1614 probability-models and concepts

ROBABILITY:P
MODELS & CONCEPTS

Chance, Consequence & Strategy:
Likelihood or Probability
Since there is little in life that occurs with
absolute certainty, probability theory has
found application in virtually every field of
human endeavor.

Why Probability Theory?
As we observe the universe about us, wonderful
Craftsmanship can be seen.
As we examine the elements of this creation we
discover that there is incredible order, but also
variation therein.
Probability theory seeks to describe the variation or
randomness within order so that underlying order
may be better understood.
Once understood, strategies can be more effectively
formulated and their risks evaluated.

Objective Assessment:
Apriori & Aposteriori Probability
 Apriori means “before the fact” and hence probability
assessments of this sort typically rely on a study of
traits of the phenomenon under consideration.
 Based on Theory.
Aposteriori means “after the fact”. This
approach to likelihood assessment is also
called the “relative frequency” approach.
Based on repeated observation.

Likelihood Concepts
EVENTS
• As we observe a phenomenon, we generally
note that varying, and sometimes “identical”
conditions do not always give rise to identical
results. As a phenomena is repeatedly
observed, the various possible results can be
thought of as “events”.

Mutually Exclusive Events
• Any number of events are
said to be mutually
exclusive if they have no
overlap or commonality.
“Nothing is impossible Mario; improbable, unlikely maybe, but
not impossible.” Luigi Mario speaking to brother, Mario Mario in the
movie, “Super Mario Bros.”

• A collection of events is exhaustive if,
taken in totality, they account for all
possible results or outcomes.
A
B A and B are mutually exclusive.
Mutually Exclusive &
Exhaustive Events

Intersection & Union of Events
The intersection of two or more events is
like the intersection of two streets --- it is
the property they share in-common.
The intersection of events A and B is
symbolized by AB.
The union of two or more events is the
totality of results captured by these events.
The union of two events A and B is
symbolized by AUB

Notation & Definitions
 The probability of the event A is given by: P(A)
 The probability of AB is P(AB) = P(A) + P(B) - P(AUB)
where P(AUB) is the probability of the union of events A and
B.
 The conditional probability of the event A given that the event B
has occurred is: P(A|B) = P(AB)/P(B)
DEPENDENCE & INDEPENDENCE
 Two events A and B are said to be independent if and only if:
P(A|B) = P(A) and P(B|A) = P(B)
 It follows from this that if A and B are independent then P(AB) =
P(A)*P(B)
 This is the multiplication rule for independent events.

A Service Sector Example:
Fast Food Clientele
A leading fast food restaurant chain routinely & randomly
surveys its customers in an effort to continually improve ability
to serve their clientele.
Two primary questions on the survey address frequency of
customer patronage and primary reason for this patronage.
Results of last month’s survey of 1,000 customers are
recorded in the following table.

Survey of 1000 Customers:
Frequency of and Reason for Patronage
occasional moderate frequent TOTALS
menu/food 60 120 30 210
customer
relations
75 180 45 300
value/cost 35 200 40 275
location/
access
60 80 25 165
other
reason
20 20 10 50
TOTALS 250 600 150 1000

Marginal Probability
• Marginal probabilities can be thought of as the
probabilities of being in the various margins of the
table. For example, the marginal probability of a
customer patronizing the restaurant chain due to
menu, regardless of frequency of patronage is:
• P(menu) = 210/1000 = .21
• The various marginal probabilities for this example
are determined and represented graphically as
follows. The graphs are “marginal probability
distributions”.

