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Statistics for Managers
5th Edition
Chapter 5
Some Important Discrete
Probability Distributions
.
Chap 5-1
Learning Objectives
In this chapter, you learn:
 The properties of a probability distribution
 To calculate the expected value, variance, and
standard deviation of a probability distribution
 To calculate probabilities from Binomial,
 Hypergeometric and Poisson distributions
 How to use the Binomial, Hypergeometric and
Poisson distributions to solve business problems
.
Chap 5-2
Introduction to Probability
Distributions
 Random Variable
 Represents a possible numerical value from
an uncertain event
Random
Variables
Ch. 5
Discrete
Random Variable
Continuous
Random Variable
Ch. 6
Chap 5-3
Discrete Random Variables
 Can only assume a countable number of values
Examples:
 Roll a die twice
Let X be the number of times 4 comes up
(then X could be 0, 1, or 2 times)
 Toss a coin 5 times.
Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)
Chap 5-4
Discrete Probability Distribution
Experiment: Toss 2 Coins.
T
T
H
H
T
H
T
H
Probability Distribution
X Value
Probability
0
1/4 = 0.25
1
2/4 = 0.50
2
1/4 = 0.25
Probability
4 possible outcomes
Let X = # heads.
0.50
0.25
0
.
1
2
X
Chap 5-5
Discrete Random Variable
Summary Measures
 Expected Value (or mean) of a discrete
distribution (Weighted Average)
N
  E(X)   Xi P( Xi )
i1
 Example: Toss 2 coins,
X = # of heads,
compute expected value of X:
X
P(X)
0
0.25
1
0.50
2
0.25
E(X) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25)
= 1.0
Chap 5-6
Discrete Random Variable
Summary Measures
(continued)
 Variance of a discrete random variable
N
σ 2   [X i  E(X)]2 P(Xi )
i1
 Standard Deviation of a discrete random variable
σ  σ2 
N
2
[X

E(X)]
P(Xi )
 i
i1
where:
E(X) = Expected value of the discrete random variable X
Xi = the ith outcome of X
P(Xi) = Probability of the ith occurrence of X
Chap 5-7
Discrete Random Variable
Summary Measures
(continued)
 Example: Toss 2 coins, X = # heads,
compute standard deviation (recall E(X) = 1)
σ
 [X

