Chapter Six
Normal Curves and
Sampling Probability
Distributions
Chapter 6
Section 2
Standard Units and
Areas Under the
Standard Normal
Distribution
Z Score
• The z value or z score tells the number
of standard deviations the original
measurement is from the mean.
• The z value is in standard units.
Formula for z score
z
x

Calculating z-scores
The amount of time it
takes for a pizza delivery
is approximately normally
distributed with a mean of
25 minutes and a standard
deviation of 2 minutes.
Convert 21 minutes to a
z-score.
z
z
x 

21  25
2
z
4
2
z   2 .0 0
Calculating z-scores
z
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.
z
x 

2 9 .7  2 5
2
z
4 .7
2
z  2 .3 5
Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.6.
z
x
1.6 

x  25
2
3.2  x  25
The delivery time is 28.2
minutes.
28.2  x
Standard Normal Distribution:
μ =0
σ =1
4
3
2
1
1
2
3
Values are converted to z scores where
z 
x 

4
Importance of the Standard
Normal Distribution:
Standard
Normal
Distribution:
0
Any Normal
Distribution:
1
The areas are equal.

1
Use of the Normal
Probability Table
(Table 4) - Appendix I
Entries give the probability that a
standard normally distributed random
variable will assume a value between the
mean (zero) and a given z-score.
To find the area between
z = 0 and z = 1.34
Z-Scores
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
1.1
0.3643
0.3665
0.3686
0.3708
0.3729
0.3749
0.3770
1.2
0.3849
0.3869
0.3888
0.3907
0.2925
0.3944
0.3962
1.3
0.4032
0.4049
0.4066
0.4082
0.4099
0.4115
0.4131
1.4
0.4192
0.4207
0.4222
0.4236
0.4251
0.4365
0.4279
Patterns for Finding
Areas Under the
Standard Normal Curve
To find the area between
a given z value and zero:
Use Table 4 (Appendix I) directly.
0
z
Patterns for Finding
Areas Under the
Standard Normal Curve
To find the area between
z values on either side of zero:
Add area from z1 to zero to
area from zero to z2 .
z1
0
z2
Patterns for Finding
Areas Under the
Standard Normal Curve
To find the area between
z values on the same side of zero:
Subtract area from zero to z1 from
the area from zero to z2.
0
z1
z2
Patterns for Finding
Areas Under the
Standard Normal Curve
To find the area to the right of a positive z value or
to the left of a negative z value:
Subtract the area from zero to z from 0.5000 .
0.5000
0
z
Patterns for Finding
Areas Under the
Standard Normal Curve
To find the area to the left of a positive z value
or to the right of a negative z value:
Add 0.5000 to the area from zero to z .
0.5000
table
z
0
Use of the Normal
Probability Table
a.
0.3925
P(0 < z < 1.24) = _________________
0.4452
b. P(0 < z < 1.60) = _________________
c.
0.4911
P( - 2.37 < z < 0) = ________________
Normal Probability
d. P( - 3 < z < 3 ) = ____________________
0.9974
e. P( - 2.34 < z < 1.57 ) = ______________
0.9322
f. P( 1.24 < z < 1.88 ) = ________________
0.0774
g. P(-3.52<z< -0.98) = __________________
0.1633
Normal Probability
h. P(z < 1.64) = _________________
0.9495
i. P(z > 2.39) = __________________
0.0084
0.9115
j. P(z > -1.35) = _________________
0.2611
k. P(z < -0.64) = _________________
Application of the
Normal Curve
The amount of time it takes for a pizza delivery is
approximately normally distributed with a mean of 25
minutes and a standard deviation of 2 minutes. If you
order a pizza, find the probability that the delivery time
will be:
a. between 25 and 27 minutes.
a. ____________
0.3413
b. less than 30 minutes.
b. ____________
0.9938
c. less than 22.7 minutes.
c. ____________
0.1251
Application of the
Normal Curve
The amount of time it takes for a pizza delivery is
approximately normally distributed with a mean of 25
minutes and a standard deviation of 2 minutes. If you
order a pizza, find the probability that the delivery
time will be:
a.
between 25 and 27 minutes.
27  25 
 25  25
P
 z



2
2
2
0
P  z  
2
2
P 0  z  1 
0.3413
0
1
Application of the
Normal Curve
The amount of time it takes for a pizza delivery is approximately
normally distributed with a mean of 25 minutes and a standard
deviation of 2 minutes. If you order a pizza, find the probability
that the delivery time will be:
b.
30  25 

Pz 



2
less than 30 minutes.
5

Pz  

2
P  z  2.5 
0.5000  0.4938
2.5
0.9938
Application of the
Normal Curve
The amount of time it takes for a pizza delivery is approximately
normally distributed with a mean of 25 minutes and a standard
deviation of 2 minutes. If you order a pizza, find the probability
that the delivery time will be:
c.
less than 22.7 minutes.
22.7  25 

Pz 



2
2.3 

Pz 


2 
P  z  1.15 
0.5000  0.3749
-1.15
0.1251
Homework Assignments
Chapter 6
Section 2
Pages 274 - 276
Exercises: 1 - 49, odd
Exercises: 2 - 50, even