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統計學
授課教師：統計系余清祥
日期：2010年10月26日
第五章：離散機率分配
Fall 2010
Chapter 5
Discrete Probability Distributions






Random Variables
Discrete Probability Distributions
Expected Value and Variance
Binomial Probability Distribution
Poisson Probability Distribution
Hypergeometric Probability
.40
Distribution
.30
.20
.10
0
1
2
3
4
Random Variables
A
(隨機變數) is
A random
random variable
variable(隨機變數)
is aa numerical
numerical
description
description of
of the
the outcome
outcome of
of an
an experiment.
experiment.
A
A discrete
discrete random
random variable
variable may
may assume
assume either
either aa
finite
finite number
number of
of values
values or
or an
an infinite
infinite sequence
sequence of
of
values.
values.
A
A continuous
continuous random
random variable
variable may
may assume
assume any
any
numerical
numerical value
value in
in an
an interval
interval or
or collection
collection of
of
intervals.
intervals.
Discrete Random Variables

Example1.
The certified public accountant (CPA)
examination has four parts.
Define a random variable as x = the number of
parts of the CPA examination passed and It is
a discrete random variable because it may
assume the finite number of values 0, 1, 2, 3,
or 4.
Discrete Random Variables
Example 2.
An experiment of cars arriving at a tollbooth.
The random variable is x = the number of cars
arriving during a one-day period. The possible
values for x come from the sequence of
integers 0, 1, 2, and so on. x is a discrete
random variable assuming one of the values in
this infinite sequence.
Discrete Random Variables

Examples of Discrete Random Variables
Example: JSL Appliances

Discrete random variable with a finite
number of values
Let
Let xx == number
number of
of TVs
TVs sold
sold at
at the
the store
store in
in one
one day,
day,
where
where xx can
can take
take on
on 55 values
values (0,
(0, 1,
1, 2,
2, 3,
3, 4)
4)
Example: JSL Appliances

Discrete random variable with an
infinite sequence of values
Let
Let xx == number
number of
of customers
customers arriving
arriving in
in one
one day,
day,
where
where xx can
can take
take on
on the
the values
values 0,
0, 1,
1, 2,
2, .. .. ..
We can count the customers arriving, but there
is no finite upper limit on the number that might
arrive.
Random Variables
Question
Family
size
Random Variable x
x = Number of dependents
reported on tax return
Type
Discrete
Distance from x = Distance in miles from
home to the store site
home to store
Continuous
Own dog
or cat
Discrete
x = 1 if own no pet;
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
Continuous Random Variables
Example 1.
Experimental outcomes based on measurement
scales such as time, weight, distance, and
temperature can be described by continuous
random variables.
Continuous Random Variables
Example 2.
An experiment of monitoring incoming
telephone calls to the claims office of a major
insurance company. Suppose the random
variable of interest is x = the time between
consecutive incoming calls in minutes. This
random variable may assume any value in
the interval x ≥ 0.
Continuous Random Variables

Example of Continuous Random Variables
5.2 Discrete Probability Distributions
The
機率分配) for
The probability
probability distribution
distribution ((機率分配)
for aa random
random
variable
variable describes
describes how
how probabilities
probabilities are
are distributed
distributed
over
over the
the values
values of
of the
the random
random variable.
variable.
We
We can
can describe
describe aa discrete
discrete probability
probability distribution
distribution
with
with aa table,
table, graph,
graph, or
or equation.
equation.
Discrete Probability Distributions
The
The probability
probability distribution
distribution is
is defined
defined by
by aa probability
probability
function
機率函數), denoted
(x), which
function ((機率函數),
denoted by
by ff(x),
which provides
provides
the
the probability
probability for
for each
each value
value of
of the
the random
random variable.
variable.
The
The required
required conditions
conditions for
for aa discrete
discrete probability
probability
function
function are:
are:
ff(x)
(x) > 0

f(x) = 1
f(x)
Discrete Probability Distributions

Example: Probability Distribution for the
Number of Automobiles Sold During a Day at
Dicarlo Motors.
Discrete Probability Distributions


Using past data on TV sales, …
a tabular representation of the probability
distribution for TV sales was developed.
Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
x
0
1
2
3
4
f(x)
.40
.25
.20
.05
.10
1.00
80/200
Discrete Probability Distributions
Graphical Representation of Probability Distribution
.50
Probability