E(X)]
P(X
)
i
i
2
σ  (0  1)2 (0.25)  (1  1)2 (0.50)  (2  1)2 (0.25)  0.50  0.707
Possible number of heads
= 0, 1, or 2
Chap 5-8
Example of Expected Value
A company is considering a proposal to develop a new
product. The initial cash outlay would be $1 million, and
development time would be three years. If successful, the firm
anticipates that net profit (revenue minus initial cash outlay)
over the 5 year life cycle of the product will be $1.5 million. If
moderately successful, net profit will reach $1.2 million. If
unsuccessful, the firm anticipates zero cash inflows. The firms
assigns the following probabilities to the 5-year prospects for
this product: successful, .60; moderately successful, .30; and
unsuccessful, .10. What is the expected net profit? Ignore time
value of money.
Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc.
Chap 5-9
Calculation of Expected Value
To calculate the expected value
E(X) = Σ [(X) P(X)]
= (1.5)(.6) + (1.2)(.3) + (-1)(.1)
= + 1.16
Since the expected value Is positive this project
passes the initial approval process
Chap 5-10
Investment Returns
The Mean
Consider the return per $1000 for two types of
investments.
Investment
Economic
P(XiYi) Condition
Passive Fund X Aggressive Fund Y
0.2
Recession
- $25
- $200
0.5
Stable Economy
+ $50
+ $60
0.3
Expanding Economy
+ $100
+ $350
Chap 5-11
Investment Returns
The Mean
E(X) = μX = (-25)(.2) +(50)(.5) + (100)(.3) = 50
E(Y) = μY = (-200)(.2) +(60)(.5) + (350)(.3) = 95
Interpretation: Fund X is averaging a $50.00 return
and fund Y is averaging a $95.00 return per $1000
invested.
Chap 5-12
Chap 5-12
Investment Returns
Standard Deviation
σ X  (-25  50) 2 (.2)  (50  50) 2 (.5)  (100  50) 2 (.3)
 43.30
σ Y  (-200  95) 2 (.2)  (60  95) 2 (.5)  (350  95) 2 (.3)
 193.71
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 5-13
Chap 5-13
Investment Returns
Covariance
σ XY  (-25  50)(-200  95)(.2)  (50  50)(60  95)(.5)
 (100  50)(350  95)(.3)
 8250
Interpretation: Since the covariance is large and
positive, there is a positive relationship between the
two investment funds, meaning that they will likely
rise and fall together.
Chap 5-14
Chap 5-14
The Sum of Two Random
Variables: Measures
 Expected Value: E ( X  Y )  E ( X )  E (Y )
 Variance: Var( X  Y )  σ 2X Y  σ 2X  σY2  2σ XY
 Standard
deviation:
σ X Y  σ 2X Y
Chap 5-15
Chap 5-15
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.
 Investment Objective: Maximize return (mean)
while minimizing risk (standard deviation).
Chap 5-16
Chap 5-16
Portfolio Expected Return and
Expected Risk
 Portfolio expected return (weighted average
return):
E(P)  w E ( X )  (1  w) E (Y )
 Portfolio risk (weighted variability)
σ P  w 2σ 2X  (1  w )2 σ 2Y  2w(1 - w)σ XY
where w = portion of portfolio value in asset X
(1 - w) = portion of portfolio value in asset Y
Chap 5-17
Chap 5-17
Portfolio Expect Return and
Expected Risk
Recall:
Investment X:
Investment Y:
E(X) = 50 σX = 43.30
E(Y) = 95 σY = 193.21
σXY = 8250
Suppose 40% of the portfolio is in Investment X and 60% is in
Investment Y:
E(P)  .4 (50)  (.6) (95)  77
σ P  (.4) 2 (43.30) 2  (.6)2 (193.21) 2  2(.4)(.6)( 8250)
 133.04
 The portfolio return is between the values for investments X and Y
considered individually.
Chap 5-18
Chap 5-18
Binomial Probability Distribution
 A fixed number of observations, n
 e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse
 Two mutually exclusive and collectively exhaustive
categories
 e.g., head or tail in each toss of a coin; defective or not defective
light bulb
 Generally called “success” and “failure”
 Probability of success is p, probability of failure is 1 – p
 Constant probability for each observation
 e.g., Probability of getting a tail is the same each time we toss
the coin
Chap 5-19
Binomial Probability Distribution
(continued)
 Observations are independent
 The outcome of one observation does not affect the
outcome of the other
 Two sampling methods
 Infinite population without replacement
 Finite population with replacement
Chap 5-20
Possible Binomial Distribution
Settings
 A manufacturing plant labels items as
either defective or acceptable
 A firm bidding for contracts will either get a
contract or not
 A marketing research firm receives survey
responses of “yes I will buy” or “no I will
not”
 New job applicants either accept the offer
or reject it
Chap 5-21
Binomial Distribution Formula
n!
X
nX
P(X) 
p (1-p)
X ! (n  X)!
P(X) = probability of X successes in n trials,
with probability of success p on each trial
X = number of ‘successes’ in sample,
(X = 0, 1, 2, ..., n)
n = sample size (number of trials
or observations)
p = probability of “success”
Example: Flip a coin four
times, let x = # heads:
n=4
p = 0.5
1 - p = (1 - 0.5) = 0.5
X = 0, 1, 2, 3, 4
Chap 5-22
Example:
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of success is .1?
X = 1, n = 5, and p = 0.1
n!
P(X  1) 
p X (1  p)n X
X!(n  X)!
5!

