Download Unit 2: Modeling Random Behavior Week 3 : Probability

Unit 2: Modeling Random Behavior
Week 3 : Probability
By now you should appreciate that real engineering data exhibit variability.
Probability is the tool used to model such variability. Statisticians use probability in much the
same way as chemical engineers use chemistry. Chemistry provides the chemical engineer
with the language necessary to solve typical problems in chemical engineering in much the
same way as probability is the language of uncertainty. Ultimately, probability represents a
standardised measure of chance – with zero meaning no chance and unity meaning something
is certain to occur.
Example: Nickel-Hydrogen batteries
A Nickel–Hydrogen battery (NiH2 or Ni–H2) is a rechargeable electrochemical power
source based on Nickel and Hydrogen. It differs from a Nickel–metal hydride (NIMH)
battery by the use of hydrogen in a pressurized cell at up to 8.5 MPa. The screen shot below
gives more details on the structure of this type of battery. The blistering of sintered Nickel
electrodes within this type of battery has been of continuing concern to users of long life
batteries containing these electrodes. Blistering can be affected by many parameters that are
sensitive to both the manufacturing process and to the electrodes' operating conditions. Some
of these parameters are the active material loading level and the ability of the electrode to
vent gases due to overcharge. Also, the active material impregnation process can enhance
blister formation.
Consider a manufacturer of these Nickel-Hydrogen batteries, who has run into
problems with cells shorting out prematurely. The manufacturer therefore cuts open 100
recently manufactured cells (that each have 60 nickel plates within them) and observed the
following results.
Did the cell short?
Yes
35
15
Yes
No
Did the cell have blistered plates?
No
5
45
So, for example, 35 of the 100 cells shorted and on further inspection these cells also
had one or more blisters on the Nickel plates. But 15 of the cells were fine even though some
of the plates had blisters on them. This tends to suggest that there must be a critical (i.e.
minimum) number of blisters required before a cell shorts out. 45 cells without any blistering
were fine, but 5 cells with no blisters did short - presumably because other failure
mechanisms are at work within the manufactured batteries.
A.
Defining Probability
A random experiment is one that can result in different outcomes. In the above
example there are four possible outcomes for the tested cell: the cell shorted (call this Event
A), the cell did not short (Event B), the cell had blistered plates (Event C) and the cell did not
have blistered plates (Event D).
The set of all possible outcomes of a random experiment is called the sample space,
S. In this example the sample space is made up of four outcomes
S = {Event A, Event B, Event C, Event D}.
Probability is a pure number between 0 and 1 that reflects the chances of a
particular event occurring. If an event is absolutely certain to occur that event has a
probability of one associated with it. If on the other hand an event can’t possibly occur,
that event has a probability of zero. If an experiment is repeated N times and a particular
event is observed in a fraction n of these experiments, then the ratio n/N is defined to be the
probability P of that event occurring (provided N is very large)
P(Event) = n/N
Of the N = 100 cells tested, n = 35 + 5 = 40 of them had shorted. Thus the probability
of event A occurring is:
P(A) = 40/100 = 0.40 (or 40%)
Also
P(B) = (15+45)100 = 0.6 (or 60%)
Notice that events A and B cannot occur together (either the cell has failed or it has
not) and so these events are said to be mutually exclusive. Mutually exclusive events cannot
occur at the same time. Further,
P(C) = (35+15)/100 = 0.50 (or 50%)
and
P(D) = (5+45)100 = 0.5 (or 50%)
Again, events C and D are mutually exclusive.
B.
Combinations of events (AND, OR events)
Events A and C are not mutually exclusive as 35 of the cells that shorted also had
blistered plates. Similarly, Event B and D are also not mutually exclusive. It is therefore
possible to calculate a probability of two or more events like these occurring together. For
example, the probability of observing a cell that has shorted and has blistered plates is
P(A AND C) = 35/100 = 0.35 (or 35 %)
Again the probability of observing a cell that has not shorted and has no blistered plates
is
P(B AND D) = 45/100 = 0.45 (or 45 %)
It is also possible to look at the probability of one of a number of events occurring. For
example, Event A or C. (Given that if A and C occur, we have the result that A or C has
