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Discrete Probability
Distributions

Chap 5-1

Learning Objectives
In this chapter, you learn:
 The properties of a probability distribution
 To compute the expected value and variance of a
probability distribution
 To calculate the covariance and understand its use
in finance
 To compute probabilities from binomial,
hypergeometric, and Poisson distributions
 How the binomial, hypergeometric, and Poisson
distributions can be used to solve business
problems
Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
g
...




Continuous variables produce outcomes that
come from a measurement (e
...
your annual
salary, or your weight)
...


Chap 5-3

Types Of Variables
Types Of
Variables

Ch
...
6

Chap 5-4

Discrete Random Variables


Can only assume a countable number of values
Examples:


Roll a die twice
Let X be the number of times 4 occurs
(then X could be 0, 1, or 2 times)



Toss a coin 5 times
...


Chap 5-5

Probability Distribution For A
Discrete Random Variable


A probability distribution for a discrete random
variable is a mutually exclusive listing of all
possible numerical outcomes for that variable and
a probability of occurrence associated with each
outcome
...
20
0
...
24
0
...


Chap 5-6

Example of a Discrete Random
Variable Probability Distribution
Experiment: Toss 2 Coins
...
25

1

2/4 = 0
...
25

Probability

4 possible outcomes

Let X = # heads
...
50
0
...


1

2

X
Chap 5-7

Discrete Variables
Expected Value (Measuring Center)


Expected Value (or mean) of a discrete
variable (Weighted Average)
N

µ = E(X) = ∑ X i P( X = X i )
i =1



Example: Toss 2 coins,
X = # of heads,
compute expected value of X:

X

P(X=Xi)

0

0
...
50

2

0
...
25) + (1)(0
...
25))
= 1
...


Chap 5-8

Discrete Random Variables
Measuring Dispersion


Variance of a discrete random variable
N

σ 2 = ∑ [X i − E(X)]2 P(X = X i )
i =1



Standard Deviation of a discrete random variable

σ = σ2 =

N

2
−
[X
E(X)]
P(X = X i )
∑ i
i =1

where:
E(X)
= Expected value of the discrete random variable X
Xi
= the ith outcome of X
P(X=Xi) = Probability of the ith occurrence of X
Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
25) + (1− 1)2 (0
...
25) = 0
...
707
Possible number of heads
= 0, 1, or 2

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...




A positive covariance indicates a positive
relationship
...


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...


Chap 5-12

Investment Returns
The Mean
Consider the return per $1000 for two types of
investments
...


Passive Fund X

Aggressive Fund Y

0
...
5

Stable Economy

+ $50

+ $60

0
...


Chap 5-13

Investment Returns
The Mean
E(X) = μX = (-25)(
...
5) + (100)(
...
2) +(60)(
...
3) = 95
Interpretation: Fund X is averaging a $50
...
00 return per $1000
invested
...


Chap 5-14

Investment Returns
Standard Deviation
σ X = (-25 − 50) 2 (
...
5) + (100 − 50) 2 (
...
30
σ Y = (-200 − 95) 2 (
...
5) + (350 − 95) 2 (
...
71

Interpretation: Even though fund Y has a higher
average return, it is subject to much more variability
and the probability of loss is higher
...


Chap 5-15

Investment Returns
Covariance
σ XY = (-25 − 50)(-200 − 95)(
...
5)
+ (100 − 50)(350 − 95)(
...


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...


Chap 5-17

Portfolio Expected Return and
Expected Risk


Investment portfolios usually contain several
different funds (random variables)



The expected return and standard deviation of
two funds together can now be calculated
...


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...


Chap 5-19

Portfolio Example
Investment X:
Investment Y:

μX = 50 σX = 43
...
21
σXY = 8250

Suppose 40% of the portfolio is in Investment X and
60% is in Investment Y:
E(P) = 0
...
6) (95) = 77

σP =

(0
...
30) 2 + (0
...
71) 2 + 2(0
...
6)(8,250)

= 133
...


Chap 5-20

Probability Distributions
Probability
Distributions
Ch
...


Ch
...
g
...
g
...


Chap 5-22

Binomial Probability Distribution
(continued)


Observations are independent




The outcome of one observation does not affect the
outcome of the other
Two sampling methods deliver independence



Infinite population without replacement
Finite population with replacement

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...


Chap 5-24

The Binomial Distribution
Counting Techniques


Suppose the event of interest is obtaining heads on the
toss of a fair coin
...

In how many ways can you get two heads?



Possible ways: HHT, HTH, THH, so there are three
ways you can getting two heads
...
We need to be able to
count the number of ways for more complicated
situations
...


Chap 5-25

Counting Techniques
Rule of Combinations


The number of combinations of selecting X
objects out of n objects is

n!
n Cx =
X!(n − X)!
where:
n! =(n)(n - 1)(n - 2)
...
(2)(1)
0! = 1 (by definition)
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...


