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16 Mathematics of Normal Distributions
16.1 Approximately Normal Distributions
of Data
16.2 Normal Curves and Normal
Distributions
16.3 Standardizing Normal Data
16.4 The 68-95-99.7 Rule
16.5 Normal Curves as Models of RealLife Data Sets
16.6 Distribution of Random Events
16.7 Statistical Inference
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 2
Standardizing the Data
We have seen that normal curves don’t all
look alike, but this is only a matter of
perception. In fact, all normal distributions
tell the same underlying story but use
slightly different dialects to do it. One way to
understand the story of any given normal
distribution is to rephrase it in a simple
common language–a language that uses
the mean  and the standard deviation  as
its only vocabulary. This process is called
standardizing the data.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 3
z-value
To standardize a data value x, we measure
how far x has strayed from the mean 
using the standard deviation  as the unit of
measurement. A standardized data value is
often referred to as a z-value.
The best way to illustrate the process of
standardizing normal data is by means of a
few examples.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 4
Example 16.4 Standardizing Normal
Data
Let’s consider a normally distributed data set
with mean  = 45 ft and standard deviation
 = 10 ft. We will standardize several data
values, starting with a couple of easy cases.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 5
Example 16.4 Standardizing Normal
Data
■
x1 = 55 ft is a data point located 10 ft above
(A in the figure) the mean  = 45 ft.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 6
Example 16.4 Standardizing Normal
Data
■
Coincidentally, 10 ft happens to be exactly
one standard deviation. The fact that
x1 = 55 ft is located one standard deviation
above the mean can be rephrased by
saying that the standardized value of
x1 = 55 is z1 = 1.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 7
Example 16.4 Standardizing Normal
Data
■
x2 = 35 ft is a data point located 10 ft (i.e.,
one standard deviation) below the mean (B
in the figure). This means that the
standardized value of x2 = 35 is z2 = –1.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 8
Example 16.4 Standardizing Normal
Data
■
x3 = 50 ft is a data point that is 5 ft (i.e.,
half a standard deviation) above the mean
(C in the figure). This means that the
standardized value of x3 = 50 is z3 = 0.5.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 9
Example 16.4 Standardizing Normal
Data
■
x4 = 21.58 is ... uh, this is a slightly more
complicated case. How do we handle this
one? First, we find the signed distance
between the data value and the mean by
taking their difference (x4 – ). In this case
we get 21.58 ft – 45 ft = –23.42 ft. (Notice
that for data values smaller than the mean
this difference will be negative.)
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 10
Example 16.4 Standardizing Normal
Data
■
If we divide this difference by  = 10 ft, we
get the standardized value z4 = –2.342. This
tells us the data point x4 is –2.342 standard
deviations from the mean  = 45 ft (D in the
figure).
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 11
Standardizing Values
In Example 16.4 we were somewhat
fortunate in that the standard deviation was
 = 10, an especially easy number to work
with. It helped us get our feet wet. What do
we do in more realistic situations, when the
mean and standard deviation may not be
such nice round numbers? Other than the
fact that we may need a calculator to do the
arithmetic, the basic idea we used in
Example 16.4 remains the same.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 12
STANDARDIZING RULE
In a normal distribution with mean  and
standard deviation , the standardized
value of a data point x is z = (x – )/.
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Excursions in Modern Mathematics, 7e: 16.3 - 13
Example 16.5 Standardizing Normal
Data: Part 2
This time we will consider a normally
distributed data set with mean  = 63.18 lb
and standard deviation  = 13.27 lb. What is
the standardized value of x = 91.54 lb?
This looks nasty, but with a calculator, it’s a
piece of cake:
z = (x – )/ = (91.54 – 63.18)/13.27
= 28.36/13.27 ≈ 2.14
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Excursions in Modern Mathematics, 7e: 16.3 - 14
Example 16.5 Standardizing Normal
Data: Part 2
One important point to note is that while the
original data is given in pounds, there are no
units given for the z-value.
The units for the z-value are standard
deviations, and this is implicit in the very fact
that it is a z-value.
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Excursions in Modern Mathematics, 7e: 16.3 - 15
Finding the Value of a Data Point
The process of standardizing data can also
be reversed, and given a z-value we can go
back and find the corresponding x-value. All
we have to do is take the formula
z = (x – )/ and solve for x in terms of z.
When we do this we get the equivalent
formula x =  +  •z. Given , , and a value
for z, this formula allows us to
“unstandardize” z and find the original data
value x.
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Excursions in Modern Mathematics, 7e: 16.3 - 16
Example 16.6 “Unstandardizing” a
z-Value
Consider a normal distribution with mean
 = 235.7 m and standard deviation
 = 41.58 m.
What is the data value x that corresponds to
the standardized z-value z = –3.45?
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 16.3 - 17
Example 16.6 “Unstandardizing” a
z-Value
We first compute the value of –3.45 standard
deviations:
–3.45 = –3.45  41.58 m = –143.451 m.
The negative sign indicates that the data
point is to be located below the mean.
Thus, x = 235.7 m – 143.451 m = 92.249 m.
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Excursions in Modern Mathematics, 7e: 16.3 - 18