Chapter 6
Normal Probability
Distributions
1
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Chapter 6
Normal Probability Distributions
6-1 Overview
6-2 The Standard Normal Distribution
1
6-3 Applications of Normal Distributions
6-2 and 66-3: Normal Distributions: Finding
Values
6-5 The Central Limit Theorem
2
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Review
x
(# of correct)
0
1
2
3
4
5
0.5
P(x)
.05
.10
.25
.40
.15
.05
Probability
Distribution
0.4
.40
P(x)
0.3
.25
0.2
.15
0.1
0.0
.05
0
.1
1
2
3
4
.05
5
# of correct answers
Probability Histogram
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Overview
v Continuous random variable
v Normal distribution
4
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Overview
v Continuous random variable
v Normal distribution
2
Curve is bell shaped
and symmetric
Figure 5 -1
µ
Score
5
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Overview
v Continuous random variable
v Normal distribution
Curve is bell shaped
and symmetric
Figure 6 -1
µ
Score
Formula 66-1
y=
e
1
2
σ
( xσ- µ )
2
2π
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6- 2
The Standard Normal
Distribution
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Definitions
v Density Curve
(or probability density function)
the graph of a continuous probability
distribution
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8
Definitions
v Density Curve
(or probability density function)
the graph of a continuous probability
distribution
1. The total area under the curve must
equal 1.
2. Every point on the curve must have a
vertical height that is 0 or greater.
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Because the total area under
the density curve is equal to 1,
there is a correspondence
between area and probability
probability..
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Definition
v Uniform Distribution
a probability distribution in which the
continuous random variable values are
spread evenly over the range of
possibilities; the graph results in a
rectangular shape.
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11
Uniform Distributions
P( 1º < x < 4º ) =
P(x )
3
0.2
x
0
0
1
2
3
4
5
Temperature (degrees Celsius)
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Uniform Distributions
Shaded Area = L • W = 3 • 0.2 = 0.6
P( 1º < x < 4º ) = 0.6
P(x )
3
0.2
x
0
0
1
2
3
4
5
Temperature (degrees Celsius)
13
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Uniform Distributions
P( 1º < x < 4º ) = 0.6
P(x )
P( x > 1º ) =
5
P(x )
3
0.2
0.2
x
0
0
1
2
3
4
x
0
5
0
Temperature (degrees Celsius)
1
2
3
4
5
Temperature (degrees Celsius)
14
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Uniform Distributions
Shaded Area = L • W = 4 • 0.2 = 0.8
P( 1º < x < 4º ) = 0.6
P(x )
P( x > 1º ) = 0.8
P(x )
3
0.2
0.2
x
0
0
1
2
3
4
5
Temperature (degrees Celsius)
x
0
0
1
2
3
4
5
Temperature (degrees Celsius)
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Heights of Adult Men and Women
Women:
µ = 63.6
σ = 2.5
Men:
µ = 69.0
σ = 2.8
63.6
Figure 5 -4
69.0
Height (inches)
16
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Normal Distributions
Different Locations
µ = 25
µ = 100
µ = 100
σ=8
σ =8
σ=8
Same Shape
0
25
50
Same
Location
µ = 100
σ = 15
75
100
125
6
Different
Shapes
150
175
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Definition
Standard Normal Distribution
a normal probability distribution that has a
mean of 0 and a standard deviation of 1,
1 , and the
total area under its density curve is equal to 1.
-3
-2
-1
0
1
2
3
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Example:
If thermometers have an average
(mean) reading of 0 degrees and a standard deviation
of 1 degree for freezing water and if one thermometer
is randomly selected, find the probability that it reads
freezing water is less than 1.58 degrees.
µ =0
σ=1
P(zz < 1.58) = ?
