situation, but 
what if there are only 20 people in the room?  

Percentages rescale the actual population.  That is, the percentage 
tells you how many people in the room WOULD have brown hair if there 
WERE 100 people, while still having the same proportion as the 
original group.

A z-score does something not just similar, but EXACTLY like this to a 
normal distribution to rescale it to a standard normal distribution.  
As percentages rescale a proportion to a group of 100, z-scoring 
rescales any normal distribution to a standard normal distribution.

Let's do an example of a z-score rescaling:

Let's say we have a normal distribution with mean 10 and standard 
deviation 2.  We'd like to know the probability that a random sampling 
will produce a value greater than or equal to 11.5.

The standard normal distribution is a bell curve that peaks at zero 
when graphed, and has a standard deviation of 1.  We need to rescale 
our distribution do the same.  The z-score conversion formula is

        y - u
    z = -----
          s

where y (you might also see it as x, or even another symbol) is our 
sample value (11.5 in this case), u (the mean, usually denoted by the 
Greek letter mu, but unfortunately the keyboard confines me to the 
letters in our alphabet only!  In this case, u is 10.), and s is the 
standard deviation, usually denoted by the Greek letter sigma, which 
has a value of 2 in this case.

Let's find our z-score:

       11.5 - 10
   z = ---------  = 0.75
           2

So our z-score is 0.75.  What does this tell us?  This value says that 
if we were to obtain a value greater than or equal to 11.5 by sampling 
our original distribution, the probability of doing so is the same as 
the probability of obtaining a value greater than or equal to 0.75 by 
sampling the standard normal distribution.  

If we look at the table for this particular z-score, we see that the 
probability of a random sampling of the normal distribution yielding 
a value greater than 0.75, which is equal to the probability of a 
random sampling of the original distribution yielding a value greater 
than 11.5, is equal to 0.2266.  

That is, 22.66% of random samples from the standard normal 
distribution will yield values greater than or equal to 0.75, and 
correspondingly, 22.66% of the random samples of the original 
distribution will yield values greater than or equal to 11.5.

The reason that we do this is because statistics texts have a table 
of z-scores and their corresponding probabilities.  That handy table
of values gives us the probability of z (value obtained by a random
sampling of the standard normal distribution) being greater than the
z-score obtained by the above formula.  That probability is exactly
equal to the probability that the example problem asked us to find!  

This process enables us to find the probability that a random sample 
from ANY normal distribution (there are other distributions also, and 
this formula does NOT work for them, by the way) is greater than a 
certain given value.  

If you want to find the probability that a sample produces a value 
LESS than a certain value, keep in mind that the standard normal 
distribution is symmetric about zero, and that its total area is 1.  
You can use these two properties to adjust the process to yield the 
desired answer.  For example, the probability of a random sampling of 
our original distribution yielding a value less than 11.5 is equal to 
the probability that we found earlier subtracted from 1, or 0.7734.

I hope that this will help you get started.  Please feel free to 
write again if you need further assistance, or if you have any other 
questions.  Thanks for using Dr. Math!

- Doctor Minter, The Math Forum
  http://mathforum.org/dr.math/