Frequency of Patronage:
Marginal Probability Distribution
• Occasional Patrons:
P(occasional) = 250/1000 = .25
• Moderate Patronage:
P(moderate) = 600/1000 = .60
• Frequent Patrons:
P(frequent) = 150/1000 = .15
0
0.1
0.2
0.3
0.4
0.5
0.6
occasion
moderate
frequent

Reasons for Patronage:
Marginal Probability Distribution
• Menu: P(Menu) = 210/1000 = .21
• C.Rel.: P(CR) = 300/1000 = .30
• Value: P(Value) = 275/1000 = .275
• Location: P(Loc) =165/1000 = .165
• Other: P(Other) = 50/1000 = .05
0
0.05
0.1
0.15
0.2
0.25
0.3
menu
cust.rel.
value
location
other

Joint Probability
• Consider the cross-tabulation relating the two traits:
– frequency of patronage, and
– primary reason for patronage
• Joint probabilities are probabilities of intersections of the
categories (or events) of two traits. As an example, the joint
probability that a customer is moderate in their patronage and
their primary reason for patronage is the menu is given by
– P(moderate ∩ menu) = P(AB) = 120/1000 = .120.
• A graphical representation of the complete joint probability
distribution follows.

Reasons & Frequency of Patronage
Joint Probabilities
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
occasion moderate frequent
menu
cust.rel.
value
location
other

Conditional Probability
 Conditional probability can be thought of as probability
determined in the mode of either “what if” or “given that”
 For example, we might ask, “what is the probability that a
customer’s primary reason for patronage is the value (A), given
that the customer is frequent (B) in their patronage?”
 This is symbolized by P(A|B) and is calculated as P(AB)/P(B)
where the vertical line, “|” is read as “given that”.
 Thus a “conditional probability” is equal to the probability of
the appropriate intersection, divided by the marginal
probability of the given.

Reasons for Frequent Patrons:
Conditional Probabilities
0
0.2
0.4
menu
cust.rel.
value
location
other
• P(value | frequent) = P(value ∩ frequent)/P(frequent) =
(40/1000) / (150/1000) = .04/.15 = .267
• This is represented by the “red” bar above. The entire
“conditional probability distribution of reasons for patronage by
frequent customers” is displayed above.

Independence & Dependence
 Recall that two events, A & B, are mutually independent if
and only if
 P(A|B) = P(A) and P(B|A) = P(B)
 Are the events A & B independent where:
 A = Primary patronage reason is customer relations
 B = Customer is a frequent in patronage
 Recall that P(A) = .3, that P(B) = .15 and that P(AB) =
45/1000 = .045 so that
 P(A|B) = P(AB)/P(B) = .045/.15 = .30 = P(A)
 P(B|A) = P(AB)/P(A) = .045/.30 = .15 = P(B)
 Indeed, A & B are independent.,

Independence - Key Concept
If two events, A & B, are independent then the
occurrence of one of the two events does not
change the LIKELIHOOD or probability that the
other of the two events will occur.
Occurrence of one of the two events does alter the
MANNER in which the other of the two events
may occur.

Dependence
 If two events A & B are dependent then P(A|B) will not equal
P(A) and, similarly, P(B|A) will not equal P(B).
 Let A = primary reason for patronage is menu.
 Let B = frequency of patronage is moderate.
 We have P(A|B) = 120/600 = .20 and is not equal to P(A)
= 210/1000 = .21.
 P(B|A) = 120/210 = .57 which is not equal to P(B) =
600/1000 = .60.
 In this case, even though values are comparable they are not
equal => dependence.

Dependence - Key Concept
 If two events A & B are dependent, then occurrence of
one of the two events will alter the likelihood and the
manner in which the other of the two events may occur.
 In the case of mutually exclusive events, occurrence of
one of the two events will preclude occurrence of the
other event.
 Mutually exclusive events are always dependent.

Probability Models
 Probability models are mathematical descriptions of the
behavior of one or more variables. The ability to
somewhat anticipate the behavior of a variable can be
useful in risk assessment and strategy formulation.
 Three commonly used models, the binomial, Poisson,
and normal models, are introduced.
 Random variables described by these models may be
either ‘discrete’ or ‘continuous’.

Mean, Variance and Standard
Deviation of a Random Variable
 The mean of a random variable (r.v.) Y is denoted by
µY. For a discrete r.v. Y this is calculated as: µY =
ΣyiP(yi) This is the weighted average of the values of
Y. For continuous random variables, integration
replaces summation.
 The variance and standard deviation of the r.v. Y are
represented by σ2
Y and σY, respectively. For a discrete
r.v. Y, these are:
σ2
Y = ΣP(yi)(yi - µY)2
and σY = σ2
Y

The Poisson Model
Napoleon had a problem: many of
his men were killed when kicked in
the head by their own horse or
mule.
Napoleon had to plan for this
problem.
The Poisson model helped him to
do so.