(0.1)1(1  0.1) 51
1! (5  1)!
 (5)(0.1)(0 .9) 4
 0.32805
Chap 5-23
Binomial Distribution
Characteristics
 Mean
μ  E(x)  np
 Variance and Standard Deviation
σ  np(1 - p)
2
σ  np(1 - p)
Where n = sample size
p = probability of success
(1 – p) = probability of failure
Chap 5-24
Binomial Example
A doctor prescribes Nexium to 60% of his patients
who have gastro-intestinal problems. What is the
probability that out of the next 10 patients:
1.
Six of them are given a prescription of
Nexium
2.
At least 6 are given a prescription of
Nexium?
Chap 5-25
Binomial Probabilities Table
X
P(X)
6
0.250822656
7
0.214990848
8
0.120932352
9
0.040310784
10
0.006046618
Data
Sample size
Probability of success
10
0.6
P(<=X)
0.617719398
0.832710246
0.953642598
0.993953382
1
P(<X)
0.366896742
0.617719398
0.832710246
0.953642598
0.993953382
P(>X)
0.382280602
0.167289754
0.046357402
0.006046618
0
P(>=X)
0.633103258
0.382280602
0.167289754
0.046357402
0.006046618
Chap 5-26
The Poisson Distribution
 It is appropriate to use the Poisson distribution
when:
 You have an event that occurs randomly
through time and space
 You know the average number of successes
you expect to observe over a given time frame
or in a given space
Chap 5-27
Poisson Distribution Formula
 x
e 
P( X) 
X!
where:
X = number of events in an area of opportunity
 = expected number of events
e = base of the natural logarithm system (2.71828...)
Chap 5-28
Poisson Example
On average there are three thread defects in a
10 yard bolt of fine wool fabric. What is the
probability of finding no more than two thread
defects in a randomly chosen 10 bolt lot?
Chap 5-29
Solution
Poisson Probabilities
Data
Average/Expected number of successes:
Poisson Probabilities Table
X
P(X)
P(<=X)
0
0.049787 0.049787
1
0.149361 0.199148
2
0.224042 0.423190
3
0.224042 0.647232
4
0.168031 0.815263
P(<X)
0.000000
0.049787
0.199148
0.423190
0.647232
3
P(>X)
0.950213
0.800852
0.576810
0.352768
0.184737
P(>=X)
1.000000
0.950213
0.800852
0.576810
0.352768
Chap 5-30
Poisson Distribution
Characteristics
 Mean
μλ
 Variance and Standard Deviation
σ λ
2
σ λ
where  = expected number of events
Chap 5-31
The Hypergeometric Distribution
 “n” trials in a sample taken from a finite
population of size N
 Sample taken without replacement
 Outcomes of trials are dependent
 Concerned with finding the probability of “X”
successes in the sample where there are “A”
successes in the population
Chap 5-32
Hypergeometric Distribution
Formula
 A  N  A 
 

 X  n  X 

P( X)   
N
 
n 
 
Where
N = population size
A = number of successes in the population
N – A = number of failures in the population
n = sample size
X = number of successes in the sample
n – X = number of failures in the sample
Chap 5-33
Properties of the
Hypergeometric Distribution
 The mean of the hypergeometric distribution is
nA
μ  E(x) 
N
 The standard deviation is
nA(N - A) N - n
σ

2
N
N -1
Where
N-n
is called the “Finite Population Correction Factor”
N -1
from sampling without replacement from a
finite population
Chap 5-34
Using the
Hypergeometric Distribution
■ Example: 3 different computers are checked from 10 in
the department. 4 of the 10 computers have illegal
software loaded. What is the probability that 2 of the 3
selected computers have illegal software loaded?
N = 10
A=4
n=3
X=2
 A  N  A   4  6 
 
   
 X  n  X   2 1  (6)(6)
     
P(X  2)   
 0.3
120
 N
10 
 
 
n 
3 
 
 
The probability that 2 of the 3 selected computers have illegal
software loaded is .30, or 30%.
.
Chap 5-35
Hypergeometric Distribution
in PHStat
(continued)
 Complete dialog box entries and get output …
N = 10
A=4
n=3
X=2
P(X = 2) = 0.3
Chap 5-36
Chapter Summary
 Addressed the probability of a discrete random
variable
 Discussed the Binomial distribution
 Discussed the Poisson distribution
 Discussed the Hypergeometric distribution
Chap 5-37