.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
Discrete Uniform Probability Distribution
The
均勻) probability
The discrete
discrete uniform
uniform ((均勻)
probability distribution
distribution
is
is the
the simplest
simplest example
example of
of aa discrete
discrete probability
probability
distribution
distribution given
given by
by aa formula.
formula.
The
The discrete
discrete uniform
uniform probability
probability function
function is
is
ff(x)
(x) = 1/
n
1/n
the values of the
random variable
are equally likely
where:
n = the number of values the random
variable may assume
Discrete Uniform Probability Distribution

Example:
An experiment of rolling a die we define the
random variable x to be the number of dots on
the upward face. There are n = 6 possible
values for the random variable; x = 1, 2, 3, 4, 5,
6. The probability function for this discrete
uniform random variable is f (x) = 1/6 x = 1, 2,
3, 4, 5, 6.
x
1
2
3
4
5
6
f (x)
1/6
1/6
1/6
1/6
1/6
1/6
Discrete Uniform Probability Distribution

Example:
Consider the random variable x with the
following discrete probability distribution.
x
1
2
3
4
f (x)
1/10
2/10
3/10
4/10
This probability distribution can be defined by
the formula f (x) = x/ 10 for x = 1, 2, 3, or 4.
5.3 Expected Value and Variance
The
The expected
expected value
value,, or
or mean,
mean, of
of aa random
random variable
variable
is
is aa measure
measure of
of its
its central
central location.
location.
E
(x) =  = 
xf(x)
E(x)
xf(x)
The
The variance
variance summarizes
summarizes the
the variability
variability in
in the
the
values
values of
of aa random
random variable.
variable.
Var(
x) =  2 = 
(x - )2ff(x)
(x)
Var(x)
(x
The
The standard
standard deviation
deviation,, ,, is
is defined
defined as
as the
the positive
positive
square
square root
root of
of the
the variance.
variance.
Expected Value and Variance

Example: Calculation of the Expected Value
for the Number of Automobiles Sold During A
Day at Dicarlo Motors.
Expected Value and Variance

Example: Calculation of the Variance for the
Number of Automobiles Sold During A Day at
Dicarlo Motors.
The standard deviation is   1.25  1.118
Expected Value and Variance

Expected Value
x
0
1
2
3
4
f(x)
xf(x)
.40
.00
.25
.25
.20
.40
.05
.15
.10
.40
E(x) = 1.20
expected number of
TVs sold in a day
Expected Value and Variance

Variance and Standard Deviation
x
x-
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
(x - )2
f(x)
(x - )2f(x)
1.44
0.04
0.64
3.24
7.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
Variance of daily sales = 2 = 1.660
TVs
squared
Standard deviation of daily sales = 1.2884 TVs
5.4 Binomial(二項) Distribution

Four Properties of a Binomial Experiment
1.
1. The
The experiment
experiment consists
consists of
of aa sequence
sequence of
of nn
identical
identical trials.
trials.
2.
and failure
failure,, are
are possible
possible
2. Two
Two outcomes,
outcomes, success
success and
on
on each
each trial.
trial.
3.
3. The
The probability
probability of
of aa success,
success, denoted
denoted by
by pp,, does
does
not
not change
change from
from trial
trial to
to trial.
trial.
stationarity
assumption
4.
The
trials
are
independent.
4. The trials are independent.
Binomial Distribution
Our
Our interest
interest is
is in
in the
the number
number of
of successes
successes
occurring
occurring in
in the
the nn trials.
trials.
We
We let
let xx denote
denote the
the number
number of
of successes
successes
occurring
occurring in
in the
the nn trials.
trials.
Binomial Distribution

Binomial Probability Function
n!
f (x) 
p x (1  p )( n  x )
x !(n  x )!
where:
f(x) = the probability of x successes in
n trials,
n = the number of trials,
p = the probability of success on any
one trial.
Binomial Distribution

Binomial Probability Function
n!
f (x) 
p x (1  p )( n  x )
x !(n  x )!
n!
x !( n  x )!
Number
Number of
of experimental
experimental
outcomes
outcomes providing
providing exactly
exactly
xx successes
successes in
in nn trials
trials
p x (1  p)( n  x )
Probability
Probability of
of aa particular
particular
sequence
sequence of
of trial
trial outcomes
outcomes
with
with xx successes
successes in
in nn trials
trials
Binomial Distribution


Example:
The experiment of tossing a coin five times and
on each toss observing whether the coin lands
with a head or a tail on its upward face. we
want to count the number of heads appearing
over the five tosses.
Does this experiment show the properties of a
binomial experiment?
Binomial Distribution

