example: z 3 = 1.5 tells us that the value 3 is 1.5 standard deviations above the mean. z 3 = –2 tells
us that the value 3 is two standard deviations below the mean.
example: For the first test of the year, Harvey got a 68. The class average (mean) was 73, and the
standard deviation was 3. What was Harvey’s z -score on this test?solution:
Thus, Harvey was 1.67 standard deviations below the mean.Suppose we have a set of data with mean and standard deviation s . If we subtract from every
term in the distribution, it can be shown that the new distribution will have a mean of . If we divide
every term by s , then the new distribution will have a standard deviation of s/s = 1. Conclusion: If you
compute the z -score for every term in a distribution, the distribution of z scores will have a mean of 0
and a standard deviation of 1.
Calculator Tip: We have used 1-Var Stats a number of times so far. Each of the statistics generated
by that command is stored as a variable in the VARS menu. To find, say, , after having done 1-Var
Stats on L1 , press VARS and scroll down to STATISTICS . Once you press ENTER to access the
STATISTICS menu, you will see several lists. is in the XY column (as is Sx ). Scroll through the other
menus to see what they contain. (The EQ and TEST menus contain saved variables from procedures
studied later in the course.)
To demonstrate the truth of the assertion about a distribution of z -scores in the previous paragraph,
do 1-Var Stats on, say, data in L1 . Then move the cursor to the top of L2 and enter (L1 – )/Sx ,
getting the and Sx from the VARS menu. This will give you the z -score for each value in L1 . Now do
1-Var Stats L2 to verify that = 0 and Sx = 1. (Well, for , you might get something like 5.127273E–- That’s the calculator’s quaint way of saying 5.127 × 10–14 , which is 0.00000000000005127.
That’s basically 0.)
You need to be aware that only the most recently calculated set of statistics will be displayed in
the VARS Statistics menu—it changes each time you perform an operation on a set of data.
Normal Distribution
We have been discussing characteristics of distributions (shape, center, spread) and of the individual
terms (percentiles, z -scores) that make up those distributions. Certain distributions have particular
interest for us in statistics, in particular those that are known to be symmetric and mound shaped. The
following histogram represents the heights of 100 males whose average height is 70′′ and whose standard
deviation is 3′′.