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John
Loucks
St.
Edward’s Univ.
2
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Chapter 3, Part B
Probability Distributions
Uniform Probability Distribution
Normal Probability Distribution
Exponential Probability Distribution
f
(
x
)
x
Uniform
x
f
(
x
)
Normal
x
f
(
x
)
Exponential
3
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Continuous Random Variables
Examples of continuous random variables include the
following:
•
The
number of ounces
of soup placed in a can
labeled “8 ounces”
•
The
flight time
of an airplane traveling from
Chicago to New York
•
The
lifetime
of the picture tube in a new television
set
•
The
drilling depth
required to reach oil in an
offshore drilling operation
4
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Continuous Probability Distributions
A
continuous random variable
can assume any value
in an interval on the real line or in a collection of
intervals.
It is not possible to talk about the probability of the
random variable assuming a particular value.
Instead, we talk about the probability of the random
variable assuming a value within a given interval.
5
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Continuous Probability Distributions
The probability of the random variable assuming a
value within some given interval from
x
1
to
x
2
is
defined to be the
area under the graph
of the
probability density function
between
x
1
and
x
2
.
f
(
x
)
x
Uniform
x
1
x
2
x
f
(
x
)
Normal
x
1
x
2
x
1
x
2
Exponential
x
f
(
x
)
x
1
x
2
6
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2013
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A random variable is
uniformly distributed
whenever
the probability
that the variable will assume a value in
any interval of equal length is the same for each
interval.
The
uniform probability density function
is:
Uniform Probability Distribution
where:
a
= smallest value the variable can assume
b
= largest value the variable can assume
f
(
x
) = 1/(
b
–
a
) for
a
<
x
<
b
= 0 elsewhere
7
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Example: Flight Time
Uniform Probability Distribution
Let
x
denote the flight time of an airplane
traveling from Chicago to New York. Assume that the
minimum time is 2 hours and that the maximum time
is 2 hours 20 minutes.
Assume that sufficient actual flight data are
available to conclude that the probability of a flight
time between 120 and 121 minutes is the same as the
probability of a flight time within any other 1

minute
interval up to and including 140 minutes.
8
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Uniform Probability Density Function
Example: Flight Time
f
(
x
) = 1/20 for 120
<
x
<
140
= 0 elsewhere
where:
x
= flight time in minutes
9
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Uniform Probability Distribution for Flight Time
f
(
x
)
x
120
130
140
1/20
Flight Time (mins.)
Example: Flight Time
10
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or duplicated, or posted to a publicly accessible website, in whole or in part.
f
(
x
)
x
120
130
140
1/20
Flight Time (mins.)
Example: Flight Time
P(135
<
x
<
140) = 1/20(5) = .25
What is the probability that a flight will take
between 135 and 140 minutes?
135
11
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f
(
x
)
x
120
130
140
1/20
Flight Time (mins.)
Example: Flight Time
P(124
<
x
<
136) = 1/20(12) = .6
What is the probability that a flight will take
between 124 and 136 minutes?
136
124
12
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Normal Probability Distribution
The
normal probability distribution
is the most
important distribution for describing a continuous
random variable.
It is widely used in statistical inference.
13
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Heights
of people
Normal Probability Distribution
It has been used in a wide variety of applications:
Scientific
measurements
Test
scores
Amounts
of rainfall
14
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Normal Probability Distribution
Normal Probability Density Function
2 2
( )/2
1
( )
2
x
f x e
= mean
= standard deviation
= 3.14159
e
= 2.71828
where:
15
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
The distribution is
symmetric
, and is
bell

shaped
.
Normal Probability Distribution
Characteristics
x
16
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2013
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The entire family of normal probability
distributions is defined by its
mean
and its
standard deviation
.
Normal Probability Distribution
Characteristics
Standard Deviation
Mean
x
17
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The
highest point
on the normal curve is at the
mean
, which is also the
median
and
mode
.
Normal Probability Distribution
Characteristics
x
18
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Normal Probability Distribution
Characteristics

10
0
20
The mean can be any numerical value: negative,
zero, or positive.
x
19
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Normal Probability Distribution
Characteristics
= 15
= 25
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
x
20
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Probabilities for the normal random variable are
given by
areas under the curve
. The total area
under the curve is 1 (.5 to the left of the mean and
.5 to the right).
Normal Probability Distribution
Characteristics
.5
.5
x
21
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Normal Probability Distribution
Characteristics
of values of a normal random variable
are within of its mean.
68.26%
+/

1 standard deviation
of values of a normal random variable
are within of its mean.
95.44%
+/

2 standard deviations
of values of a normal random variable
are within of its mean.
99.72%
+/

3 standard deviations
22
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Normal Probability Distribution
Characteristics
x
–
3
–
1
–
2
+ 1
+ 2
+ 3
68.26%
95.44%
99.72%
23
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Standard Normal Probability Distribution
A
random variable having a normal distribution
with
a mean of 0 and a standard deviation of 1 is
said
to have a
standard normal probability
distribution
.
24
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1
0
z
The letter
z
is used to designate the standard
normal random variable.
Standard Normal Probability Distribution
25
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Converting to the Standard Normal Distribution
Standard Normal Probability Distribution
z
x
We can think of
z
as a measure of the number of
standard deviations
x
is from
.
26
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Example: Pep Zone
Standard Normal Probability Distribution
Pep Zone sells auto parts and supplies including
a popular multi

grade motor oil. When the stock of
this oil drops to 20 gallons, a replenishment order is
placed.
27
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Example: Pep Zone
Standard Normal Probability Distribution
The store manager is concerned that sales are
being lost due to stockouts while waiting for an
order. It has been determined that demand during
replenishment lead

time is normally distributed with
a mean of 15 gallons and a standard deviation of 6
gallons.
The manager would like to know the probability
of a stockout,
P
(
x
> 20).
28
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
z
= (
x