Poisson Conditions
The Poisson model (or distribution) is
commonly applicable when:
 We are modeling events which occur only “rarely”,
where “rare” means “rare relative to opportunity for
occurrence”.
 Our random variable will be the “number of occurrences
of the event over the region of opportunity for
occurrence”.

Poisson Conditions:
Region of Opportunity
Examples of region of opportunity include:
number of customers arriving per minute
(or any other time unit);
number of phone calls arriving at a switch
board per unit time;
number of scars on the surface of a compact
disk.
Generally “region of opportunity” is defined
either temporally or spatially.

The Poisson Model is Integral to
the Study of Queueing Theory

The Poisson Model
• Defining our random variable as Y = “number of
occurrences of the event over the region of opportunity”,
y = 0, 1, 2, 3, ... we have the Poisson probability model:
• P(y) = µy
e-µ
/y! for y = 0, 1, 2, 3, ...
• Where µ is the mean or average number of occurrences
of the event over the region of opportunity and e =
2.7183 is the natural base.

Estimation of the Process Mean, µ
• The mean of the Poisson process is µ,
• The variance of the process is also µ, that is
σ2
= µ
• so that the standard deviation is σ = µ
• In the following example we proceed as though µ is of
known value. When this is not the case we simply
estimate µ with X, the mean of the sample.

First Federal Bank of Centerville
A Queueing Example
 First Federal Bank (FFB) of Centreville has an automatic teller
machine (ATM) near the entrance of the bank.
 Long lines at the ATM have sometimes led to congestion and
perhaps a diminishing clientele. With a view toward improved
customer service, FFB is considering the addition of one or
more ATMs or, possibly, relocation of the current ATM.
 During peak hours ATM users arrive in a manner described by a
Poisson distribution with a mean of 1.7 customers per minute.

First Federal Bank of Centreville
The Probability Distribution
• What is the probability that no customers arrive in one-
minute during a peak business period?
• Solution: P(0) = 1.70
e-1.7
/0! = .1827
• Similarly, P(1) = 1.71
e-1.7
/1! = .3106
• Determine probabilities for 2, 3, ...., 9 customers. The
probability distribution appears on the next slide.

FFB of Centreville
Probability Distribution
x P(X = x)
0 0.1827
1 0.3106
2 0.2640
3 0.1496
4 0.0636
5 0.0216
6 0.0061
7 0.0015
8 0.0003
9 0.0001
10 0.0000
x P(X LESS < x)
0 0.1827
1 0.4932
2 0.7572
3 0.9068
4 0.9704
5 0.9920
6 0.9981
7 0.9996
8 0.9999
9 1.0000
PoissonProbabilitieswithµ=1.7
PoissonCumulativeProbabilities
withµ=1.7

ATM Customer Probabilities
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0
1
2
3
4
5
6
7
8
9

CDF Graph
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
6
7
8
9
The cdf graph above was constructed by adding the appropriate Poisson probabilities.

Key Considerations
Key factors that FFB should address prior to
making a decision include:
What is the service rate (how quickly do customers
complete their ATM transactions)?
If long lines are forming during peak hours, the service
rate may be less than customer arrival rate and addition
of one or more ATMs may be necessary.
If the problem is congestion, rather than excessive wait
to use the ATM, the solution may be to simply move the
ATM.

Model Adequacy:
Chi-Square Goodness of Fit Testing

DOES THIS MODEL FIT?
Chi-Square Goodness-of-Fit Tests
 The purpose of χ2
goodness-of-fit tests is to
evaluate whether a particular probability distribution
does an adequate job of modeling the behavior of the
process under consideration. This sort of test can be
applied to any model.
 A “skeleton” or template for the chi-square
 goodness-of-fit test follows.