Note that:
1. The experiment consists of five identical trials; each
trial involves the tossing of one coin.
2. Two outcomes are possible for each trial: a head or
a tail. We can designate head a success and tail a failure.
3. The probability of a head and the probability of a tail
are the same for each trial, with p = .5 and 1- p = .5.
4. The trials or tosses are independent because the outcome
on any one trial is not affected by what happens on
other trials or tosses.
Binomial Distribution

Example: Evans Electronics
Evans is concerned about a low retention rate for
employees. In recent years, management has seen a
turnover of 10% of the hourly employees annually.
Thus, for any hourly employee chosen at random,
management estimates a probability of 0.1 that the
person will not be with the company next year.
Binomial Distribution

Using the Binomial Probability Function
Choosing 3 hourly employees at random, what
is the probability that 1 of them will leave the
company this year?
Let: p = .10, n = 3, x = 1
n!
f ( x) 
p x (1  p ) ( n  x )
x !( n  x )!
3!
f (1) 
(0.1)1 (0.9)2  3(.1)(.81)  .243
1!(3  1)!
Binomial Distribution

Tree Diagram
1st Worker
2nd Worker
Leaves (.1)
Leaves
(.1)
3rd Worker
L (.1)
x
3
Prob.
.0010
S (.9)
2
.0090
L (.1)
2
.0090
S (.9)
1
.0810
L (.1)
2
.0090
S (.9)
1
.0810
1
.0810
0
.7290
Stays (.9)
Leaves (.1)
Stays
(.9)
L (.1)
Stays (.9)
S (.9)
Binomial Distribution

Using Tables of Binomial Probabilities
p
n
x
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
3
0
1
2
3
.8574
.1354
.0071
.0001
.7290
.2430
.0270
.0010
.6141
.3251
.0574
.0034
.5120
.3840
.0960
.0080
.4219
.4219
.1406
.0156
.3430
.4410
.1890
.0270
.2746
.4436
.2389
.0429
.2160
.4320
.2880
.0640
.1664
.4084
.3341
.0911
.1250
.3750
.3750
.1250
Binomial Distribution

Expected Value
E(x) =  = np

Variance
Var(x) =  2 = np(1  p)

Standard Deviation
  np(1  p )
Binomial Distribution

Expected Value
E(x) =  = 3(.1) = .3 employees out of 3

Variance
Var(x) =  2 = 3(.1)(.9) = .27

Standard Deviation
  3(.1)(.9)  .52 employees
5.5 Poisson(布阿松)Distribution
A
A Poisson
Poisson distributed
distributed random
random variable
variable is
is often
often
useful
useful in
in estimating
estimating the
the number
number of
of occurrences
occurrences
over
over aa specified
specified interval
interval of
of time
time or
or space
space
ItIt is
is aa discrete
discrete random
random variable
variable that
that may
may assume
assume
an
an infinite
infinite sequence
sequence of
of values
values (x
(x == 0,
0, 1,
1, 2,
2, .. .. .. ).).
Poisson Distribution
Examples
Examples of
of aa Poisson
Poisson distributed
distributed random
random variable:
variable:
the
the number
number of
of knotholes
knotholes in
in 14
14 linear
linear feet
feet of
of
pine
pine board
board
the
the number
number of
of vehicles
vehicles arriving
arriving at
at aa
toll
toll booth
booth in
in one
one hour
hour
Bell
Bell Labs
Labs used
used the
the Poisson
Poisson distribution
distribution to
to model
model the
the
arrival
arrival of
of phone
phone calls.
calls.
Poisson Distribution

Two Properties of a Poisson Experiment
1.
1. The
The probability
probability of
of an
an occurrence
occurrence is
is the
the same
same
for
for any
any two
two intervals
intervals of
of equal
equal length.
length.
2.
2. The
The occurrence
occurrence or
or nonoccurrence
nonoccurrence in
in any
any
interval
interval is
is independent
independent of
of the
the occurrence
occurrence or
or
nonoccurrence
nonoccurrence in
in any
any other
other interval.
interval.
Poisson Distribution

Poisson Probability Function
f ( x) 
 x e
x!
where:
f(x) = probability of x occurrences in
an interval,
 = mean number of occurrences in
an interval,
e = 2.71828.
Poisson Distribution

Example: We are interested in the number of
arrivals at the drive-up teller window of a
bank during a 15-minute period on weekday
mornings.
Assume that the probability of a car arriving is
the same for any two time periods of equal
length and that the arrival or nonarrival of a
car in any time period is independent of the
arrival or nonarrival in any other time period.
Poisson Distribution