31!
31! 31 • 30 • 29 • 28!
= 31 • 5 • 29 = 4,495
=
=
31 C 3 =
3!(31 − 3)! 3!28!
3 • 2 • 1 • 28!

Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
, n)
n

= sample size (number of trials
or observations)
π = probability of “event of interest”
Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
5
1 - π = (1 - 0
...
5
X = 0, 1, 2, 3, 4

Chap 5-28

Example:
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of an event of
interest is 0
...
1

n!
P(X = 1 | 5,0
...
1)1 (1 − 0
...
1)(0
...
32805
Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
02
...
02
n!
π x (1 − π ) n − x
P(X = 2 | 10, 0
...
02) 2 (1 −
...
0004)(
...
01531
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...
1

P(X=x|5, 0
...
6

...
2
0
0

1

2

3

4

5

x

2

3

4

5

x

P(X=x|5, 0
...
5


...
4

...


1

Chap 5-31

The Binomial Distribution Using
Binomial Tables (Available On Line)
n = 10
x

…

π=
...
25

π=
...
35

π=
...
45

π=
...
1074
0
...
3020
0
...
0881
0
...
0055
0
...
0001
0
...
0000

0
...
1877
0
...
2503
0
...
0584
0
...
0031
0
...
0000
0
...
0282
0
...
2335
0
...
2001
0
...
0368
0
...
0014
0
...
0000

0
...
0725
0
...
2522
0
...
1536
0
...
0212
0
...
0005
0
...
0060
0
...
1209
0
...
2508
0
...
1115
0
...
0106
0
...
0001

0
...
0207
0
...
1665
0
...
2340
0
...
0746
0
...
0042
0
...
0010
0
...
0439
0
...
2051
0
...
2051
0
...
0439
0
...
0010

10
9
8
7
6
5
4
3
2
1
0

…

π=
...
75

π=
...
65

π=
...
55

π=
...
35, x = 3:

P(X = 3|10, 0
...
2522

n = 10, π = 0
...
75) = 0
...


Chap 5-32

Binomial Distribution
Characteristics


Mean



Variance and Standard Deviation

μ = E(X) = nπ

σ = nπ (1 - π )
2

σ = nπ (1 - π )
Where

n = sample size
π = probability of the event of interest for any trial
(1 – π) = probability of no event of interest for any trial

Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
1) = 0
...
1)(1 −
...
6708

μ = nπ = (5)(
...
5
σ = nπ (1 - π ) = (5)(
...
5)
= 1
...
1)

...
4

...
5)

...
4

...


0

1

Chap 5-34

Using Excel For The
Binomial Distribution

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...

An area of opportunity is a continuous unit or
interval of time, volume, or such area in which
more than one occurrence of an event can
occur
...


Chap 5-36

The Poisson Distribution


Apply the Poisson Distribution when:


You wish to count the number of times an event
occurs in a given area of opportunity



The probability that an event occurs in one area of
opportunity is the same for all areas of opportunity



The number of events that occur in one area of
opportunity is independent of the number of events
that occur in the other areas of opportunity



The probability that two or more events occur in an
area of opportunity approaches zero as the area of
opportunity becomes smaller



The average number of events per unit is λ (lambda)

Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
71828
...


Chap 5-38

Poisson Distribution
Characteristics


Mean



Variance and Standard Deviation

μ=λ
σ2 = λ
σ= λ

where λ = expected number of events

Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
10

0
...
30

0
...
50

0
...
70

0
...
90

0
1
2
3
4
5
6
7

0
...
0905
0
...
0002
0
...
0000
0
...
0000

0
...
1637
0
...
0011
0
...
0000
0
...
0000

0
...
2222
0
...
0033
0
...
0000
0
...
0000

0
...
2681
0
...
0072
0
...
0001
0
...
0000

0
...
3033
0
...
0126
0
...
0002
0
...
0000

0
...
3293
0
...
0198
0
...
0004
0
...
0000

0
...
3476
0
...
0284
0
...
0007
0
...
0000

0
...
3595
0
...
0383
0
...
0012
0
...
0000

0
...
3659
0
...
0494
0
...
0020
0
...
0000

Example: Find P(X = 2 | λ = 0
...
50 (0
...
50) =
=
= 0
...


Chap 5-40

Using Excel For The
Poisson Distribution

Statistics for Managers Using Microsoft Excel® 7e Copyright ©2014 Pearson Education, Inc
...
50
X

λ=
0
...
6065
0
...
0758
0
...
0016
0
...
0000
0
...
50) = 0
...


Chap 5-42

Poisson Distribution Shape


The shape of the Poisson Distribution
depends on the parameter λ :
λ = 3
...
50

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