Basic Business Statistics
10th Edition
Chapter 5
Some Important Discrete
Probability Distributions
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc..
Chap 5-1
Learning Objectives
In this chapter, you learn:
 The properties of a probability distribution
 To calculate the expected value and variance of a
probability distribution
 To calculate the covariance and its use in finance
 To calculate probabilities from binomial, Poisson
distributions and hypergeometric
 How to use the binomial, hypergeometric, and
Poisson distributions to solve business problems
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
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)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
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
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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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-6
Discrete Random Variable
Summary Measures
(continued)
 Variance of a discrete random variable
N
σ 2   [Xi  E(X)]2 P(X i )
i1
 Standard Deviation of a discrete random variable
σ  σ2 
N
2
[X

E(X)]
P(X i )
 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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
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 )
2
i
i
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
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Chap 5-8
二元隨機變數之機率分配
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-9
事件機率的定義
事件機率
設事件A定義於隨機實驗的樣本空間，其發生之機率P(A)
為事件A之基本出象的機率總和，即 P( A)   P( Ei )，Ei  A。
 聯合機率的定義
二個或二個以上事件同時發生的機率稱為聯合機率。
 邊際機率的定義
在有二個或二個以上類別的樣本空間中，若僅考慮某一類
別個別發生的機率者稱為邊際機率。
 條件機率的定義
令A、B為定義於樣本空間的事件，已知發生事件B之後再
發生事件A的機率，稱為事件A的條件機率。
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-10
表5.6 一般化的聯合機率分配表
B1
…
A1
P( A1  B1 )
…




Ar
A\B
c
P( A1  Bc )
 P( A1 , B j )  P( A1 )






P( Ar  B1 )
…
P( Ar  Bc )
 P( Ar , B j )  P( Ar )
 P( Ai , B1 )  P( B1 )
…
 P( Ai , Bc )  P( Bc )
1
r
P( B j )
P( Ai )
Bc
i 1
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
j 1
c
j 1
r
i 1
Chap 5-11
Expected Value
Z  A B
E ( A  B )   ( Ai  B j ) P( Ai , B j )
i
j
  Ai P( Ai , B j )   B j P( Ai , B j )
i
j
i
j
  Ai  P( Ai , B j )   B j  P ( Ai , B j )
i
j
j
i
  Ai P( Ai )   B j P( B j )
i
j
 E ( A)  E ( B )
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-12
The Covariance
 The covariance measures the strength of the
linear relationship between two variables
 The covariance:
N
σ XY   [ Xi  E( X)][(Yi  E( Y )] P( Xi Yi )
i1
where:
X = discrete variable X
Xi = the ith outcome of X
Y = discrete variable Y
Yi = the ith outcome of Y
P(XiYi) = probability of occurrence of the
ith outcome of X and the ith outcome of Y
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Chap 5-13
Computing the Mean for
Investment Returns
Return per $1,000 for two types of investments
P(XiYi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
0.2
Recession
- $ 25
- $200
0.5
Stable Economy
+ 50
+ 60
0.3
Expanding Economy
+ 100
+ 350
E(X) = μX = (-25)(0.2) +(50)(0.5) + (100)(0.3) = 50
E(Y) = μY = (-200)(0.2) +(60)(0.5) + (350)(0.3) = 95
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Chap 5-14
Computing the Standard Deviation
for Investment Returns
P(XiYi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
0.2
Recession
- $ 25
- $200
0.5
Stable Economy
+ 50
+ 60
0.3
Expanding Economy
+ 100
+ 350
σ X  (-25  50)2 (0.2)  (50  50)2 (0.5)  (100  50)2 (0.3)
 43.30
σ Y  (-200  95)2 (0.2)  (60  95)2 (0.5)  (350  95)2 (0.3)
 193.71
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-15
Computing the Covariance
for Investment Returns
P(XiYi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
0.2
Recession
- $ 25
- $200
0.5
Stable Economy
+ 50
+ 60
0.3
Expanding Economy
+ 100
+ 350
σ XY  (-25  50)(-200  95)(0.2)  (50  50)(60  95)(0.5)
 (100  50)(350  95)(0.3)
 8250
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Chap 5-16
Interpreting the Results for
Investment Returns
 The aggressive fund has a higher expected
return, but much more risk
μY = 95 > μX = 50
but
σY = 193.71 > σX = 43.30
 The Covariance of 8250 indicates that the two
investments are positively related and will vary
in the same direction
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-17
The Sum of
Two Random Variables
 Expected Value of the sum of two random variables:
E(X  Y)  E( X)  E( Y)
 Variance of the sum of two random variables:
Var(X Y)  σ2XY  σ2X  σ2Y  2σ XY
 Standard deviation of the sum of two random variables:
σ X Y  σ
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2
X Y
Chap 5-18
Portfolio Expected Return
and Portfolio 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
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Chap 5-19
Portfolio Example
Investment X:
Investment Y:
μX = 50
σX = 43.30
μY = 95
σY = 193.21
σXY = 8250
Suppose 40% of the portfolio is in Investment X and
60% is in Investment Y:
E(P)  0.4 (50)  (0.6)(95)  77
σP  (0.4)2 (43.30)2  (0.6)2 (193.71)2  2(0.4)(0.6)(8250)
 133.30
The portfolio return and portfolio variability are between the values
for investments X and Y considered individually
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Chap 5-20
Probability Distributions
Probability
Distributions
Ch. 5
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Normal
Poisson
Uniform
Hypergeometric
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Ch. 6
Exponential
Chap 5-21
The Binomial Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Hypergeometric
Poisson
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Chap 5-22
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
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Chap 5-23
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
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Chap 5-24
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
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Chap 5-25
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)
X! = (X)(X - 1)(X - 2) . . . (2)(1)
0! = 1
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(by definition)
Chap 5-26
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”
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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 5-27
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
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Chap 5-28
Binomial Distribution
 The shape of the binomial distribution depends on the
values of p and n
Mean
 Here, n = 5 and p = 0.1
P(X)
.6
.4
.2
0
X
0
P(X)
 Here, n = 5 and p = 0.5
.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
Chap 5-29
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
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Chap 5-30
二項機率分配之期望值與變異數
設 X 為一二項隨機變數，其機率函數為：
f ( x)  Cxn p x q n x
則其期望值為：
x  0,1,2,, n
E ( X )  np
變異數為：
V ( X )  npq
證明 茲證明二項分配的期望值與變異數如下：
(1)期望值
n
n
n
x 0
x 0
x 0
E ( X )   xf ( x)   xC xn p x q n  x   x
n!
p x q n x
(n  x )! x!
n 1
n(n  1)!
(n  1)!
p x q n  x  np 
p y q n 1 y
x 1 ( n  x )! ( x  1)!
y  0 ( n  1  y )! y!
n

