Download Find the indicated z score

The Normal Distribution
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1
Properties of The Normal
Distribution

The curve is bell-shaped with the
highest point over the mean, .
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2
Properties of The Normal
Distribution

The curve is symmetrical about a
vertical line through .
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3
Properties of The Normal
Distribution

The curve approaches the horizontal
axis but never touches or crosses it.
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4
Properties of The Normal
Distribution
–


The transition points between cupping
upward and downward occur
above  +  and  –  .
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5
The Empirical Rule
Approximately 68% of the data values lie is
within one standard deviation of the mean.
68%

One standard deviation from the mean.
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6
The Empirical Rule
Approximately 95% of the data values lie within
two standard deviations of the mean.
95%
x
Two standard deviations from the mean.
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7
The Empirical Rule
Almost all (approximately 99.7%) of the data
values will be within three standard deviations of
the mean.
99.7%
x
Three standard deviations from the mean.
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8
Application of the Empirical
Rule
The life of a particular type of light bulb
is normally distributed with a mean of
1100 hours and a standard deviation of
100 hours.
What is the probability that a light bulb of
this type will last between 1000 and 1200
hours?
Approximately 68%
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9
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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10
Formula for z score
x
z

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11
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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12
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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13
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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14
Standard Normal Distribution:

=0

=1
-1
0
1
Values are converted to z

scores wherexz =

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15
Importance of the Standard
Normal Distribution:
Standard
Normal
Distribution:
Any Normal
Distribution:
0
1
Areas will be equal.

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1
16
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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17
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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18
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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19
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
20
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
21
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on either
side of zero:
Subtract area to left of z1 from area to left
of z2 .
z1
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0
z2
22
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on the
same side of zero:
Subtract area to left of z1 from area to left
of z2 .
0
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z1
z2
23
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
24
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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25
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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26
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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27
Normal Probability
j.
.9236
P ( z > - 1.43 ) = __________
k.
.0034
P( z < - 2.71 ) = __________
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28
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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29
Inverse Normal Distribution
Finding z scores when probabilities
(areas) are given
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30
Find the indicated z score:
Find the indicated z score:
.8907
0
Copyright (C) 2006 Houghton Mifflin Company. All rights reserved .
z=
1.23
31
Find the indicated z score:
.6331
.3669
z = – 0.34
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32
Find the indicated z score:
.3560
.8560
0
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z=
1.06
33
Find the indicated z score:
.4792
.0208
z = – 2.04
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0
34
Find the indicated z score:
.4900
0
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z=
2.33
35
Find the indicated z score:
.005
z = – 2.575
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0
36
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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37