χ2
Goodness of Fit Test - General Layout.
1) H0: p1 = p10, p2 = p20, ... , pk = pk0
HA: at least one pi ≠ pi0
2) n = _______ α = _______
3) DR: Reject H0 in favor of HA iff χ2
calc > χ2
crit = ___.
Otherwise, FTR H0.
4) χ2
calc = Σ(Oi - npio)2
/npio = Σ(Oi - Ei)2
/Ei
5) Interpretation: Should relate to whether the hypothesized
model adequately describes behavior of the process under
consideration.

Generic Example: A computer manufacturer produces a disk
drive which has three major causes of failure (A, B, C) and a
variety of minor failure causes (D).
Suppose that historic failure rates are:
Due to A: .20 Due to B: .35 Due to C: .30 Due to D: .15
The manufacturer has worked on A, B, and C and believes that failures due
to these causes has been reduced, so that, while fewer failure will occur, it is
more likely that when one occurs, it will be due to D. To examine this claim
the manufacturer will sample 200 failed disk drives manufactured since
process changes were made.
IF THE CHANGES HAD NO IMPACT then the number of these failed drives
that were due to causes A, B, C, and D that would be EXPECTED would be:
EA = npA0 = 200(.20) = 40 EB = npB0 = 200(.35) = 70
EC = npC0 = 200(.30) = 60 ED = npD0 = 200(.15) = 30
Upon observation, suppose that we had OA = 28, OB = 66, OC = 46, OD =
60. Test the appropriate hypothesis at the α = .05 level.
CONTINUED NEXT PAGE

Failure Mode Profile Example - Continued
1) H0: pA = .20, pB = .35, pC = .30, pD = .15
HA: at least one pi ≠ pi0 for i = A, B, C, D
2) n = 200 α = .05
c > χ2
T = 7.8147. Otherwise, FTR H0.
Note: There are (k-1) = 3 degrees of freedom.
4) χ2
c = Σ(Oi - npio)2
/npio = Σ(Oi - Ei)2
/Ei
= (28-40)2
/40 + (66-70)2
/70 + (46-60)2
/60 + (60-30)2
/30
= 3.6000 + 0.2286 + 3.2667 + 30.0000 = 37.0953
5) Interpretation: Since χ2
c exceeds χ2
T, we can conclude that the historic failure
mode distribution no longer applies (reject H0 in favor of HA). So how has the
distribution changed? The answer is embedded in the individual category
contributions to χ2
calc ... larger contributions indicate where the changes have
occurred: reductions in A and C, no obvious change in B, the various failures that
make-up D now comprise a (proportionally) larger amount of the failures.

Chi-Square Goodness of Fit Test
for the Poisson Distribution
A sample of 120 minutes selected during rush periods at FFB
gave the following number of customers arriving during each of
those 120 minutes. Is this data consistent with a Poisson
distribution with a mean of 1.7 customers per minute, as
previously stated? Test the appropriate hypothesis at the α = .10
level of significance.
Number of 0 1 2 3 4 or more
Customers
Frequency 25 42 35 9 9

FFB of Centreville
Poisson Goodness of Fit Test
Customers/ Prob. Obs (O) Exp (E) (O-E)2
/E
minute
0 0.1827 25 21.924 0.4316
1 0.3106 42 37.272 0.5998
2 0.2640 35 31.680 0.3479
3 0.1496 9 17.952 4.4640
> 4 0.0932 9 11.184 0.4265
1.00 120 120 6.2698 = χ2
calc
with α = .10 and (k-1) = 4 df, the critical value is 7.7794

FFB of Centreville - Continued
1) H0: the number of customers arriving per minute is Poisson distributed
with a mean of 1.7. OR
p(0) = .1827 p(1) = .3106 p(2) = .2640 p(3) = .1496 p(4+) = .0932
HA: the number of customers arriving per minute is not Poisson with µ = 1.7
2) n = 120 and α = .10
calc > χ2
crit = 7.7794. Otherwise, FTR H0.
(NOTE - THERE ARE 4 DF)
4) χ2
calc = 6.2698 (calculations on previous slide)
5) FTR H0. In this case, the number of customers arriving per minute during the business
rush at FFB of Centreville is reasonably well-modeled by a Poisson distribution with a mean
of 1.7.
As a modification --- if we had not had information about the mean number of customers
arriving per minute, we would have had to estimate this value with the sample mean and
then determined the estimated probabilities. This would have cost an additional degree of
freedom (e.g. df = (k-1) - 1 = 3.