「式中 ( y  x  1) 」
 np( p  q ) n1
(利用 (n  1) 次方的二項展開式)
 np
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Chap 5-31
(2)變異數
n
n
x 0
x 0
V ( X )   ( x   ) 2 f ( x)   x 2 f ( x)   2
n
n
x 0
x 0

  x( x  1) f ( x)   xf ( x)   2
其中
n
n
x 0
x 0
 x( x  1) f ( x)   x( x
 1)Cxn p x q n x
n
  x ( x  1)
x 0
n!
p x q n x
(n  x )! x!
(n  2)!
px 2qn x
x 2 ( n  2  x )!( x  2)!
 n( n  1) p2 ( p  q ) n2
n 2
 n(n  1) p 
2
(利用 (n  2) 次方的二次展開式)
 n( n  1) p2

將代入可得：
V ( X )  n( n  1) p2  np  ( np)2  np2  np  np (1  p )  npq
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-32
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Chap 5-33
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-34
The Bernoulli Distribution
 Suppose that a random experiment can give rise to just two possible
mutually exclusive and collectively exhaustive outcomes, which for
convenience we will label “success” and “failure.” Let p denote the
probability of success, so that the probability of failure is (1-p). Now define
the random variable X so that X takes the value 1 if the outcome of the
experiment is success and 0 otherwise.
 The probability function of this random variable is then
Px(0) = (1-p), Px(1) = p,
This distribution is known as the Bernoulli distribution. Its mean and variance
are
E(X) = 0*(1-p)+1*(p) = p
V(X) = (0-p)2(1-p)+(1-p)2p = p(1-p)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-35
Binomial Distribution
 Experiment involves n identical trials
 Each trial has exactly two possible outcomes: success
and failure
 Each trial is independent of the previous trials
 p is the probability of a success on any one trial
 q = (1-p) is the probability of a failure on any one trial
 p and q are constant throughout the experiment
 X is the number of successes in the n trials
 Applications
 Sampling with replacement
 Sampling without replacement -- n < 5% N
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-36
Binomial Characteristics
Examples
μ  np  (5)(0.1) 0.5
Mean
σ  np(1- p)  (5)(0.1)(1 0.1)
 0.6708
P(X)
.6
.4
.2
0
X
0
μ  np  (5)(0.5) 2.5
σ  np(1- p)  (5)(0.5)(1 0.5)
 1.118
P(X)
.6
.4
.2
0
1
2
3
4
5
n = 5 p = 0.5
X
0
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
n = 5 p = 0.1
1
2
3
4
5
Chap 5-37
Using Binomial Tables
n = 10
x
…
p=.20
p=.25
p=.30
p=.35
p=.40
p=.45
p=.50
0
1
2
3
4
5
6
7
8
9
10
…
…
…
…
…
…
…
…
…
…
…
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0001
0.0000
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0000
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0001
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
10
9
8
7
6
5
4
3
2
1
0
…
p=.80
p=.75
p=.70
p=.65
p=.60
p=.55
p=.50
x
Examples:
n = 10, p = 0.35, x = 3:
P(x = 3|n =10, p = 0.35) = 0.2522
n = 10, p = 0.75, x = 2:
P(x = 2|n =10, p = 0.75) = 0.0004
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-38
The Hypergeometric
Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Hypergeometric
Poisson
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-39
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-40
Hypergeometric Distribution
Formula
 A  N  A 
 