P(
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Methods for Finding Normal
Distribution Areas
vTable AA-2 (sometimes called zzdistribution or zz -table)
7
vSTATDISK
vOther computer programs
vTI
TI--83/84 Calculator
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Table AA -2
vFormulas and Tables card
v Appendix
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NEGATIVE Z Scores Table AA-2
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Table AA -2
v Designed only for standard normal distribution
v Is on two pages: negative z -scores and
positive z-scores
8
v Body of table is a cumulative area from the left
up to a vertical boundary
v Avoid confusion between zz-scores and areas
v Z-score hundredths is across the top row
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Table AA-2
Standard Normal Distribution
Negative z-scores: cumulative from left
x
z
0
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Table AA-2
Standard Normal Distribution
Positive zz -scores: cumulative from left
X
z
25
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Table AA-2
Standard Normal Distribution
µ=0
σ=1
9
z=x-µ
σ
X
z
26
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Table AA-2
Standard Normal Distribution
µ=0
σ=1
z=x-0
1
X
z
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Table AA-2
Standard Normal Distribution
µ=0
σ=1
z=x
X
z
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Table AA-2
Standard Normal Distribution
µ=0
σ=1
10
z=x
Area =
Probability
X
z
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To find:
z Score
the distance along horizontal scale of the
standard normal distribution; refer to the
leftmost column and top row of Table AA-2
Area
the region under the curve; refer to the
values in the body of Table AA-2
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Example:
If thermometers have an average
(mean) reading of 0 degrees and a standard deviation
of 1 degree for freezing water and if one thermometer
is randomly selected, find the probability that it reads
freezing water is less than 1.58 degrees.
µ =0
σ=1
P(zz < 1.58) = ?
P(
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POSITIVE z Scores
11
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POSITIVE z Scores
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Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water is
less than 1.58 degrees.
µ =0
σ=1
P(zz < 1.58) =
P(
0.9429
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34
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water is
less than 1.58 degrees.
µ =0
σ=1
12
P(zz < 1.58) =
P(
0.9429
The probability that the chosen thermometer will measure freezing
freezing
water less than 1.58 degrees is 0.9429.
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35
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water is
less than 1.58 degrees.
µ =0
σ=1
P(zz < 1.58) =
P(
0.9429
94.29% of the thermometers will read freezing water less than 1.58
1.58
degrees.
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = ?
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = ?
13
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = ?
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = 0.8907
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = 0.8907
14
The probability that the chosen thermometer with a
reading above - 1.23 degrees is 0.8907.
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Example:
If we are using the same thermometers,
and if one thermometer is randomly selected, find the
probability that it reads (at the freezing point of water)
above –1.23 degrees.
P (z
(z > – 1.23) = 0.8907
The percentage of thermometers with a reading
above - 1.23 degrees is 89.07%.
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (z
(z < – 2.00) = 0.0228
15
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P ((zz < – 2.00) = 0.0228
P (z
(z < 1.50) = 0.9332
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P ((zz < – 2.00) = 0.0228
P (z
(z < 1.50) = 0.9332
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (z
(z < – 2.00) = 0.0228
P (z
(z < 1.50) = 0.9332
P (–
(– 2.00 < z < 1.50) =
0.9332 – 0.0228 = 0.9104
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47
Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (z
(z < – 2.00) = 0.0228
P (z
(z < 1.50) = 0.9332
P (–
(– 2.00 < z < 1.50) =
0.9332 – 0.0228 = 0.9104
The probability that the chosen thermometer has a
reading between – 2.00 and 1.50 degrees is 0.9104.
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Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (z
(z < – 2.00) = 0.0228
P (z
(z < 1.50) = 0.9332
P (–
(– 2.00 < z < 1.50) =
0.9332 – 0.0228 = 0.9104
The percentage of thermometers that read
between – 2.00 and 1.50 degrees is 91.04%.
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The Empirical Rule
Standard Normal Distribution: µ = 0 and σ = 1
17
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The Empirical Rule
Standard Normal Distribution: µ = 0 and σ = 1
68% within
1 standard deviation
x
-
1s
x
x + 1s
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The Empirical Rule
Standard Normal Distribution: µ = 0 and σ = 1
95% within
2 standard deviations
68% within
1 standard deviation
x
-
x
2s
-
1s
x + 1s
x
x + 2s
52
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The Empirical Rule
Standard Normal Distribution: µ = 0 and σ = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
18
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x
-
3s
x
-
2s
13.5%
x
-
1s
x
x + 1s
x + 2s
x + 3s
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Notation
P(a < z < b)
denotes the probability that the z score is
between a and b
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Notation
P(a < z < b)
denotes the probability that the z score is
between a and b
P(zz > a)
P(
denotes the probability that the z score is
greater than a
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Notation
P(a < z < b)
denotes the probability that the z score is
between a and b
19
P(zz > a)
P(
denotes the probability that the z score is
greater than a
P (z
(z < a)
denotes the probability that the z score is
less than a
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Notation
P(a < z < b)
between a and b
betweena
P(zz > a)
P(
greater than, at least, more than,
not less than
P (z
(z < a)
less than, at most, no more than,
not greater than
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20
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