Transcript Business Statistics: A Decision

Statistics
Please Stand By….
Chap 4-1
Chapter 4: Probability and Distributions





Randomness
General Probability
Probability Models
Random Variables
Moments of Random Variables
Chap 4-2
Randomness

The language of probability

Thinking about randomness

The uses of probability
Chap 4-3
Randomness
Chap 4-4
Randomness
Chap 4-5
Randomness
Chap 4-6
Randomness
Technical Analysis 1 | More Examples | Another (Tipping Pts)
Chap 4-7
Chapter Goals
After completing this chapter, you should be
able to:
 Explain three approaches to assessing
probabilities
 Apply common rules of probability
 Use Bayes’ Theorem for conditional probabilities
 Distinguish between discrete and continuous
probability distributions
 Compute the expected value and standard
deviation for a probability distributions
Chap 4-8
Important Terms




Probability – the chance that an uncertain event
will occur (always between 0 and 1)
Experiment – a process of obtaining outcomes
for uncertain events
Elementary Event – the most basic outcome
possible from a simple experiment
Randomness –


Does not mean haphazard
Description of the kind of order that emerges only in
the long run
Chap 4-9
Important Terms (CONT’D)


Sample Space – the collection of all possible
elementary outcomes
Probability Distribution Function



Maps events to intervals on the real line
Discrete probability mass
Continuous probability density
Chap 4-10
Sample Space
The Sample Space is the collection of all
possible outcomes (based on an probabilistic
experiment)
e.g., All 6 faces of a die:
e.g., All 52 cards of a bridge deck:
Chap 4-11
Events

Elementary event – An outcome from a sample
space with one characteristic


Example: A red card from a deck of cards
Event – May involve two or more outcomes
simultaneously

Example: An ace that is also red from a deck of
cards
Chap 4-12
Elementary Events

A automobile consultant records fuel type and
vehicle type for a sample of vehicles
2 Fuel types: Gasoline, Diesel
3 Vehicle types: Truck, Car, SUV
6 possible elementary events:
e1
Gasoline, Truck
e2
Gasoline, Car
e3
Gasoline, SUV
e4
Diesel, Truck
e5
Diesel, Car
e6
Diesel, SUV
e1
Car
e2
e3
e4
Car
e5
e6
Chap 4-13
Independent vs. Dependent Events

Independent Events
E1 = heads on one flip of fair coin
E2 = heads on second flip of same coin
Result of second flip does not depend on the result of
the first flip.

Dependent Events
E1 = rain forecasted on the news
E2 = take umbrella to work
Probability of the second event is affected by the
occurrence of the first event
Chap 4-14
Probability Concepts

Mutually Exclusive Events

If E1 occurs, then E2 cannot occur

E1 and E2 have no common elements
E1
Black
Cards
E2
Red
Cards
A card cannot be
Black and Red at
the same time.
Chap 4-15
Coming up with Probability

Empirically





From the data!
Based on observation, not theory
Probability describes what happens in very
many trials.
We must actually observe many trials to pin
down a probability
Based on belief (Bayesian Technique)
Chap 4-16
Assigning Probability

Classical Probability Assessment
P(Ei) =

Number of ways Ei can occur
Total number of elementary events
Relative Frequency of Occurrence
Number of times Ei occurs
Relative Freq. of Ei =
N

Subjective Probability Assessment
An opinion or judgment by a decision maker about
the likelihood of an event
Chap 4-17
Calculating Probability

Counting Outcomes
Number of ways Ei can occur
Total number of elementary events

Observing Outcomes in Trials
Chap 4-18
Counting
Chap 4-19
Counting
a b c d e ….
___ ___ ___ ___
2.
N take n
N take k
3.
Order not important – less than permutations
1.
___ ___ ___
___ ___
Chap 4-20
Counting
Chap 4-21
Rules of Probability



S is sample space
Pr(S) = 1
Events measured in numbers result in a
Probability Distribution
Chap 4-22
Rules of Probability
Rules for
Possible Values
and Sum
Individual Values
Sum of All Values
k
0 ≤ P(ei) ≤ 1
 P(e )  1
For any event ei
i1
i
where:
k = Number of elementary events
in the sample space
ei = ith elementary event
Chap 4-23
Addition Rule for Elementary Events

The probability of an event Ei is equal to the
sum of the probabilities of the elementary
events forming Ei.

