
The distribution of heights of adult
American men is approximately normal
with mean 69 inches and standard
deviation 2.5 inches. Use the 68-95-99.7
rule to answer the following questions:




What percent of men are taller than 74
inches?
Between what heights do the middle 95% of
men fall?
What percent of men are shorter than 66.5
inches?
A height of 71.5 inches corresponds to what
percentile of adult male American heights?
2.2
More on Normal Distributions
and
Standard Normal Calculations
Standardizing



The standardized value
is called a z-score. x is
the given value.
This tells you how many
standard deviations you
are from the mean.
This also allows you to
find the percent of data
under a given part of the
curve.
z
x

Who’s Taller? (relatively speaking)


Verne is 67” tall. Assume the heights of
women her age are normally distributed
with a mean μ = 64 inches and standard
deviation σ = 2.5 inches.
Hank is 72” tall. Assume the heights of
men his age are normally distributed with
a mean μ = 69.5 inches and standard
deviation σ = 2.25 inches.
Example

The score for each student on a quiz is
determined and a histogram is created
from the data. It is bell-shaped and
symmetric with a mean of 80 and
standard deviation of 10. Interpret a zscore for a student who scored a 87 on
the quiz.
Standardized Normal Curves

Recall our formula for standardizing normal
curves.
z

x

Since any normal curve can be standardized, we
can find areas under the curve using one table,
Table A. This table is found in the front of your
book or in your folder.
Table A


It is very important to remember that
Table A gives the area under the curve to
the LEFT!!!
Also, standardized normal curves have a
mean of 0 and a standard deviation of 1.
Reading Table A

Use Table A to find the proportion of
observations that have a z-score less than
1.4 (this is 1.4 standard deviations from
the mean).
Find the
hundredths
digit across the
top of the
table. In this
case, the
hundredths
digit is 0.
Find the
ones and
the tenths
The answer
is in
the
digits
intersection:
this .9192
column.
P(Z<1.4)=.9192
Reading Table A:
“Greater Than” Problems

Use Table A to find the proportion of
observations greater than a z-score of
-2.15.
Table A gives us .0158 for
the area to the LEFT. We
want the area to the RIGHT
(greater than -2.15), so
subtract from 1.
P(Z>-2.15) = 1-.0158 =
.9842
Steps in Finding Normal Proportions


Step 1: Draw a picture of the distribution and
shade the area of interest . Label the curve with
the values given (center and important points).
Step 2: Standardize x by using the formula.
z
x

Label your picture with the standardized values.
 Step 3: Use Table A to find the area under the
curve.
 Step 4: State your conclusion in words in the
context of the problem.
Now to the actual problems…

A commonly used IQ “cut-off” score for
AIG identification is 125. IQ scores on the
WISC-IV are normally distributed with a
mean = 100 and a standard deviation =
15. Find the proportion of people whose
IQ score is at least 125.
“Between” Problems

IQs between 140 and 170 are commonly
referred to as “moderately profoundly
gifted.” What proportion of the population
have IQ scores between 140 and 170?
Working Backwards

Scores on the SAT Verbal approximately
follow the N(505,110) distribution. How
high must a student score to be in the top
10% of all students taking the SAT?
Caution about Test Items

Many test items
ask students to
distinguish
between types of
density curves.
Once the hear the
word, students
have a tendency
to call everything
“normal.” Be
careful!
Density Curves
Skewed
Bimodal – Two
peaks… Looks
kinda like a camel
Just because a
curve is symmetric,
has one peak and is
bell shaped does not
mean it is a
NORMAL curve!!!
Symmetric
Unimodal
Bell Shaped
Normal Curves
68-95-99.7 Rule
What if they don’t tell me whether
the data are from a normal
population?
If you’re given the data, you have several
ways to assess normality.
Start by looking at a histogram, stemplot,
dotplot, or box-and-whisker plot. Does
the data appear symmetrical, with most of
the data being near the center?
And of course there is always the
Empirical Rule…
Another method is to check the 68-95-99.7
rule. First, find the mean and standard
deviation. Then count what percent of the
observations fall within one standard
deviation of the mean. Is it close to 68%?
Repeat for 2 and 3 standard deviations
away from the mean.
Normal Probability Plots


Another (and easier ) method is to
construct a normal probability plot using
your calculator.
If the plot is approximately linear, it is
safe to assume the data are from a normal
distribution.
Constructing Normal Prob. Plots


Type your data in your calculator (it is
probably already there, because I know
you have looked at your histogram or boxand-whisker plot!).
Go to StatPlot. Choose the last graph
option. This represents Normal
Probability.
Lets see what we can come up
with!
Let’s look at the example on page 125
Homework
Chapter 2#53-56, 60