Binomial Conditions
Suppose that there are two possible outcomes to an
experiment which are mutually exclusive and exhaustive
(refer to these generically as “success” and “failure”);
a predetermined sample size, n;
the probability of “success” is a constant, p, and the
probability of “failure” is a constant, (1-p);
the condition of one item is not influenced by the
condition of any other item (this is called independence).
Collectively, these are the binomial conditions.

Binomial Probability Model
• When the binomial conditions are present, and the
random variable Y is defined as the number of
“successes” out of n items sampled, then the model
which determines probabilities for the various values of
Y is given by:
• P(Y = y) = [nCy]py
(1-p)n-y
• where nCy = n!/[y!(n-y)!] is read as the number of
combinations of n things selected y-at-a-time.
• with any integer x! being x(x-1)(x-2)...(1)
• so that, for example, 5! = 5(4)(3)(2)(1) = 120

Binomial Mean, Variance
and Standard Deviation
• Although the formulas previously presented can
be used to determine the values of µY, σ2
Y and σY,
the following results are more easily applied in
the binomial case:
µY = np
σ2
Y = np(1-p)
σY = np(1-p)

Estimation of p
 The binomial parameter, p, is thought of as the “probability that
any single item sampled is identified as a ‘success’ “.
 Frequently this value will be unknown and will need to be
estimated from sample information.
 p is estimated as simply x/n where x is the number of ‘successes’
in the sample of n items.
 This estimate is often denoted by p.
 Similarly, the estimate of (1-p) is (1-p).
^
^

The Electronix Store
In a competitive local retail electronics market, the
probability that a randomly selected “customer”
browsing in The Electronix Store will make a
purchase is .2.
 If 6 “customers” are randomly selected, what is the
probability that exactly 2 of these individuals will
make a purchase?
This and similar questions can be addressed via the
binomial distribution.

We identify:
n = 6 customers
p = .2 = probability that a customer buys
Y = number of the six customers who buy
Thus we see that:
µY = np = 6(.2) = 1.2 Customers
σ2
Y = np(1-p) = 6(.2)(.8) = .96
σY = √ .96 = .98 Customers
^
^
^ ^

• We have:
– P(0) = .2621 P(1) = .3932
– P(2) = .2458 P(3) = .0819
– P(4) = .0154 = {6!/[4!2!]}(.2)4
(.8)2
– = 15(.0016)(.64)
– P(5) = .0015 P(6) = .0001 (or .000064)

0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0
1
2
3
4
5
6
Customer Purchase Probabilities

We may require answers to such questions as:
“What is the probability that no more than two of six
customers make a purchase?”
“What is the probability that at least four of six
customers make a purchase?”
“How many cash registers are needed?”
Answers to these and similar questions can be
investigated through study such as we have
undertaken.

Cumulative Probabilities
0
0.5
1
0
1
2
3
4
5
6
• A tabulation of the “less than or equal to” probabilities
is called a “cumulative distribution function” or cdf.
The Electronix Store cdf appears above.

Application of this
information might
spark discussion on:
staffing decisions,
sales representative
specialization
focus on merchandise,
value, and customer
service.
The Electronix Store:
Strategy

χ2
Goodness-of-Fit Test:
Binomial Example
Oil & Gas Exploration is both expensive and risky. The average cost of a “dry
hole” is in excess of $20 million. New technologies are always under development
in an effort to reduce the likelihood of drilling a “dry hole” with the result being
increased profitability. Suppose an experimental technology has been developed
that claims to have an 80% success rate (e.g. only 20% dry holes). This technology
was tested by drilling four holes and counting the number of productive wells.
This was done 100 times, each time counting the number of productive wells. The
data is recorded below:
Number of
productive wells 0 1 2 3 4
Observed 3 6 22 41 28
Frequency
Test the appropriate hypothesis at the α = .01 level of significance.

