Download Continuous Probability Distributions

Introduction to Business Statistics, 6e
Kvanli, Pavur, Keeling
Chapter 6 –
Continuous
Probability
Distributions
Slides prepared by Jeff Heyl, Lincoln University
Thomson/South-Western Learning™
1
©2003 South-Western/Thomson
Probability for a
Continuous Random Variable
Curve describing
the population
Area = P(20 < X < 60)
20
60
X
Figure 6.1
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Properties of a
Normal Distribution
 Continuous random variable
 Symmetrical in shape (bell shaped)
 The probability of any given range of
numbers is represented by the area
under the curve for that range
 Probabilities for all normal
distributions are determined using
the standard normal distribution
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520
500
480
460
440
420
400
380
360
340
320
300
280
Relative frequency
Histogram for Lightbulb Life
X
Figure 6.2
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Distribution of Lightbulb Life
This distance is 
= 450 - 400
= 50
Inflection point
P
|
|
|
|
|
µ = 400
|
|
|
450
|
X
Figure 6.3
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Probability Density Function
for Normal Distribution
1
1
f(x) =
e 2
 2π
x-µ

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Relative frequency
Normal Curves with Unequal Means
and Equal Standard Deviations
Females
Males
|
|
Average
Average
female height male height
Height
Figure 6.4
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Normal Curves with Equal Means and
Unequal Standard Deviations
Relative frequency
Company A
Company B
Average age
in both companies
Age
Figure 6.5
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Area Under the Normal Curve
Area = .5
Area = .5
X
µ
Total area = 1
Figure 6.6
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Normal Curve for Lightbulbs
 = 50
360
|
300
|
350
|
400
|
450
|
500
X
Figure 6.7
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Standard Normal Curve
=1
|
|
-2
(µ - 2)
-1
(µ - )
|
µ=0
|
1
20
(µ + ) (µ + 2)
Z
Figure 6.8
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Standard Normal Curve
Area = .4474
0
1.62
Z
Figure 6.9
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Determining the Probability
for a Standard Normal
Random Variable
P(Z > 1.62) = .5 - .4474 = .0526
Area = .5
Area = .5
0
1.62
Z
Figure 6.10
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Determining the Probability
for a Standard Normal
Random Variable
P(Z < 1.62) = .5 + .4474 = .9474
A1 = .5
A2 = .4474
0
1.62
Z
Figure 6.11
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Determining the Probability
for a Standard Normal
Random Variable
P(1.0 < Z < 2.0) = P(0 < Z < 2.0) - P(0 < Z < 1.0)
= .4772 - .3413 = .1359
A1 = .3413
A2 = .4772
0
1
2
Z
Figure 6.12
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Determining the Probability
for a Standard Normal
Random Variable
P(-1.25 < Z < 1.15) = P(-1.25 < Z < 0) + P(0 < Z < 1.15)
= A1 + A2 = .3944 + .3749 = .7693
A1 = .3944
-1.25
A2 = .3749
0
1.15
Z
Figure 6.13
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Determining the Probability
for a Standard Normal
Random Variable
P(-1.25 < Z < 1.15) = P(-1.25 < Z < 0) + P(0 < Z < 1.15)
= A1 + A2 = .3944 + .3749 = .7693
These areas are the same
-1.25
0
1.25
Z
Figure 6.14
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Determining the Probability
for a Standard Normal
Random Variable
P(Z < -1.45) = P(Z < 0) - P(-1.45 < Z < 0)
= .5 - .4265 = .0735
A1 = .5 - .4265
A2 = .4265
-1.25
0
Z
Figure 6.15
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Determining the Probability
for a Standard Normal
Random Variable
P(Z ≥ z) = .03 = .5 - .03 = .47
Area = .5 - .03
Area = .03
0
z
Z
Figure 6.16
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Determining the Probability
for a Standard Normal
Random Variable
P(Z ≤ z) = .2 = .5 - .2 = .3
Area = .5 - .2
Area = .2
z
0
Z
Figure 6.17
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Areas Under Any
Normal Curve
-120
-100
-80
-60
-40
-20
0
20
40
60
80
100
120
µ = 0,  = 50
Y

Figure 6.18
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Areas Under Any
Normal Curve
-2.4
-2.0
-1.6
-1.2
-.8
-.4
0
.4
.8
1.2
1.6
2.0
2.4
µ = 0,  = 1
Y

