Download Chap. 4: Discrete Random Variables

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Behavioral Statistics
Discrete Random Variables
Chapter 4
4-1
Learning Objectives
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1. Distinguish Between the Two Types of
Random Variables
2. Compute the Expected Value & Variance
of Discrete Random Variables
3. Describe the Binomial and Poisson
4. Calculate Probabilities for Discrete
Random Variables
4-2
Thinking Challenge
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You’re taking a 33
question multiple choice
test. Each question has
4 choices. Clueless on 1
question, you decide to
guess. What’s the
chance you’ll get it right?
If you guessed on all 33
questions, what would be
your grade? pass?
4-3
Data Types
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Data
Numerical
Discrete
4-4
Continuous
Qualitative
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Discrete Random Variables
4-5
Discrete
Random Variable
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1. Random Variable


A Numerical Outcome of an Experiment
Example: Number of Tails in 2 Coin Tosses
2. Discrete Random Variable



Whole Number (0, 1, 2, 3 etc.)
Obtained by Counting
Usually Finite Number of Values

4-6
Poisson Random Variable Is Exception ()
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Discrete Random
Variable Examples
Experiment
Random
Variable
Make 100 Sales Calls # Sales
Inspect 70 Radios
Possible
Values
0, 1, 2, ..., 100
# Defective 0, 1, 2, ..., 70
Answer 33 Questions # Correct
0, 1, 2, ..., 33
Count Cars at Toll
# Cars
Between 11:00 & 1:00 Arriving
0, 1, 2, ..., 
4-7
Discrete
Probability Distribution
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1. List of All possible [x, p(x)] pairs


x = Value of Random Variable (Outcome)
p(x) = Probability Associated with Value
2. Mutually Exclusive (No Overlap)
3. Collectively Exhaustive (Nothing Left Out)
4. 0  p(x)  1
5.  p(x) = 1
4-8
Discrete Probability
Distribution Example
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Experiment: Toss 2 Coins. Count # Tails.
Probability Distribution
Values, x Probabilities, p(x)
© 1984-1994 T/Maker Co.
4-9
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Visualizing Discrete
Probability Distributions
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Listing
Table
# Tails
f(x)
Count
p(x)
0
1
2
1
2
1
.25
.50
.25
{ (0, .25), (1, .50), (2, .25) }
p(x)
.50
.25
.00
Graph
Equation
p ( x) 
x
0
4 - 10
1
2
n!
p x (1  p) n  x
x !(n  x)!
Summary Measures
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1. Expected Value



Mean of Probability Distribution
Weighted Average of All Possible Values
 = E(X) = x p(x)
2. Variance


Weighted Average Squared Deviation
about Mean
2 = E[ (x (x  p(x)
4 - 11
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x
p(x)
Total
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Summary Measures
Calculation Table
x p(x )
x p(x )
x-
(x -)
2
2
(x -) p( x )
 (x -) p( x )
2
Thinking Challenge
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You toss 2 coins. You’re
interested in the number
of tails. What are the
expected value &
standard deviation of
this random variable,
number of tails?
© 1984-1994 T/Maker Co.
4 - 13
© 2001 prentice-Hall, Inc.
Expected Value &
Variance Solution*
2
x
p(x)
x p(x )
x-
(x -)
0
.25
0
-1.00
1.00
.25
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
 = 1.0
4 - 14
2
(x -) p( x )
 = .50
2
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Discrete Probability
Distribution Function
4 - 15
Discrete Probability
Distribution Function
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1.
Type of Model

2.
Representation of Some
Underlying phenomenon
Mathematical Formula
3.
Represents Discrete
Random Variable
4.
Used to Get Exact
Probabilities
4 - 16
P (X  x )
 e
x
x!
-

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Discrete Probability
Distribution Models
Discrete
Probability
Distribution
Binomial
4 - 17
Poisson
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Binomial Distribution
4 - 18
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Discrete Probability
Distribution Models
Discrete
Probability
Distribution
Binomial
4 - 19
Poisson
Binomial Distribution
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1. Number of ‘Successes’ in a Sample of
n Observations (Trials)

# Reds in 15 Spins of Roulette Wheel

# Defective Items in a Batch of 5 Items

# Correct on a 33 Question Exam

# Customers Who Purchase Out of 100
Customers Who Enter Store
4 - 20
Binomial Distribution
Properties
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1. Two Different Sampling Methods


Infinite Population Without Replacement
Finite Population With Replacement
2. Sequence of n Identical Trials
3. Each Trial Has 2 Outcomes