occurred, so when we say A or C we actually mean A or C or both).
Some n = 35 + 5 + 15 = 55 out of the N = 100 cells had either shorted or had
blistered plates and so
P(A OR C) = 55/100 = 0.55 (or 55 %)
Similarly, some n = 15 + 45 + 5 = 65 out of the N = 100 cells had not shorted or did not have
blistered plates and so
P(B OR D) = 65/100 = 0.65 (or 65 %)
C.
The Addition Rule
The Addition rule of probability states
P(A OR C) = P(A) + P(C) – P(A AND C)
This can be proved using the above illustrations
P(A OR C) = 0.40 + 0.50 – 0.35 = 0.55
which is exactly the same answer as that given above using the n/N definition of probability.
For mutually exclusive events P(A AND C) would equal zero so that the addition rule
simplifies to:
P(A OR C) = P(A) + P(C)
for mutually exclusive event
D.
Conditional Probability
Often two or more events will be related. Conditional probability quantifies the chances
of one even occurring given that the others have occurred. Dependent events have
conditional probabilities associated with them, whereas independent events do not. In the
battery example, the occurrence of event A is dependent upon the presence of blistered plates,
as more cells short in the presence of blistered plates. The conditional probability of event A
occurring, given that event C has already occurred is written as P(A|C). Given that event C
has already occurred the reduced number of observations is N = 35 + 15 = 50. Out of these
some 35 cells shorted, thus
P(A|C) = 35/50 = 0.7 (or 70%)
This probability differs from P(A) and so events A and C are said to be dependent. If
A and C were independent then P(A) would be the same irrespective of whether event C had
occurred or not. So for independent events P(A|C) = P(A).
E.
The Product Rule
The Product rule of probability states
P(A|C) = P(A AND C) / P(C)
This can be proved using the above illustrations
P(A|C) = 0.35/0.5 = 0.70
which is exactly the same answer as that given above using the n/N definition of probability.
The product rule is more often expressed in the following way by rearranging the last
equation
P(A AND C) = P(A|C) P(C)
For independent events P(A|C) = P(A) so that the product rule simplifies to
P(A AND C) = P(A) P(C)
for independent events only.
All the above calculations and more are contained in the screen shot below.
F.
Systems Reliability
Many physical systems (e.g. bridges, car engines, air-conditioning systems, biological
and ecological systems) and non-physical systems (e.g. chains of command in civilian or
military organizations and quality control systems in manufacturing plants) may be viewed as
assemblies of many interacting elements. The elements are often arranged in mechanical or
logical series or parallel configurations and the reliability of a system is easily calculated
from the reliability of its components using the above rules of probability.
Series systems function properly only when all their components function properly.
Thus the addition rule of probability applies. Examples are chains made out of links,
highways that may be closed to traffic due to accidents at different locations, the food chains
of certain animal species, and layered company organizations in which information is passed
from one hierarchical level to the next.
Let Bi be the event that component i fails. Also denote by P(Bi) the probability that
component i fails. The probability of failure of a system with k components arranged in series
is then
P(Systems failure) = P(B1 or B2 …. or Bk)
For components that can’t fail together (i.e. mutually exclusive components) this
simplifies to
P(Systems failure) = P(B1) + P(B2) + …. + P(Bk)
Example: Welded parts
Consider the screen shot below showing a simple component manufactured by
welding together 3 separate parts. The welds are arranged in series as the structure will
clearly fail if any one of the welds break. Suppose the probability of an individual weld
failing is 0.01 and the welds are independent of each other. The screen shot shows how to
calculate the probability of the part failing.
Note that the probability of the part failing is close to twice that for a single weld.
Thus probability theory tells us that simple series systems (i.e. those with fewer parts (e.g.
fewer welds)) will be more reliable than those with many parts.
A parallel system fails only if all its components fail. Thus the product rule of
probability applies. For example, if an office has k copy machines, it is possible to copy a
document if at least one machine is in good working condition.
The probability of failure of a system with k components in parallel is then
P(Systems failure)= P(B1 and B2 …. and Bk)
If components fail independently of each other this simplifies to