)/
= (20

15)/6
= .83
Solving for the Stockout Probability
Example: Pep Zone
Step
1: Convert
x
to the standard normal distribution.
Step
2: Find the area under the standard normal
curve
between the mean and
z
= .83.
see next slide
29
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Probability Table for the
Standard Normal Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
.5
.1915
.1695
.1985
.2019
.2054
.2088
.2123
.2157
.2190
.2224
.6
.2257
.2291
.2324
.2357
.2389
.2422
.2454
.2486
.2517
.2549
.7
.2580
.2611
.2642
.2673
.2704
.2734
.2764
.2794
.2823
.2852
.8
.2881
.2910
.2939
.2967
.2995
.3023
.3051
.3078
.3106
.3133
.9
.3159
.3186
.3212
.3238
.3264
.3289
.3315
.3340
.3365
.3389
.
.
.
.
.
.
.
.
.
.
.
Example: Pep Zone
P
(0
<
z
<
.83)
30
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
P
(
z
> .83) = .5
–
P
(0
<
z
<
.83)
= 1

.2967
= .2033
Solving for the Stockout Probability
Example: Pep Zone
Step
3: Compute the area under the standard normal
curve
to the right of
z
= .83.
Probability
of a stockout
P
(
x
> 20)
31
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Solving for the Stockout Probability
Example: Pep Zone
0
.83
Area = .2967
Area = .5

.2967
= .2033
z
32
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Standard Normal Probability Distribution
If the manager of Pep Zone wants the probability
of a stockout to be no more than .05, what should the
reorder point be?
Example: Pep Zone
33
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Solving for the Reorder Point
0
Area = .4500
Area = .0500
z
z
.05
Example: Pep Zone
34
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2013
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Solving for the Reorder Point
Example: Pep Zone
Step
1: Find the
z

value that cuts off an area of .05
in
the right tail of the standard normal
distribution
.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
1.5
.4332
.4345
.4357
.4370
.4382
.4394
.4406
.4418
.4429
.4441
1.6
.4452
.4463
.4474
.4484
.4495
.4505
.4515
.4525
.4535
.4545
1.7
.4554
.4564
.4573
.4582
.4591
.4599
.4608
.4616
.4625
.4633
1.8
.4641
.4649
.4656
.4664
.4671
.4678
.4686
.4693
.4699
.4706
1.9
.4713
.4719
.4726
.4732
.4738
.4744
.4750
.4756
.4761
.4767
.
.
.
.
.
.
.
.
.
.
.
We look up the area
(.5

.05 = .45)
35
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Solving for the Reorder Point
Example: Pep Zone
Step
2: Convert
z
.05
to the corresponding value of
x
.
x
=
+
z
.05
= 15 + 1.645(6)
= 24.87 or 25
A reorder point of 25 gallons will place the probability
of a stockout during leadtime at (slightly less than) .05.
36
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2013
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Solving for the Reorder Point
Example: Pep Zone
By raising the reorder point from 20 gallons to
25 gallons on hand, the probability of a stockout
decreases from about .20 to .05.
This is a significant decrease in the chance that Pep
Zone will be out of stock and unable to meet a
customer’s desire to make a purchase.
37
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Exponential Probability Distribution
The exponential probability distribution is useful in
describing the time it takes to complete a task.
The exponential random variables can be used to
describe:
Time between
vehicle arrivals
at a toll booth
Time required
to
complete a
questionnaire
Distance between
major defects
in a highway
38
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2013
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Density Function
Exponential Probability Distribution
where:
= mean
e
= 2.71828
f
x
e
x
(
)
/
1
for
x
>
0,
> 0
39
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Cumulative Probabilities
Exponential Probability Distribution
P
x
x
e
x
(
)
/
0
1
o
where:
x
0
= some specific value of
x
40
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Exponential Probability Distribution
The time between arrivals of cars
at Al’s full

service gas pump follows an exponential probability
distribution with a mean time between arrivals of
3 minutes. Al would like to know the
probability that the time between two successive
arrivals will be 2 minutes or less.
Example: Al’s Full

Service Pump
41
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2013
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Exponential Probability Distribution
x
f
(
x
)
.1
.3
.4
.2
1 2 3 4 5 6 7 8 9 10
Time Between Successive Arrivals (mins.)
Example: Al’s Full

Service Pump
P
(
x
<
2) = 1

2.71828

2/3
= 1

.5134 = .4866
42
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2013
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or duplicated, or posted to a publicly accessible website, in whole or in part.
Relationship between the Poisson
and Exponential Distributions
The Poisson distribution
provides an appropriate description
of the number of occurrences
per interval
The exponential distribution
provides an appropriate description
of the length of the interval
between occurrences
43
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End of Chapter 3, Part B
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