‘Success’ (Desired Outcome) or ‘Failure’
4. Constant Trial Probability
5. Trials Are Independent
4 - 21
© 2001 prentice-Hall, Inc.
Binomial Probability
Distribution Function
 n  x n x
n!
x
n x
p( x)    p q 
p (1  p)
x!(n  x)!
 x
p(x) = Probability of x ‘Successes’
n = Sample Size
p = Probability of ‘Success’
x = Number of ‘Successes’ in Sample
(x = 0, 1, 2, ..., n)
4 - 22
Binomial Probability
Distribution Example
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Experiment: Toss 1 Coin 5 Times in a Row.
Note # Tails. What’s the Probability of 3 Tails?
n!
x
n x
p( x ) 
p (1  p )
x !(n  x )!
5!
3
5 3
p(3) 
.5 (1 .5)
3 !(5  3)!
 0.3125
4 - 23
© 2001 prentice-Hall, Inc.
Binomial Probability
Table (Portion)
n=5
p
k
.01
…
0.50
…
.99
0
.951
…
.031
…
.000
1
.999
…
.188
…
.000
2
1.000
…
.500
…
.000
3
1.000
…
.812
…
.001
4
1.000
…
.969
…
.049
Cumulative Probabilities
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Binomial Distribution
Characteristics
Mean
  E ( x )  np
P(X)
.6
.4
.2
.0
  np (1  p)
X
0
Standard Deviation
P(X)
.6
.4
.2
.0
1
2
3
4
5
n = 5 p = 0.5
X
0
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n = 5 p = 0.1
1
2
3
4
5
© 2001 prentice-Hall, Inc.
Binomial Distribution
Thinking Challenge
You’re a telemarketer selling
service contracts for Macy’s.
You’ve sold 20 in your last
100 calls (p = .20). If you
call 12 people tonight,
what’s the probability of
A.
B.
C.
D.
No sales?
Exactly 2 sales?
At most 2 sales?
At least 2 sales?
4 - 26
© 2001 prentice-Hall, Inc.
Binomial Distribution
Solution*
Using the Binomial Tables:
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2)
= .0687 + .2062 + .2835
= .5584
D. p(at least 2) = p(2) + p(3)...+ p(12)
= 1 - [p(0) + p(1)]
= 1 - .0687 - .2062
= .7251
4 - 27
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Poisson Distribution
4 - 28
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Discrete Probability
Distribution Models
Discrete
Probability
Distribution
Binomial
4 - 29
Poisson
Poisson Distribution
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1. Number of Events that Occur in an
Interval

Events Per Unit

Time, Length, Area, Space
2. Examples



# Customers Arriving in 20 minutes
# Strikes Per Year in the U.S.
# Defects Per Lot (Group) of VCR’s
4 - 30
Poisson Process
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1.
Constant Event
Probability

Average of 60/Hr Is 1/Min
for 60 1-Minute Intervals
2.
One Event Per
Interval

3.
Don’t Arrive Together
Independent Events
Arrival of 1 Person Does
Not Affect Another’s
Arrival
4 - 31

© 1984-1994 T/Maker Co.
© 2001 prentice-Hall, Inc.
Poisson Probability
Distribution Function
x -
p (x) 
e
x!
p(x) = Probability of x Given 
 = Expected (Mean) Number of ‘Successes’
e = 2.71828 (Base of Natural Logs)
x = Number of ‘Successes’ Per Unit
4 - 32
Poisson Distribution
Characteristics
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Mean
  E(x)  

= 0.5
P(X)
N
 x p( x )
.6
.4
.2
.0
X
0
1
2
3
4
5
i 1
= 6
P(X)
Standard Deviation
 
.6
.4
.2
.0
X
0
4 - 33
2
4
6
8
10
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Poisson Distribution
Example
Customers arrive at a
rate of 72 per hour.
What is the
probability of 4
customers arriving in
3 minutes?
4 - 34
© 1995 Corel Corp.
© 2001 prentice-Hall, Inc.
Poisson Distribution
Solution
72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval
x -
e
p( x) 
x!
4 -3.6

3.6 e
p(4) 
 0.1912
4!
4 - 35
© 2001 prentice-Hall, Inc.

.02
:
3.4
3.6
3.8
:
Poisson Probability
Table (Portion)
0
.980
:
.033
.027
.022
:
…
…
:
…
…
…
:
x
3
4
:
:
.558 .744
.515 .706
.473 .668
:
:
…
9
:
…
…
…
:
:
.997
.996
.994
:
Cumulative Probabilities
4 - 36
Thinking Challenge
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You work in Quality
Assurance for an
investment firm. A
clerk enters 75 words
per minute with 6
errors per hour. What
is the probability of 0
errors in a 255-word
bond transaction?
© 1984-1994 T/Maker Co.
4 - 37
© 2001 prentice-Hall, Inc.
Poisson Distribution
Solution: Finding *
75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
6 errors/hr = 6 errors/4500 words
= .00133 errors/word
In a 255-word transaction (interval):
 = (.00133 errors/word )(255 words)
= .34 errors/255-word transaction
4 - 38
© 2001 prentice-Hall, Inc.
Poisson Distribution
Solution: Finding p(0)*
x -
e
p( x) 
x!
0 -.34

.34 e
p(4) 
 0.7118
0!
4 - 39
Conclusion
© 2001 prentice-Hall, Inc.
1. Distinguished Between the Two Types of
Random Variables
2. Computed the Expected Value &
Variance of Discrete Random Variables
3. Described the Binomial and Poisson
Distributions
4. Calculated Probabilities for Discrete
Random Variables
4 - 40
End of Chapter
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