P(Systems failure) = P(B1) x P(B2) x…. x P(Bk)
Example: Safety valves
Consider the pressure vessel with three safety valves shown in the following
screenshot. The use of three safety valves is an example of a triply redundant system
provided that destructive over pressure can only occur if all the valves fail. One valve is
sufficient to release pressure. If this is the case the three valves are said to be arranged in
parallel. Suppose also that the probability of a safety valve failing is 0.15 and that these
valves are independent of each other.
Notice that the chances of destructive over pressure are a lot less than the chances of
an individual safety valve failing. This is the benefit derived from designing in parallel. If
only two safety valves had been built into the pressure vessel the probability of destructive
over pressure would rise to:
P(Systems failure) = P(B1) x P(B2) = 0.15 x 0.15 = 0.0225
Building additional safety valves has cost implications and a sensible balance needs
to be struck between economic cost and safety.
Weeks 4 & 5: Random Variables
A.
Random Variables
In most applications of statistics to engineering problems, the researcher is only
concerned with one or maybe just a few numbers that are associated with the outcomes of the
experiment being carried out. For example, in the inspection of a manufactured product the
engineer may be interested only in the number of defectives or in the study of the
performance of a miniature rechargeable battery the engineer may be interested only in its
power and lifetime.
VOICE OF EXPERIENCE
Random variables provide a way to model the
behavior of real data.
In general, each outcome of an
experiment may be associated with a
number by specifying a rule of
association. Such a rule of association
is called a random variable – a variable because different numerical values are possible and
random because all these values are associated with situations involving an element of chance
or uncertainty. An upper case letter denotes a random variable and a lower case letter
denotes a specific value for this variable.
Example: Failure of a 9 Cr steel
Suppose four specimens of a 9Cr steel alloy are put on test at a load of 150 MPa and a
temperature of 5500 C. The engineer is interested in how many of these specimens will fail
before 1,000 hours of testing at this condition. This creates a random variable
X = number of failed specimens
which can take the values x = 0, 1, 2, 3 and 4.
The actual outcome of this experiment is uncertain – it will not be known until the end of the
test (i.e. after 1,000 hours) what value X actually takes. For example, if at the end of the test
two specimens are observed to have failed, then x = 2.
A random variable can either be discrete in nature or continuous. A discrete random
variable is one that can assume at most a countable number of values (e.g. the number of
defects in a new car or the number of defective silicon chips in a lot). A continuous random
variable is one that can assume any real value over a specified interval (e.g. the outside
diameter of a pen barrel or the times between breakdowns of a stamping press at a car
assembly plant.
One way of specifying the uncertainty associated with a random variable is through
the function
F(x) = P(X ≤ x)
F(x) is the probability that the random variable X will take on the value x or less at the end of
the experiment. F(x) is called the cumulative distribution function (or cdf for short). When X
takes on its smallest possible value, the probability of observing a smaller value than this is
zero. Then when X takes on its largest possible value, the probability of observing this value
or less is one. Thus F(x) varies from zero through too one as X increases.
The cdf also provides a convenient way of finding the probability that a random
variable lies within a certain range
P(x1 ≤ X ≤ x2) = P(X ≤ x2) - P(X ≤ x1) = F(x2) – F(x1)
Another approach commonly used to represent a random variable is through the use
of the probability density function (or pdf for short). The pdf of a random variable X is
defined as the derivative of the cdf
f(x) 
d F(x)
dx
As integrals define areas under curves, it follows that P(x1 ≤ X ≤ x2) can also be
found by calculating the area under the pdf between x1 and x2. As the maximum value for
F(x) is one, it follows that the total area under the pdf must also sum to one.
Example: Battery failures
Suppose that a battery failure time, measured in hours, has a cdf given by
F(x)  1 
1
(x  1) 2
The probability that the battery lasts between x1 = 1 and x2 = 2 hours is therefore
P(x1 ≤ X ≤ x2) = P(1 ≤ X ≤ 2) = F(x2) – F(x1) = F(2) – F(1)
F(2)  1 
1
8
1
3