The Poisson probability function is applicable.
Suppose that the average number of cars
arriving in a 15-minute period of time is 10; in
this case, the following probability function
applies.
10
10 e
f ( x) 
x!
x
The random variable here is
x = number of cars arriving in any 15-minute
period.
Poisson Distribution

Example: Mercy Hospital
M ERCY
Patients arrive at the
emergency room of
Mercy Hospital at
the average rate of
6 per hour on
weekend evenings.
What is the probability
of 4 arrivals in 30 minutes on a weekend
evening?
Poisson Distribution

Using the Poisson Probability Function
 = 6/hour = 3/half-hour, x = 4
34 (2.71828)3
f (4) 
 .1680
4!
M ERCY
Poisson Distribution

M ERCY
Using Poisson Probability Tables

x
0
1
2
3
4
5
6
7
8
2.1
.1225
.2572
.2700
.1890
.0992
.0417
.0146
.0044
.0011
2.2
.1108
.2438
.2681
.1966
.1082
.0476
.0174
.0055
.0015
2.3
.1003
.2306
.2652
.2033
.1169
.0538
.0206
.0068
.0019
2.4
.0907
.2177
.2613
.2090
.1254
.0602
.0241
.0083
.0025
2.5
.0821
.2052
.2565
.2138
.1336
..0668
.0278
.0099
.0031
2.6
.0743
.1931
.2510
.2176
.1414
.0735
.0319
.0118
.0038
2.7
.0672
.1815
.2450
.2205
.1488
.0804
.0362
.0139
.0047
2.8
.0608
.1703
.2384
.2225
.1557
.0872
.0407
.0163
.0057
2.9
.0550
.1596
.2314
.2237
.1622
.0940
.0455
.0188
.0068
3.0
.0498
.1494
.2240
.2240
.1680
.1008
.0504
.0216
.0081
M ERCY
Poisson Distribution
Poisson Distribution of Arrivals
Poisson Probabilities
0.25
Probability

0.20
actually,
the sequence
continues:
11, 12, …
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
Number of Arrivals in 30 Minutes
10
Poisson Distribution
A
A property
property of
of the
the Poisson
Poisson distribution
distribution is
is that
that
the
the mean
mean and
and variance
variance are
are equal.
equal.
 = 2
Poisson Distribution

Variance for Number of Arrivals
During 30-Minute Periods
 ==  22 == 33
M ERCY
5.6 Hypergeometric(超幾何)Distribution
The
The hypergeometric
hypergeometric distribution
distribution is
is closely
closely related
related
to
to the
the binomial
binomial distribution.
distribution.
However,
However, for
for the
the hypergeometric
hypergeometric distribution:
distribution:
the
the trials
trials are
are not
not independent,
independent, and
and
the
the probability
probability of
of success
success changes
changes from
from trial
trial
to
to trial.
trial.
Hypergeometric Distribution

Hypergeometric Probability Function
 r  N  r 

 
x  n  x 

f ( x) 
N
 
n
for 0 < x < r
where: f(x) = probability of x successes in n trials,
n = number of trials,
N = number of elements in the population,
r = number of elements in the population
labeled success.
Hypergeometric Distribution

Hypergeometric Probability Function
f (x) 
r  N r
x  n  x 
  

N
n
 
number of ways
x successes can be selected
from a total of r successes
in the population
for 0 < x < r
number of ways
n – x failures can be selected
from a total of N – r failures
in the population
number of ways
a sample of size n can be selected
from a population of size N
Hypergeometric Probability
Distribution

Hypergeometric Probability Function
The probability function ff(x)
(x) on the previous slide
is usually applicable for values of x = 0, 1, 2, … nn..
However, only values of x where: 1) x < r and
2) n – x < N – r are valid.
If these two conditions do not hold for a value of xx,,
the corresponding ff(x)
(x) equals 0.
Hypergeometric Distribution



Example: A Quality Control Application.
Electric fuses produced by Ontario Electric are
packaged in boxes of 12 units each. Suppose an
inspector randomly selects 3 of the 12 fuses in a
box for testing.
If the box contains exactly 5 defective fuses,
what is the probability that the inspector will
find exactly 1 of the 3 fuses defective?
In this application, n = 3 and N = 12. With r = 5
defective fuses in the box.
Hypergeometric Distribution

The probability of finding x = 1 defective fuse is
 5  7 
 5!   7! 
   



1
2
( 5 )( 21 )
1!4!   2!5! 

 . 4773
f (1)       
220
 12 ! 
 12 


 
 3!9! 
 3 


What is the probability of finding at least 1 defective
fuse?
The probability of x = 0 is
 5  7 
 5!   7 ! 
   