That is, if:
Ei = {e1, e2, e3}
then:
P(Ei) = P(e1) + P(e2) + P(e3)
Chap 4-24
Complement Rule

The complement of an event E is the collection of
all possible elementary events not contained in
event E. The complement of event E is
represented by E.
E

Complement Rule:
P( E )  1 P(E)
E
Or,
P(E)  P( E )  1
Chap 4-25
Addition Rule for Two Events
■
Addition Rule:
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
E1
+
E2
=
E1
E2
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
Don’t count common
elements twice!
Chap 4-26
Addition Rule Example
P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)
= 26/52 + 4/52 - 2/52 = 28/52
Type
Color
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Don’t count
the two red
aces twice!
Chap 4-27
Addition Rule for
Mutually Exclusive Events

If E1 and E2 are mutually exclusive, then
P(E1 and E2) = 0
E1
E2
So
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
= P(E1) + P(E2)
Chap 4-28
Conditional Probability

Conditional probability for any
two events E1 , E2:
P(E 1 andE2 )
P(E 1 | E2 ) 
P(E 2 )
w here P(E 2 )  0
Chap 4-29
Conditional Probability Example


Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.
What is the probability that a car has a CD
player, given that it has AC ?
i.e., we want to find P(CD | AC)
Chap 4-30
Conditional Probability Example
(continued)

Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a CD player (CD).
20% of the cars have both.
CD
No CD
Total
AC
No AC
Total
1.0
P(CD and AC) .2
P(CD | AC) 
  .2857
P(AC)
.7
Chap 4-31
Conditional Probability Example
(continued)

Given AC, we only consider the top row (70% of the cars). Of these,
20% have a CD player. 20% of 70% is about 28.57%.
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
P(CD and AC) .2
P(CD | AC) 
  .2857
P(AC)
.7
Chap 4-32
For Independent Events:

Conditional probability for
independent events E1 , E2:
P(E1 | E2 )  P(E1 )
w here P(E 2 )  0
P(E 2 | E1 )  P(E 2 )
w here P(E1 )  0
Chap 4-33
Multiplication Rules

Multiplication rule for two events E1 and E2:
P(E1 andE2 )  P(E1)P(E 2 | E1)
Note: If E1 and E2 are independent, then P(E 2 | E1 )  P(E 2 )
and the multiplication rule simplifies to
P(E1 andE2 )  P(E1)P(E 2 )
Chap 4-34
Tree Diagram Example
P(E1 and E3) = 0.8 x 0.2 = 0.16
Car: P(E4|E1) = 0.5
Gasoline
P(E1) = 0.8
Diesel
P(E2) = 0.2
P(E1 and E4) = 0.8 x 0.5 = 0.40
P(E1 and E5) = 0.8 x 0.3 = 0.24
P(E2 and E3) = 0.2 x 0.6 = 0.12
Car: P(E4|E2) = 0.1
P(E2 and E4) = 0.2 x 0.1 = 0.02
P(E3 and E4) = 0.2 x 0.3 = 0.06
Chap 4-35
Get Ready….

More Probability
Examples

Random Variables

Probability Distributions
Chap 4-36
Introduction to Probability
Distributions


Random Variable – “X”
 Is a function from the sample space to
another space, usually Real line
 Represents a possible numerical value from
a random event
Each r.v. has a Distribution Function – FX(x),
fX(x) based on that in the sample space
 Assigns probability to the (numerical)
outcomes (discrete values or intervals)
Chap 4-37
Random Variables

Not Easy to Describe
Chap 4-38
Random Variables
Chap 4-39
Random Variables

Not Easy to Describe
Chap 4-40
Introduction to Probability
Distributions

Random Variable
 Represents a possible numerical value from
a random event
Random
Variables
Discrete
Random Variable
Continuous
Random Variable
Chap 4-41
Discrete Probability Distribution

A list of all possible [ xi , P(xi) ] pairs
xi = Value of Random Variable (Outcome)
P(xi) = Probability Associated with Value

xi’s are mutually exclusive
(no overlap)
xi’s are collectively exhaustive
(nothing left out)
0  P(xi)  1 for each xi

S P(xi) = 1


Chap 4-42
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 4-43
Discrete Probability Distribution
Experiment: Toss 2 Coins.
T
T
H
H
T
H
T
H
Probability Distribution
x Value
Probability
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Probability
4 possible outcomes
Let x = # heads.
.50
.25
0
1
2
x
Chap 4-44
X )  1 as X  
The Distribution Function