Figure 6.19
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Areas Under Any
Normal Curve
P(X < 360) = ?
360
400
X = lifetime of Everglo bulb
Same
area
Figure 6.20A
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Areas Under Any
Normal Curve
Same
area
From Table A.4, this
area is .2881. So shaded
area = .5 - .2881
= .2119
-.8
0
Z = standard
normal
Figure 6.20B
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Interpreting Z
 In Example 6.2 Z = - 0.8 means that the
value 360 is .8 standard deviations
below the mean
 A positive value of Z designates how
may standard deviations () X is to the
right of the mean (µ)
 A negative value of Z designates how
may standard deviations () X is to the
left of the mean (µ)
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ATM Example
 = $625
$2000
$3700
$5000
X
Figure 6.21A
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ATM Example
Area = .4967
-2.72
Area = .4812
0
2.08
X
Figure 6.21B
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Policyholder Lifetimes
 = 4.4 yr
75.0 –
(2)
70
70.6 –
65
66.2 –
61.8 –
57.4 –
(1)
X = age at death
(years)
Figure 6.22
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Policyholder Lifetimes
A1 + A2 = P(Z > -.27)
= .1064 + .5
= .6064
A1 = .1064
A2 = .5
-.27
Z
Figure 6.23
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Policyholder Lifetimes
A1 = .3051 (Table A.4)
A2 = .5 - .3051
= .1949
.86
Z
Figure 6.24
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Policyholder Lifetimes
A3 = A2 - A1
= .4772 - .3051
= .1721
Z
.86
2.00
A1 = .3051
A2 = .4772
Figure 6.25
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Everglo Lightbulbs
After how many hours will 80%
of the Everglo bulbs burn out?
P(X < x0) = .8
A2 = .3
A1 = .5
A1 + A2 = .8
400
Figure 6.26A
x0
X = lifetime of
Everglo bulb
Example 6.5
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Everglo Lightbulbs
After how many hours will 80%
of the Everglo bulbs burn out?
A1 = .5
P(X < x0) = .8
P(Z < .84)
= .5 + .2995
= .7995  .8
A2 = .3
x0 - 400
= .84
50
.84
Figure 6.26B
Example 6.5
x0 - 400
= (50)(.84)
= 42
x0 = 400 + 42
= 442
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Bakery Example
µ = 35
=8
Area = .9
35
x0
X = demand for French
bread (loaves)
Figure 6.27A
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Bakery Example
A2 = .4
A1 = .5
A1 + A2 = .9
1.28
Z
Figure 6.27B
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Empirical Rule
1. Approximately 68% of the data should lie
between X - s and X + s
2. Approximately 95% of them should lie between
X - 2s and X + 2s
3. Approximately 99.7% of them should lie
between X - 3s and X + 3s
These numbers are generated
directly from Table A.4
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Empirical Rule
A1 + A2 = .3413 + .3413
= .6826
A1 = .3413
A2 = .3413
-1
1
Z
Figure 6.28
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Determining Areas and
Values With Excel
Figure 6.29A
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Determining Areas and
Values With Excel
Figure 6.29B
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Allied Manufacturing
Figure 6.30
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Allied Manufacturing
Figure 6.31A
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Allied Manufacturing
Figure 6.31B
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Normal Approximation to
the Binomial Distribution
Poisson approximation:
Use when n > 20 and np ≤ 7
Normal approximation:
Use when np > 5 and n(1 - p) > 5
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Normal Approximation to
the Binomial Distribution
= Solution to 1
= Solution to 2
A normal curve
with µ = 6 and  = 1.732
0
1
2
3
4
5
6 7
8 9 10 11 12
X
4.5 5.5
Figure 6.32
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How to Adjust for Continuity
If X is a binomial random variable with n trials and probability
of success = p, then:
b + .5 - µ
1. P(X ≤ b)  P Z ≤

a - .5 - µ
2. P(X ≥ a)  P Z ≥

a - .5 - µ
b + .5 - µ
3. P(a ≤ X ≤ b)  P
≤Z≤


4. Be sure to convert a < probability to a ≤ and to convert
a > probability to a ≥ before switching to the normal
approximation
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Continuous Uniform
Distribution
The probability of a given range of values is
proportional to the width of the range
a+b
µ=
2
=
b-a
12
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Relative frequency
Continuous Uniform
Distribution
6
7
8
Content of cup (fluid ounces)
Figure 6.33
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Continuous Uniform
Distribution
Total area =
1
(2) = 1
2
1
–
2
6
|
7
8
X = amount of
soda (ounces)
Figure 6.34
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Continuous Uniform
Distribution
Total area = (b - a)
1
= .5 –
b-a
a=6
|
µ=7
b=8
1
= (2)(.5) = 1
b-a
X = amount of
soda (ounces)
Figure 6.34
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Continuous Uniform
Distribution
Area = (8 - 7.5)(.5)
= .25
.5 –
6
|
6.5
|
7
|
7.5
8
X = amount of soda (ounces)
Figure 6.36
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Continuous Uniform
Distribution
Area = (7.5 - 6.5)(.5)
= .5
.5 –
6
|
6.5
|
7
|
7.5
8
X = amount of soda (ounces)
Figure 6.37
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Exponential Distribution
 Time between arrivals to a queue -
time between people arriving at a
line to check out in a department
store (people, machines, or
telephone calls may wait in a queue)
 Lifetime of components in a
machine
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Exponential Distribution
If the random variable Y, representing
the number of arrivals over a specified
time period T, follows a Poisson
distribution, then X, representing the
time between successive arrivals, will be
an exponential random variable
P(X ≥ x0) = e-Ax 0
µ = 1/A
 = 1/A
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Exponential Distribution
Total area = 1.0
0
X
Figure 6.38
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Exponential Distribution
A–
Area = P(X ≥ x0)
= e-Ax0
0
x0
X
Figure 6.39
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Exponential Distribution
A–
Area = e-(4)(.5) = e-2 = .135
0
.5
X = time
between
arrivals
Figure 6.40
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Exponential Distribution
.001 –
Area = 1 - e-(.001)(1000) = 1 - e-1 = .632
LongLife
0
1000
X = battery
lifetime
Figure 6.41
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Exponential Distribution
.001 –
Area = 1 - e-(.001)(365) = 1 - e-.365 = .306
LongLife
0
1000
X = battery
lifetime
Figure 6.41
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