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Probability Models
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UNIVERSITY OF PRETORIA
DEPARTMENT OF INDUSTRIAL AND SYSTEMS ENGINEERING
ENGINEERING STATISTICS BES221
PROBABILITY MODELS
The Chancellor of the Exchequer is a man whose
duties make him more or less of a taxing machine.
He is entrusted with a certain amount of misery which
it is his duty to distribute as fairly as he can.
Robert Lowe, Viscount Sherbrooke, British Liberal politician
Histograms
Probabilities
Distributions
Distributions
Probability Models
Sampling
DISCRETE DISTRIBUTIONS : CONCEPTS
Random variables – Variates
Output from a stochastic process
Discrete and continuous variates and distributions
Histograms and distributions
Stochastic models – mathematical
The raison d’être of probability models
BINOMIAL DISTRIBUTION (Discrete)
Variate : Number of “successes” in n Bernoulli trails
Number of “successes” in a “small” sample of size n
taken from a “large” population of size N
Jean (or Johann) Bernoulli (1667 – 1748)
Jacques (or Jakop) Bernoulli (1654 – 1705)(brother of Jean)
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Daniel Bernoulli
(1700 – 1782)
Johann Bernoulli
(1667 – 1748)
(Son of Johann)
Assumptions :
1. Only two possible outcomes – “success” and “failure”
2. Probability of a “success” stays the same and constant = p
3. There are n trails with n a constant
4. Trails are mutually independent
P(X=x) = b(x;n,p) = nCx px (1-p)n-x
x = 0,1,2,…, n
0p1
n > 0 and integer
Mean :
 =np
Variance : 2 = n p (1-p)
Symmetrical Binomial Distributions (n large and p  0,5)
The Normal approximation to the Binomial distribution
(In general, if np > 5 and n(1-p) > 5 this approximation may be
used)
Cumulative probabilities
Binomial probability tables
Applications of the Binomial distribution
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POISSON DISTRIBUTION (Discrete)
Siméon-Denis Poisson (1781 – 1840) (Fr : Fish)
The Poisson Process (One of the most important
stochastic processes)
Assumptions :
1. The probability of a “success” during a
small interval of time is proportional to the
length of the time interval
2. The probability of more than one “success” during a small
interval of time is negligible
3. The probability of a “success” during a small interval of
time does not depend on what happened during any of the
previous time intervals (no memory property)
Variate : A discrete number of occurrences per time/space/entity
(Goals per soccer game, Drop-goals per rugby game, Typing
errors per page, Defectives per assembly, Arrivals per hour,
Inclusions per m3, Particles per liter, Breakdowns per day, Dropouts per class, Accidents per kilometer, Sixes per cricket innings,
Freckles per nose, etc.)
P(X=x) = f(x;) = x e- / x!
for
x = 0,1,2, …
Mean :
=
Variance : 2 = 
Use Poisson as a good approximation to the Binomial if n  20
and p  0,05 or if n  100 then np  10
Relationship to the Exponential distribution – Queuing Theory
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MEAN AND VARIANCE OF DISTRIBUTIONS
Every distribution has a mean and variance that are functions of
the parameters of the distribution
In general :
 = Σ x f(x)
2 = Σ (x - )2 f(x)
Binomial :
 =np
2 = n p (1-p)
 =
2 = 
There are many other somewhat specialized discrete distributions
Poisson :
The most widely used, and therefore most important, discrete
distributions are the Binomial and Poisson
MONTE CARLO SIMULATION
Generating and transforming random numbers (using a pseudorandom generator and appropriate transformations) may be used
to simulate many statistical experiments (with the help of a
computer and appropriate software !)
(For further information you may enroll for BAN313 !)
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Probability Models
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CONTINUOUS DISTRIBUTIONS : CONCEPTS
Continuous random variables
Probability is an area :
P(a  X  b) = ∫ f(x) dx for all values of x between a and b
 = ∫ x f(x) dx
for all values of x
2 = ∫(x - )2 f(x) dx
for all values of x
Please note : P(X = x) = 0
if X is continuous !!!
THE NORMAL DISTRIBUTION (Continuous)
Developed and applied mainly by Abraham de Moivre (16671754), Carl Friedrich Gauss (1777-1855), Pierre Simon Laplace
(1747-1827) and Sir Francis Galton (1822 -1911)
Carl Friedrich Gauss
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Abraham de Moivre
Pierre Simon Laplace
Sir Francis Galton
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f(x; ,2) = [ 1 / {(2) }] [exp { –(x - )2 / 22 }]
- < x < +
This function is NOT analytical integratable !