Assigns Probability to Outcomes



F, f > 0; right-continuous
F ( X )  0 as X   and F ( X )  1 as X  
P(X<a)=FX(a)

Discrete Random Variable

Continuous Random Variable
Chap 4-45
Discrete Random Variable
Summary Measures - Moments

Expected Value of a discrete distribution
(Weighted Average)
E(x) = Sxi P(xi)

Example: Toss 2 coins,
x = # of heads,
compute expected value of x:
x
P(x)
0
.25
1
.50
2
.25
E(x) = (0 x .25) + (1 x .50) + (2 x .25)
= 1.0
Chap 4-46
Discrete Random Variable
Summary Measures

(continued)
Standard Deviation of a discrete distribution
σx 
 {x  E(x)} P(x)
2
where:
E(x) = Expected value of the random variable
x = Values of the random variable
P(x) = Probability of the random variable having
the value of x
Chap 4-47
Discrete Random Variable
Summary Measures

(continued)
Example: Toss 2 coins, x = # heads,
compute standard deviation (recall E(x) = 1)
σx 
 {x  E(x)} P(x)
2
σ x  (0  1)2 (.25)  (1  1)2 (.50)  (2  1)2 (.25)  .50  .707
Possible number of heads
= 0, 1, or 2
Chap 4-48
Two Discrete Random Variables

Expected value of the sum of two discrete
random variables:
E(x + y) = E(x) + E(y)
= S x P(x) + S y P(y)
(The expected value of the sum of two random
variables is the sum of the two expected
values)
Chap 4-49
Sums of Random Variables




Usually we discuss sums of INDEPENDENT random
variables, Xi i.i.d.
Only sometimes is f SX  f
Due to Linearity of the Expectation operator,
E(SXi) = SE(Xi) and
Var(SXi) = S Var(Xi)
CLT: Let Sn=SXi then (Sn - E(Sn))~N(0, var)
Chap 4-50
Covariance

Covariance between two discrete random
variables:
σxy = S [xi – E(x)][yj – E(y)]P(xiyj)
where:
xi = possible values of the x discrete random variable
yj = possible values of the y discrete random variable
P(xi ,yj) = joint probability of the values of xi and yj occurring
Chap 4-51
Interpreting Covariance

Covariance between two discrete random
variables:
xy > 0
x and y tend to move in the same direction
xy < 0
x and y tend to move in opposite directions
xy = 0
x and y do not move closely together
Chap 4-52
Correlation Coefficient

The Correlation Coefficient shows the
strength of the linear association between
two variables
σxy
ρ
σx σy
where:
ρ = correlation coefficient (“rho”)
σxy = covariance between x and y
σx = standard deviation of variable x
σy = standard deviation of variable y
Chap 4-53
Interpreting the
Correlation Coefficient

The Correlation Coefficient always falls
between -1 and +1
=0
x and y are not linearly related.
The farther  is from zero, the stronger the linear
relationship:
 = +1
x and y have a perfect positive linear relationship
 = -1
x and y have a perfect negative linear relationship
Chap 4-54
Useful Discrete Distributions

Discrete Uniform

Binary – Success/Fail (Bernoulli)

Binomial

Poisson

Empirical


Piano Keys
Other “stuff that happens” in life
Chap 4-55
Useful Continuous Distributions