[ A C X ][N A Cn X ]  X  n  X 
P(X) 

 N
N Cn
 
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-41
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
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-42
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 0.30, or 30%.
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-43
The Poisson Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Hypergeometric
Poisson
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-44
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  (lambda)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-45
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...)
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-46
Poisson Distribution
Characteristics
 Mean
μλ
 Variance and Standard Deviation
σ λ
2
σ λ
where  = expected number of events
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-47
Using Poisson Tables

X
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0
1
2
3
4
5
6
7
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.0000
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
0.0000
0.0000
0.7408
0.2222
0.0333
0.0033
0.0003
0.0000
0.0000
0.0000
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
0.0000
0.0000
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.0000
0.0000
0.4966
0.3476
0.1217
0.0284
0.0050
0.0007
0.0001
0.0000
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.0002
0.0000
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
0.0003
0.0000
Example: Find P(X = 2) if  = 0.50
e  λ λ X e 0.50 (0.50)2
P(X  2) 

 0.0758
X!
2!
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-48
Graph of Poisson Probabilities
0.70
Graphically:
0.60
 = 0.50
0
1
2
3
4
5
6
7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
P(x)
X
=
0.50
0.50
0.40
0.30
0.20
0.10
0.00
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
0
1
2
3
4
5
6
7
x
P(X = 2) = 0.0758
Chap 5-49
Poisson Distribution Shape
 The shape of the Poisson Distribution
depends on the parameter  :
 = 0.50
 = 3.00
0.70
0.25
0.60
0.20
0.40
P(x)
P(x)
0.50
0.30
0.15
0.10
0.20
0.05
0.10
0.00
0.00
0
1
2
3
4
5
x
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
6
7
1
2
3
4
5
6
7
8
9
10
11
12
x
Chap 5-50
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-51
證明當 n  ， np 固定，二項分配趨近於泊松分配
lim Cxn p x q n x
n

x e
x!
式中：   np
證明 證明如下：
lim Cxn p x q n  x
n
n!


( ) x (1  ) n x
n x!( n  x )! n
n
 lim
 lim
n 
 x n( n  1)...(n  x  1)
x!

nx
 (1 

n
) (1 
n

n
)
x

x e 
x!
....(  np)
因為
n(n  1)...(n  x  1)
1
2
x 1 


(
1

)(
1

)...(
1

) 1
lim

n 
n  
n
n
n 
nx
lim

lim(1  ) n  e 
n
n
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.

lim(1  )  x  1
n
n
Chap 5-52
表6.21 二項分配與泊松分
配
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-53
圖6.15 二項分配
分配
0.14
圖6.16 泊松
0.14
f (x )
f (x )
0.12
0.12
0.1
0.1
0.08
0.08
0.06
0.06
0.04
0.04
0.02
0.02
x
0
1
3
5
7
9
11
13
15
17
19
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
21
23
25
x
0
1
3
5
7
9
11
13
15
17
19
21
23
25
Chap 5-54
圖6.17 二項分配與泊松分配
0.14
f (x)
0.12
0.1
0.08
0.06
0.04
0.02
0
1
3
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
5
7
9
11
13
15
17
19
21
23
x 25
Chap 5-55
Chapter Summary
 Addressed the probability of a discrete random
variable
 Defined covariance and discussed its
application in finance
 Discussed the Binomial distribution
 Discussed the Hypergeometric distribution
 Discussed the Poisson distribution
Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.
Chap 5-56