Oil & Gas Exploration Example
1) H0: the new technology delivers success according to a binomial distribution with p = .8
or ... p(0 or 1) = .0272 p(2) = .1536 p(3) = .4096 p(4) = .4096
(NOTE - SEE NEXT PAGE FOR THESE VALUES)
HA: the new technology does not deliver success according to a binomial distribution
with p=.8.
2) n = 100 and α = .01
calc > χ2
crit = 11.3449. Otherwise, FTR H0.
4) χ2
calc = 21/4705 (calculations on next slide)
5) Reject H0 in favor of HA. In this case, note that “O” tends to be greater than “E” for
lower numbers of successful wells, and the reverse for higher numbers of successful
wells ... this indicates that the success rate of the new technology is LESS THAN THE
CLAIMED 80% rate.

Hits Prob Count Expected Combined C-Prob C-Count C-Expect (O-E)^2/E X^2calc
0 0.0016 3 0.16 0-1 0.0272 9 2.72 14.4994 21.4705
1 0.0256 6 2.56 2 0.1536 22 15.36 2.8704
2 0.1536 22 15.36 3 0.4096 41 40.96 0.0000
3 0.4096 41 40.96 4 0.4096 28 40.96 4.1006
4 0.4096 28 40.96

Modified Oil & Gas Exploration Example
(still binomial)
If p were unknown, then it would have to be estimated from the data.
There is a cost to this --- a lost degree of freedom. In general df = (k - 1)
- m
where k = number of categories
-1 because the probabilities across all categories add to one
(lacking only one probability, we can determine the other
m = the number of parameters that must be estimated.
In this case, the estimate of p is this: a total of 400 wells were drilled (100
fields at 4 wells each). The number of productive wells was (3*0 + 6*1
+ 28*2 + 41*3 + 22*4) = 273
So that our estimate of p is 273/400 = .6825. The modified calculations
follow.

Modified Oil & Gas Exploration Example
MTB > pdf;
SUBC> binomial n=4 p=.6825.
BINOMIAL WITH N = 4 P = 0.682500
K P( X = K) Observed Expected (O-E)2
/E
0 0.0102
combine these .0976 9 9.76 0.0592
1 0.0874
2 0.2817 28 28.17 0.0010
3 0.4037 41 40.37 0.0098
4 0.2170 22 21.70 0.0041
0.0742 = calculated value of χ2
MTB > invcdf .99;
SUBC> chis 2.
0.9900 9.2103 = critical value
Clearly we would FTR H0. So that if you combine the information, really, you have
not rejected the binomial distribution altogether ... though you did reject the binomial
distribution with p=.8. The binomial distribution with p=.6825 does an excellent job
of modeling the performance of this new oil & gas exploration technology.

The Normal Probability Model
The “normal” or “Gaussian” distribution is the most commonly
used of all probability models.
This distribution is known perhaps most familiarly as the “bell
curve”.
The normal distribution serves as the assumed model of
behavior for various phenomena, generally as an approximation.
It is also foundational to the development of numerous
commonly used statistical methods.

The Normal Distribution
• The normal distribution is described by the
mathematical expression:
f(x) = (1/ √ 2πσ2
)exp(-(x-µ)2
/2σ2
)
X is a random variable with mean µ and
standard deviation σ, exp = e = 2.7183 is the
natural base, raised to the power expressed in
the ( ). As will be seen, we need not work with
the formula above.

0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
A histogram
representation
of the normal
distribution
might appear
as this one.
 The normal distribution is symmetric about its mean, µ.
 It is also well-tabled as the “standard normal distribution”
 with µ = 0 and σ = 1.
The Normal Probability Model

Table Use - Relationships
 Since the normal distribution is
a probability distribution, with total area under the
curve equal to 1, and
symmetric about its mean, µ, we have:
 P(Z > Z*) = .5 - A(Z*) where Z* > 0
 A(-Z*) = A(Z*) by symmetry.
 Knowing these few relationships, any needed
probabilities can be found.
 Only positive values of Z need be tabled.