; F(1)  1 

2
2
9
4
(2  1)
(1  1)
Thus
8 3 5
P(x1 ≤ X ≤ x2) =  
9 4 36
The pdf is given by
f(x) 
d F(x)
2
 2[x  1]3 
dx
[x  1]3
Working with only the pdf
2
P(x1 ≤ X ≤ x2) = 
2
1
 -1 
 -1   -1  1 1 5
2
dx  


  
3
2
2
2
[x  1]
[x  1] 1 [2  1]  [1  1]  4 9 36
More details on this type of calculation can be seen in the screen shot below (click on
it to access the actual Excel file).
B.
Distributions
Depending on the characteristics of the random variable being studied, there are a
variety of different cdf’s available. This module will consider only some of the many cdf
available to practicing engineers. These cdf’s are described by formulas that depend on
parameter values. These are population parameters in that they should ideally be calculated
from the population of values on X.
i.
The Binomial Distribution
VOICE OF EXPERIENCE
The binomial distribution models defective/non
defective.
The simplest discrete random variable
is one than can take on just two
values. Such a random variable can
be used to model the outcome of a coin toss, whether an important safety valve is open or
shut, whether an item or component is defective or not and so on. Typically, the outcomes are
labeled 0 (for not defective) and 1 (for defective) and this random variable is defined by the
parameter p, where p is the probability that X takes on the value 1. Such a simple random
variable is called a Bernoulli random variable.
Many experiments and tasks carried out by engineers can be thought of as consisting
of a sequence of Bernoulli trials, such as the repeated examination of critical components and
parts to determine whether they are defective. In such cases, the random variable of interest is
the number of observed successes obtained within a fixed number of trials, n. Such a random
variable is called a binomial random variable. Specifically, if n independent Bernoulli trials
X1,……,Xn are performed, each with a probability p of being defective, then the random
variable
X = X1 + X2 +….+ Xn
is said to have a binomial distribution with parameters n and p, which is written as
X ~ B(n, p)
The pdf of a B(n,p) random variable is given by the formula
f(x) 
n!
p x (1  p) n x
(n  x)! x!
where ! stands for factorial (e.g. 3! = 3 x 2 x 1 = 6).
The mean and variance are in turn given by
Mean  np ; Variance  np(1  p)
These equations is easily derived from the rules of probability looked at in an earlier
unit. This derivation is best done via an example.
Example: Polyester yarn
A chemical engineer monitors a dying process for polyester yarn used in clothing by
comparing a sample of the yarn against a standard colour chart. The engineer accepts or
rejects the entire batch based on the result of this comparison. Historically, this dying process
averages 25% rejected batches, so in the terminology above p = 0.25. Each shift, the process
dyes four batches and so in the terminology above, the number of trials is n = 4.
Now take X = 1 as an example. There are only four ways in which this can happen:
Possibility 1: X = X1 + X2 + X3 + X4 = 1 + 0 + 0 + 0 = 1. (Only the first batch is defective).
The probability that this outcome will occur is the probability that X1 = 1 And X2 = 0 And X3
= 0 And X4 = 0, which of course requires the use of the product rule. Further, the outcome of
the second batch is independent of what happened to the first batch and what will happen to
the third batch and so the simpler product rule for independent events is applicable
P[X1 = 1 And X2 = 0 And X3 = 0 And X4 = 0] = p(1-p)(1-p)(1-p)
Possibility 2: X = X1 + X2 + X3 + X4= 0 + 1 + 0 + 0 = 1. (Only the second batch is defective).
P[X1 = 0 And X2 = 1 and X3 = 0 And X4 = 0] = (1-p)p(1-p)(1-p)
Possibility 3: X = X1 + X2 + X3 + X4 = 0 + 0 + 1 + 0 = 1. (Only the third batch is defective).
P[X1 = 0 And X2 = 0 and X3 = 1 And X4 = 0] = (1-p)(1-p)p(1-p)
Possibility 4: X = X1 + X2 + X3 + X4 = 0 + 0 + 0 + 1 = 1. (Only the fourth batch is defective).
P[X1 = 0 And X2 = 0 and X3 = 0 And X4 = 1] = (1-p)(1-p)(1-p)p
Notice that X = 1 can therefore occur in four separate ways and so X = 1 if possibility
1 Or possibility 2 Or possibility 3 Or possibility 4 occurs. Or events require the addition rule
of probability and because these outcomes are mutually exclusive (the researcher can only
observe one of the possibilities above during a single shift)
P[X = 1] = P(possibility 1 Or possibility 2 Or possibility 3 Or possibility 4 ]
= p(1-p)(1-p)(1-p) + (1-p)p(1-p)(1-p) + (1-p)(1-p)p(1-p) + (1-p)(1-p)(1-p)p
P[X = 1] = 4p1(1-p)3
The number 4 is this last expression is the number of ways that X can equal 1. This
n!
can be worked out quickly using the first part of the pdf for B(n,p), i.e. using
.
(n  x)! x!
When x = 1 for n = 4 this gives
4!
24

