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.
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1
Formula for z score
x
z

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2
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.
x   21  25
z

 2.00

2
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3
Calculating z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.
x   29.7  25
z

 2.35

2
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4
Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.6.
x  z     1.6( 2 )  25  28 .2
The delivery time is 28.2 minutes.
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5
Standard Normal Distribution:

=0

=1
-1
0
1
Values are converted to z
x
scores where

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6
Importance of the Standard
Normal Distribution:
Standard
Normal
Distribution:
Any Normal
Distribution:
0
1
Areas will be equal.

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1
7
Use of the Normal Probability
Table
(Table 5) - Appendix II
Entries give the probability that a
standard normally distributed
random variable will assume a
value to the left of a given negative
z-score.
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8
Use of the Normal Probability
Table
(Table 5a) - Appendix II
Entries give the probability that a
standard normally distributed
random variable will assume a
value to the left of a given positive z
value.
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9
To find the area to the left of
z = 1.34
_____________________________________
z … 0.03
0.04
0.05 ..…
_____________________________________
.
.
1.2 … .8907
.8925
.8944 ….
1.3 … .9082
.9099
.9115 ….
1.4 … .9236
.9251
.9265 ….
.
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10
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the left of a given
negative z :
Use Table 5 (Appendix II) directly.
z
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0
11
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the left of a given
positive z :
Use Table 5 a (Appendix II) directly.
0
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z
12
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values
Subtract area to left of z1 from area to left
of z2 .
z1
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0
z2
13
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values, subtract
area to left of z1 from area to left of z2 .
0
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z1
z2
14
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the right of a positive z
value or to the right of a negative z value:
Subtract from 1.0000 the area to the left of the
given z.
Area under
entire curve
= 1.000.
0
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z
15
Use of the Normal Probability
Table
a.
.8925
P(z < 1.24) = ______
b.
.4452
P(0 < z < 1.60) = _______
c.
.4911
P( - 2.37 < z < 0) = ______
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Normal Probability
d.
.9974
P( - 3 < z < 3 ) = ________
e.
.9322
P( - 2.34 < z < 1.57 ) = _____
f.
.0774
P( 1.24 < z < 1.88 ) = _______
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Normal Probability
g.
.2254
P( - 2.44 < z < - 0.73 ) = _______
h.
.9495
P( z < 1.64 ) = __________
i.
.0084
P( z > 2.39 ) = _________
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Normal Probability
j.
.9236
P ( z > - 1.43 ) = __________
k.
.0034
P( z < - 2.71 ) = __________
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19
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.
.3413
a. ___________
b.
less than 30 minutes.
.9938
b. __________
c.
less than 22.7 minutes.
.1251
c. __________
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20
Inverse Normal Distribution
Finding z scores when probabilities
(areas) are given
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21
Find the indicated z score:
Find the indicated z score:
.8907
0
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z=
1.23
22
Find the indicated z score:
.6331
.3669
z = – 0.34
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23
Find the indicated z score:
.3560
.8560
0
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z=
1.06
24
Find the indicated z score:
.4792
.0208
z = – 2.04
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0
25
Find the indicated z score:
.4900
0
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z=
2.33
26
Find the indicated z score:
.005
z = – 2.575
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0
27
Find the indicated z score:
A
= .005
–z
B
0
z
 2.575 or  2.58
If area A + area B = .01, z = __________
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28
Application of Determining z
Scores
The Verbal SAT test has a mean score of
500 and a standard deviation of 100.
Scores are normally distributed. A major
university determines that it will accept
only students whose Verbal SAT scores
are in the top 4%. What is the minimum
score that a student must earn to be
accepted?
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29
...students whose Verbal SAT
scores are in the top 4%.
Mean = 500, standard deviation = 100
.9600
= .04
z = 1.75
The cut-off score is 1.75 standard deviations
above the mean.
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Application of Determining z
Scores
Mean = 500, standard deviation = 100
.9600
= .04
z = 1.75
The cut-off score is 500 + 1.75(100) = 675.
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31