Finite Support



Uniform fX(x)=c
Beta
Infinite Support




Gaussian (Normal) N(m,)
Log-normal
Gamma
Exponential
Chap 4-56
Section Summary

Described approaches to assessing probabilities

Developed common rules of probability

Distinguished between discrete and continuous
probability distributions

Examined discrete and continuous probability
distributions and their moments (summary
measures)
Chap 4-57
Probability Distributions
Probability
Distributions
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Normal
Poisson
Uniform
Etc.
Etc.
Chap 4-58
Some Discrete Distributions
Chap 4-59
Binomial Distribution Formula
n!
x nx
P(x) 
p q
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)
p = probability of “success” per trial
q = probability of “failure” = (1 – p)
n = number of trials (sample size)
Example: Flip a coin four
times, let x = # heads:
n=4
p = 0.5
q = (1 - .5) = .5
x = 0, 1, 2, 3, 4
Chap 4-60
Binomial Characteristics
Examples
μ  np  (5)(.1)  0.5
Mean
σ  npq  (5)(.1)(1 .1)
 0.6708
P(X)
.6
.4
.2
0
X
0
μ  np  (5)(.5)  2.5
σ  npq  (5)(.5)(1 .5)
 1.118
n = 5 p = 0.1
P(X)
.6
.4
.2
0
1
2
3
4
5
n = 5 p = 0.5
X
0
1
2
3
4
5
Chap 4-61
Using Binomial Tables
n = 10
x
p=.15
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
0.1969
0.3474
0.2759
0.1298
0.0401
0.0085
0.0012
0.0001
0.0000
0.0000
0.0000
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=.85
p=.80
p=.75
p=.70
p=.65
p=.60
p=.55
p=.50
x
Examples:
n = 10, p = .35, x = 3:
P(x = 3|n =10, p = .35) = .2522
n = 10, p = .75, x = 2:
P(x = 2|n =10, p = .75) = .0004
Chap 4-62
The Poisson Distribution

Characteristics of the Poisson Distribution:

The outcomes of interest are rare relative to the
possible outcomes

The average number of outcomes of interest per time
or space interval is 

The number of outcomes of interest are random, and
the occurrence of one outcome does not influence the
chances of another outcome of interest

The probability of that an outcome of interest occurs
in a given segment is the same for all segments
Chap 4-63
Poisson Distribution Formula
( t ) e
P( x ) 
x!
x
 t
where:
t = size of the segment of interest
x = number of successes in segment of interest
 = expected number of successes in a segment of unit size
e = base of the natural logarithm system (2.71828...)
Chap 4-64
Poisson Distribution Shape

The shape of the Poisson Distribution
depends on the parameters  and t:
t = 0.50
t = 3.0
0.25
0.70
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
x
5
6
7
0
1
2
3
4
5
6
7
8
9
10
x
Chap 4-65
Poisson Distribution
Characteristics


Mean
μ  λt
Variance and Standard Deviation
σ  λt
2
where

σ  λt
 = number of successes in a segment of unit size
t = the size of the segment of interest
http://www.math.csusb.edu/faculty/stanton/m262/poisson_distribution
/Poisson_old.html
Chap 4-66
Graph of Poisson Probabilities
0.70
Graphically:
0.60
 = .05 and t = 100
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
t =
0.50
0.50
0.40
0.30
0.20
0.10
0.00
0
1
2
3
4
5
6
7
x
P(x = 2) = .0758
Chap 4-67
Some Continuous Distributions
Chap 4-68
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
ba
0
if a  x  b
otherw ise
where
f(x) = value of the density function at any x value
a = lower limit of the interval
b = upper limit of the interval
Chap 4-69
Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
.25
2
6
x
Chap 4-70
Normal (Gaussian) Distribtion
Chap 4-71
f X ( x) 

m
 1 x
2
1 e
2






2






By varying the parameters μ and σ, we
obtain different normal distributions

μ ± 1σ encloses about 68% of x’s; μ ± 2σ

covers about 95% of x’s; μ ± 3σ covers
about 99.7% of x’s
The chance that a value that far or farther
away from the mean is highly unlikely, given
that particular mean and standard deviation
Chap 4-72
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(x) P(  x  μ)  0.5
0.5
P(μ  x  )  0.5
0.5
μ
x
P(  x  )  1.0
Chap 4-73
The Standard Normal Distribution



Also known as the “z” distribution
= (x-m)/
Mean is by definition 0
Standard Deviation is by definition 1
f(z)
1
0
z
Values above the mean have positive z-values,
values below the mean have negative z-values
Chap 4-74
Comparing x and z units
μ = 100
σ = 50
100
0
250
3.0
x
z
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (x) or in standardized units (z)
Chap 4-75
Finding Normal Probabilities
(continued)
Suppose x is normal with mean 8.0
and standard deviation 5.0.
 Now Find P(x < 8.6)

P(x < 8.6)
.0478
.5000
= P(z < 0.12)
= P(z < 0) + P(0 < z < 0.12)
= .5 + .0478 = .5478
Z
0.00
0.12
Chap 4-76
Using Standard Normal Tables
Chap 4-77
Section Summary

Reviewed discrete distributions


binomial, poisson, etc.
Reviewed some continuous distributions

normal, uniform, exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems
Chap 4-78