4
(4  1)!1! (6)1
Now look at the second part of the pdf for B(n,p) above. This part, p x (1  p) nx , is simply
p1(1-p)3 when n = 4 and x = 1.
Now with p = 0.25, P[X = 1] = 4(0.25)(1-0.25)3 = 0.4219.
There is a 42.19% chance that by the end of the shift the control engineer will have
observed one batch with defective colour. The probabilities associated with the other values
for X are easily worked out using the pdf for B(n,p).
Play around with the p value in cell B7 of this Excel file and notice how the binomial
distribution is symmetric around its peak point when p = 0.5, which is a characteristic of the
very important normal distribution to be discussed below ((i.e. when p = 0.5 and n is large
there is very little difference between these two distributions).
ii.
The Uniform Distribution.
The uniform distribution can be used when any value for the random variable is equally
likely to occur. The shorthand for saying X in uniformly distributed between the limits a and
b is
X ~ U(a, b)
with parameters a and b. The cdf for a uniform random variable is written as
F(x) 
xa
(b  a)
The pdf is given by
for
a≤x≤b
f(x) 
d F(x)
1

dx
b-a
The mean and variance are in turn given by
Mean 
ab
(b  a) 2
; Variance 
2
12
All of this can be visualized in a simple cross plot of F(x) against x:
Mean = (a+b)/2
F(x)
1
b/(b-a)
Variance = (b-a)2/12
0
x
a
b
Example: A manufactured metal pins and Pearls
Suppose that a manufactured metal pin has a diameter that has a uniform distribution
between a = 4.182 mm and b = 4.185 mm. Thus X ~ U(4.182, 4.185) and so the mean pin
diameter is (4.182 + 4.185)/2 = 4.1835 mm with a standard deviation of (4.185 - 4.182)/√12 =
0.00087 mm. The probability that a pin selected at random from this manufacturing process
fits in a hole that has a diameter of 4.184 mm is
F(x) = F(4.184) = P(X ≤ 4.184) =
4.184  4.182
 0.6667
(4.185  4.182 )
(or 66.67%)
When pearl oysters are opened, pearls of various sizes are typically found. Suppose that each
oyster contains a pearl with a diameter in mm that has a U(0, 10) distribution. The mean pearl
diameter is therefore (0 + 10)/2 = 5 mm with a standard deviation of (10 - 0)/√12 = 2.89 mm.
If pearls with a diameter of at least 4 mm have commercial value, then the probability that a
randomly selected oyster from a recent catch of oysters has commercial value is
P(X ≥ 4) = 1 - F(3)  1 
30
 0.70
(10  0)
(or 70%)
More details on this type of calculation can be seen in the screen shot below (click on
it to access the actual Excel file).
iii.
The Poisson Distribution
The Poisson distribution can be used when the random variable being studied is the count of
the number of events within a specified boundary.
VOICE OF EXPERIENCE
The poisson distribution models counts such as the
number of defects.
For example, the number of hairline
fractures in a 10 m long steel girder,
or the number of blisters per battery
plate, or the number of cooling pump
failures within a UK nuclear power plant over a 10 year period, or the number of failures of
air conditioning equipment in a Boeing 720 aircraft over 300 hours of flight.
The shorthand for saying X is Poison distributed is
X ~ P()
with parameter . If the number of events are uniformly or randomly spaced within the
specified boundaries, then the cdf for X is given by
x
F(x)  e λ 
i0
λi
i!
and the pdf is given by
f(x) 
d F(x) λ x λ
 e
dx
x!
The mean determines how quickly the cdf approaches its maximum value of 1.
1
0.9
0.8
0.7
F(x)
0.6
0.5
0.4
0.3
Mean = 0.5
0.2
mean =1
0.1
Mean =2
0
0
1
2
3
4
5
x
6
7
8
9
10
The mean and variance are in turn given by
Mean  Variance  λ
Example: Steel girders
An engineer examines the edges of steel girders for hairline fractures. The girders are 10 m
long, and it is discovered that after inspecting many such girders, they have an average of 5
fractures each and the fractures are always randomly spaced out on the girders.
The probability that a randomly selected girder, on inspection, has three or less hairline
fractures is
 50
F(x) = F(3) = P(X ≤ 3) = e 5 

 0!