0
3
0
!
5
!
3
!
4
!
  (1 )( 35 )  . 1591



f (0 )     
220
 12 ! 
 12 


 
 3!9 ! 
 3 

we conclude that the probability of finding at least 1
defective fuse must be 1 - .1591 = .8409.
Hypergeometric Distribution
ZAP
Example: Neveready
Bob Neveready has removed two
dead batteries from a flashlight
and inadvertently mingled
them with the two good
batteries he intended
as replacements. The four batteries look
identical. Bob now randomly selects two of the
four batteries. What is the probability he
selects the two good batteries?
ZAP
ZAP
ZAP

Hypergeometric Distribution

Using the Hypergeometric Function
 r  N  r   2  2   2!  2! 
 x  n  x   2  0   2!0!  0!2! 
      

  1  .167
f ( x )   
6
N
 4
 4! 
n
2
 2!2! 
 
 


where:
x = 2 = number of good batteries selected
n = 2 = number of batteries selected
N = 4 = number of batteries in total
r = 2 = number of good batteries in total
Hypergeometric Distribution

Mean
 r 
E ( x)    n  
N

Variance
r  N  n 
 r 
Var ( x)    n  1  

 N  N  N  1 
22
Hypergeometric Distribution

Mean
 r 
2
  n   2   1
N
4

Variance
 2  2  4  2  1
  2   1   
   .333
 4   4   4 1  3
22
Hypergeometric Distribution
Consider a hypergeometric distribution with n trials
and let p = (r/n) denote the probability of a success
on the first trial.
If the population size is large, the term (N – n)/(N – 1)
approaches 1.
The expected value and variance can be written
E(x) = np and Var(x) = np(1 – p).
Note that these are the expressions for the expected
value and variance of a binomial distribution.
continued
Hypergeometric Distribution
When
When the
the population
population size
size is
is large,
large, aa hypergeometric
hypergeometric
distribution
distribution can
can be
be approximated
approximated by
by aa binomial
binomial
distribution
distribution with
with nn trials
trials and
and aa probability
probability of
of success
success
pp == ((r/N).
r/N).
期望值與變異數的計算：
  E ( X )   xf ( x)
x
  Var ( X )   ( x   ) f ( x)  E[( X   ) ]
2
2
2
x
 E[ X 2  2 X   2 ]
 E ( X )  2 E ( X )  
2
 E ( X )  2  
2
2
2
2
 E( X 2 )  E( X )2
或是
2
Var ( X )  E[ X ( X  1)]  E ( X )  E ( X )
二項分配的來源：
二項分配源自二項式展開，例如
10
(a  b)   ( 10x ) a x b10 x
10
x 0
因此
a3b7 的係數＝
10
＝120。
3
( )
二項分配即由這個想法衍生而成，或是
n
1  [ p  (1  p )]   ( ) p (1  p )
n
x 0
n
x
也就是令 f ( x)  ( ) p (1  p )
n
x
x
x
n x
n x
n
(1)
，則  f ( x)  1。
x 0
期望值的計算如下：
n!
E( X )   x f (x)   x 
px (1 p)nx
x
x0
x!(n  x)!
n
n!
x
nx

p (1 p)
x1 (x 1)!(n  x)!
n
n(n 1)!

p  px (1 p)(n1)( x1)
x1 (x 1)![(n 1)  (x 1)]!
n
(n 1)!
(令 t  x 1)  np 
pt (1 p)(n1)t
t 0 t![(n 1)  t]!
 np1  np
n
變異數的計算也類似：
n!
p x (1  p ) n x
E[ X ( X  1)]   x( x  1) f ( x)   x( x  1) 
x
x 0
x!(n  x)!
n
n!
 
p x (1  p ) n x
x  2 ( x  2)!( n  x )!
n2
(n  2)!
2
( n  2 ) t
t
t
x
n
n
p
p
p
(令   2)
 (  1)  
(1  )
t 0 t![( n  2)  t ]!
n
 n(n  1) p 2
因此，Var ( X )  E[ X ( X  1)]  E ( X )  E ( X ) 2
 n(n  1) p 2  np  n 2 p 2
 np (1  p )
布阿松分配的來源：
布阿松分配源自自然指數的展開式
2
3
n
t
t
t
et  1  t       
2! 3!
n!
x
 
因為機率函數為 f ( x)  e ，推導可得
x!

 f ( x)  e
x 0
同理可算出
E( X )  




x 0
x
x!


 e e 1
及 E[ X ( X  1)]   2
 Var ( X )  
End of Chapter 5