Mean is  and Standard Deviation is 
The Standard Normal distribution :  =
0 and 2 = 1
One may transform ANY Normal distribution to the Standard
Normal distribution by using the following transformation
(transforming the variate X to the variate Z) :
Z = (X - ) / 
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B
A

X
X1
X2
B
A
z1 =
(x1 -  ) / 
0
Z
z2 =
(x2 -  ) / 
The use of the Standard Normal Tables (examples)
The Normal approximation to the Binomial. Approximation is
acceptable if both np and n(1-p) are greater than 15
The Normal distribution is probably the most important
distribution in mathematical statistics !
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Probability Models
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OTHER CONTINUOUS DISTRIBUTIONS
The Uniform or Rectangular distribution
The Log-Normal distribution
The Gamma distribution
The Beta distribution
The Weibull distribution
The Exponential Distribution (special form of the Gamma)
f(x) = (1 /  ) e –(x /  )
x0
Mean is  and Standard Deviation is 
Cumulative distribution function : F(t) = 1 - e –(t /  )
Therefore P(0  x  t) = F(t)
F(t)
f(x)
0
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t
x
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LINEAR COMBINATIONS OF RANDOM VARIABLES
If X1 and X2 are two independent random variables with means 1
and 2 and standard deviations 1 and 2 respectively and
Y = a1 X1  a2 X2
Then
Y = a1 1  a2 2
(Y)2 = (a1)2 (1)2 + (a2)2 (2)2
Furthermore
If X1 and X2 are both Normally distributed then Y is also
Normally distributed
Y will approach a Normal distribution, independent of the
distributions of Xi, if i is fairly large (For i > 30 the Normality
assumption can be made with confidence, mainly because of the
Central Limit Theorem)
There are a large number of other continuous distributions of
special nature and application, but the Normal and Exponential
distributions are probably the most important
{Special relationship between the Poisson and Exponential
distributions}
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Probability Models
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WORKED EXAMPLES
The Poisson Distribution
The rate at which people arrives at the ground floor of a building,
with the purpose of using the lift, is Poisson distributed with a
mean of 8 persons per minute. The maximum capacity of the lift is
14 people. Determine the probability that more people will arrive
during a 2-minute interval than what the lift can accommodate.
1 = 8 persons per minute
therefore, during a two minute interval
2 = 16 persons per two minutes (The Poisson Distribution may
be scaled !)
P(X = x)
= x e- / x!
for
P(X  15) = 1 - P(X  14)
= 1 -  (x e- / x!)
= 1 - 0,367527
= 0,6325
x = 0,1,2, …
for x = 0 to 14 and  = 16
The Exponential Distribution
A manufacturer of colour television sets offers a guarantee
consisting of replacing the TV-tube if it fails within one year. The
lifetime of television tubes is Exponentially distributed with a mean
of 4 years.
Determine the expected percentage of all television sets sold that
will have to be repaired under the guarantee.
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Probability Models
F(t)
f(X)
x=t=1
X
 = 4 (mean) and t = 1
F(t) = 1 - e –(t /  ) = 1 – 0,7788
= 0,2212
22% of the tubes will have to be replaced under guarantee
The Normal Distribution
The time it takes for a driver of a motor vehicle to react to the
brake lights of a preceding vehicle is Normally distributed
(Ergonomics, 1993, pp. 390-395) with a mean of 1,25 seconds and
a standard deviation of 0,46 seconds. Determine the probability
that the reaction time will be more than 1,75 seconds.
 = 1,25 and  = 0,46
x = 1,75
z = (x - ) / 
= (1,75 – 1,25) / 0,46
= 1,0869
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0
z=1,0869
x=1,75
Z
P(X  1,75) = P(Z  1,0869)
= 1 - P(Z  1,09)
= 1 - 0,8621
= 0,1379
Linear Combinations of Random Variables
Two resistors are installed in series. The nominal resistance of each
component is 10 Ohm. It is known that the resistance is Normally
distributed.
Determine the value of the standard deviation of the resistance to
ensure that the probability for the total resistance of the assembly to
exceed 21 Ohm is 0,01 or less.
Y = a1 X1  a2 X2 with both a1 and a2 equal to 1
Y = a1 1  a2 2 = 10 + 10 = 20
z = (x - ) /  = (21 – 20) / Y
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0,01
0
z = 1 / Y
Z
1 / Y= 2,33 (from tables)
Y
= 0,4292
(Y)2 = (a1)2 (1)2 + (a2)2 (2)2 = 0,4292
2()2 = 0,4292 and  = 0,4632
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