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The Normal
Distribution
Lecture 20
Section 6.3.1
Mon, Oct 9, 2006
The Standard Normal
Distribution



The standard normal distribution – The normal
distribution with mean 0 and standard deviation
1.
It is denoted by the letter Z.
Therefore, Z is N(0, 1).
The Standard Normal
Distribution
N(0, 1)
z
-3
-2
-1
0
1
2
3
Areas Under the Standard
Normal Curve



What is the total area under the curve?
What proportion of values of Z will fall below 0?
What proportion of values of Z will fall above 0?
Areas Under the Standard
Normal Curve




What proportion of
What proportion of
What proportion of
What proportion of
and +1?
values will fall below +1?
values will fall above +1?
values will fall below –1?
values will fall between –1
Areas Under the Standard
Normal Curve

It turns out that the area to the left of +1 is
0.8413.
0.8413
z
-3
-2
-1
0
1
2
3
Areas Under the Standard Normal
Curve

So, what is the area to the right of +1?
Area?
0.8413
z
-3
-2
-1
0
1
2
3
Areas Under the Standard Normal
Curve

So, what is the area to the left of -1?
Area?
0.8413
z
-3
-2
-1
0
1
2
3
Areas Under the Standard Normal
Curve

So, what is the area between -1 and 1?
Area?
0.8413
0.8413
z
-3
-2
-1
0
1
2
3
Areas Under the Standard
Normal Curve

We will use two methods.
Standard normal table.
 The TI-83 function normalcdf.

The Standard Normal Table



See pages 406 – 407 or pages A-4 and A-5 in
Appendix A.
The entries in the table are the areas to the left
of the z-value.
To find the area to the left of +1, locate 1.00 in
the table and read the entry.
The Standard Normal Table
z
.00
.01
.02
…
:
:
:
:
…
0.9
0.8159
0.8186
0.8212
…
1.0
0.8413
0.8438
0.8461
…
1.1
0.8643
0.8665
0.8686
…
:
:
:
:
…
The Standard Normal Table


The area to the left of 1.00 is 0.8413.
That means that 84.13% of the population is
below 1.00.
0.8413
-3
-2
-1
0
1
2
3
The Three Basic Problems

Find the area to the left of a:


Find the area to the right of a:


Look up the value for a.
a
Look up the value for a; subtract it
from 1.
Find the area between a and b:

a
Look up the values for a and b;
subtract the smaller value from the
larger.
a
b
Standard Normal Areas

Use the Standard Normal Tables to find the
following.
The area to the left of 1.42.
 The area to the right of 0.87.
 The area between –2.14 and +1.36.

TI-83 – Standard Normal Areas



Press 2nd DISTR.
Select normalcdf (Item #2).
Enter the lower and upper bounds of the
interval.
If the interval is infinite to the left, enter -E99 as the
lower bound.
 If the interval is infinite to the right, enter E99 as
the upper bound.


Press ENTER.
Standard Normal Areas

Use the TI-83 to find the following.
The area to the left of 1.42.
 The area to the right of 0.87.
 The area between –2.14 and +1.36.

Other Normal Curves


The standard normal table and the TI-83
function normalcdf are for the standard normal
distribution.
If we are working with a different normal
distribution, say N(30, 5), then how can we find
areas under the curve?
Other Normal Curves

For example, if X is N(30, 5), what is the area to
the left of 35?
15
20
25
30
35
40
45
Other Normal Curves

For example, if X is N(30, 5), what is the area to
the left of 35?
15
20
25
30
35
40
45
Other Normal Curves

For example, if X is N(30, 5), what is the area to
the left of 35?
?
15
20
25
30
35
40
45
Other Normal Curves

For example, if X is N(30, 5), what is the area to
the left of 35?
?
X
15
20
25
30
35
40
45
-3
-2
-1
0
1
2
3
Z
Other Normal Curves




To determine the area, we need to find out how
many standard deviations 35 is above average.
Since  = 30 and  = 5, we find that 35 is 1
standard deviation above average.
Thus, we may look up 1.00 in the standard
normal table and get the correct area.
The number 1.00 is called the z-score of 35.
Other Normal Curves

The area to the left of 35 in N(30, 5).
0.8413
X
15
20
25
30
35
40
45
-3
-2
-1
0
1
2
3
Z
Z-Scores



Z-score, or standard score, of an observation –
The number of standard deviations from the
mean to the observed value.
Compute the z-score of x as
or
xx
x
z
z
s

Equivalently
x  x  zs or x    z
Areas Under Other Normal
Curves


If a variable X has a normal distribution, then
the z-scores of X have a standard normal
distribution.
If X is N(, ), then (X – )/ is N(0, 1).
Example


Let X be N(30, 5).
What proportion of values of X are below 38?
Compute z = (38 – 30)/5 = 8/5 = 1.6.
 Find the area to the left of 1.6 under the standard
normal curve.
 Answer: 0.9452.


Therefore, 94.52% of the values of X are below
38.
TI-83 – Areas Under Other Normal
Curves


Use the same procedure as before, except enter
the mean and standard deviation as the 3rd and
4th parameters of the normalcdf function.
For example,
normalcdf(-E99, 38, 30, 5) = 0.9452.
IQ Scores




IQ scores are standardized to have a mean of
100 and a standard deviation of 15.
Psychologists often assume a normal
distribution of IQ scores as well.
What percentage of the population has an IQ
above 120? above 140?
What percentage of the population has an IQ
between 75 and 125?
The “68-95-99.7 Rule”

For a normal distribution, what percentages of
the population lie
within one standard deviation of the mean?
 within two standard deviations of the mean?
 within three standard deviations of the mean?


What does this tell us about IQ scores?
The Empirical Rule


The well-known Empirical Rule is similar, but
more general.
If X has a “mound-shaped” distribution, then
Approximately 68% lie within one standard
deviation of the mean.
 Approximately 95% lie within two standard
deviations of the mean.
 Nearly all lie within three standard deviations of the
mean.