Z Table Use Examples
• Using available Z tables determine :
• A(1.33) and A(-1.33)
• The probability of being between Z = -1.33 and +1.33.
• The probability that Z is at most 1.33
• The probability that Z is at least 1.33
• The probability that Z is at most -1.33
• The probability that Z is between -.75 and +1.2
• The probability that Z is between +.50 and +1.2

Z Table - Selected Portions
Z 0.00 0.01 0.02 0.03 0.04 0.05 ......... 0.09
0.0 .0000 .0040 .0080 .0120 .0160 .0199 ......... .0359
0.5 .1915 .1950 .1985 .2019 .2054 .2088 ......... .2224
0.7 .2580 .2611 .2642 .2673 .2704 .2734 ......... .2852
1.2 .3849 .3869 .3888 .3907 .3925 .3944 ......... .4015
1.3 .4032 .4049 .4066 .4082 .4099 .4155 ......... .4177

Inverse Use of the Z Table
In application, there are two common variations
requiring opposite use of tables of the standard
normal distribution.
We have illustrated the first variation where, given
one or more values of Z, we can determine the
needed area under the curve (e.g. the needed
probability).
The “inverse” situation is one in which an area
under the curve is designated, and the
corresponding value(s) of Z are obtained.

Inverse Use of the Z Table
 The inverse approach is to:
locate the appropriate area or probability in the
body of the table,
then move to the corresponding top and left
table margins to identify the appropriate
value(s) of Z.
From this we have X = µ + Zσ

A(Z) = known
?
Application of the Inverse Normal

The Normal Distribution in General
We can determine probabilities for any normally
distributed process performance measure or
PPM, X, by determining the corresponding value
of Z, that is
 Z = (X - µ)/σ
Inversely, given an area under the curve, we can
determine a needed value of X as:
 X = µ + Zσ

The SUPER Market
The SUPER Market, a major metropolitan area
superstore chain, offers delivery service to addresses
within a defined region.
The SUPER Market guarantees delivery within two
hours of the time that the order is received. If this
guarantee is not met, the customer receives a 10%
discount for each 30 minutes late.

The SUPER Market
• Delivery time is approximately normally distributed
with an average delivery time of 1 hour and 20 minutes
and a standard deviation of 20 minutes. That is µ = 80
min. and σ = 20 min.
Guaranteed
delivery
within two hours!

The SUPER Market:
Time to Delivery
• Inverse Problems
• Given a designated
probability, what is the
corresponding value of Z and,
in turn, X = delivery time?

A Goodness of Fit Test for the
Normal Distribution
IS DELIVERY TIME NORMAL?
To determine whether delivery times for the SUPER MARKET are, within
reason, normally distributed we would select a random sample of delivery
times and apply any of a number of goodness of fit techniques.
While the chi-square goodness of fit test could be applied, a graphical
procedure, the normal probability plot, will be illustrated. This is
augmented by a more formal procedure, the Anderson-Darling test.
To proceed we will select a sample of, say, 40 delivery times. These appear
in the sequel.

40 Sampled Delivery Times
56 89 123 97 68 79 80 96 74 108 86
65 102 96 90 88 67 87 58 71 72 83
90 59 76 73 82 88 63 114 86 54 109
43 69 47 90 96 52 117
N Mean Median Std. Dev.
Del_Time 40 81.07 82.50 19.45

p-value: 0.934
A-Squared: 0.166
Anderson-Darling Normality Test
N of data: 40
Std Dev: 19.448
Average: 81.075
120110100908070605040
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Del_Time
Normal Probability Plot
SampledDelivery Times from theSUPER Market
Normally distributed values should plot VERY close to a straight line. While this
is a judgment call, a more objective approach is to examine the p-value from the
Anderson-Darling test -- if the p-value is less than α, then normality is questionable.

1614 probability-models and concepts

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1614 probability-models and concepts