51 52 53 
25 125 



   e 5 1  5 

  0.2650
1! 2! 3! 
2
6 


The probability that a randomly selected girder, on inspection, has exactly two hairline
fractures is
 50
F(3) – F(2) = 0.2650 - e 5 

 0!

51 52 

   0.2650  0.1247  0.1404
1! 2! 

More details on this type of calculation can be seen in the screen shot below (click on
it to access the actual Excel file).
iv.
The Exponential Distribution
VOICE OF EXPERIENCE
The exponential distribution models the length of
time or space between successive events.
The exponential distribution can be
used when the random variable being
studied is the amount of time or space
between successive events occurring
in a Poisson process. For example, the size of the gaps between fractures on the edges of steel
girders or the times between failures of air conditioning equipment in Boeing 720 aircraft.
The shorthand for saying X is exponentially distributed is
X ~ Exp()
with parameter . The cdf is given by
F(x)  1  e  λx
and the pdf is
f(x) 
d F(x)
 e λx
dx
The mean and variance are in turn given by
Mean 
1

; Variance 
1
2
The mean determines how quickly the cdf approaches its maximum value of 1. The Poisson
and exponential distributions are closely related as is illustrated in the next example.
1
0.9
0.8
0.7
F(x)
0.6
0.5
0.4
0.3
Mean = 0.5
0.2
Mean = 1.0
0.1
Mean = 2.0
0
0
1
2
3
4
5
6
7
8
9
10
x
Example: Steel girders revisited
An engineer examines the edges of steel girders for hairline fractures. The girders are
10 m long, and it is discovered that after inspecting many such girders, they have an average
of 5 fractures each and also that the fractures appear to be uniformly spaced on the girders. If
a girder has an average of 5 fractures, then there are an average of 6 gaps between fractures
or between the ends of the girder and the adjacent fractures. The average length of these gaps
between fractures is therefore 10/6 = 1.67 m, and so 1.67 = 1/, or = 1/1.67 = 0.6 m.
Thus the number of fractures follows a Poisson distribution with = 5, whilst the gap
between these fractures follows an exponential distribution with = 0.6 m. In this case, the
probability that a measured gap on a randomly selected 10 m girder is less than or equal to
0.8 m long is
F(x) = F(0.8) = P(X ≤ 0.8) = 1  e 0.6(0.8)  0.3812
If a 2.5 m segment of girder is selected, the average number of fractures it contains is
0.25(5) = 1.25. The probability that this segment contains at least 1 fracture is
1.250 1.251  1.25
F(x) = F(1) = P(X ≤ 1) = e 1.25 

  e 1  1.25  0.6446
1! 
 0!
v.
The Weibull Distribution
The Weibull distribution is often used to model times to certain events, such as the time
or number of miles driven for brake pads to wear out, or the lifetime of boiler components in
a power plant, or the lifetime of a particular bacterium at a certain high temperature.
However, this distribution is used to model a variety of other types or random variables
occurring in engineering - especially extreme or maximum values.
The shorthand for saying X is Weibull distributed is
X ~ W()
with parameters and . The cdf is given by
F(x)  1  e( λx)

and the pdf is
f(x) 
β
d F(x)
 βλβ xβ 1e(x)
dx
The mean and variance are in turn given by
2
 1 
1  1
1 
  2

Mean   1   ; Variance  2  1     1   
  
        
f(x)
The function [k] is the gamma function. If k is a positive integer number then [k] =
(k-1)!. Except for these special cases there is no closed form expression for the gamma
function. It can however be worked out numerically in Excel using the Gammaln() function
which calculates the natural log of the gamma function.  is called the scale parameter and 
is called the shape parameter. The Weibull distribution is very useful in applied engineering
research as its pdf can exhibit a wide variety of shapes, depending on the choice of values for
the parameters. Notice in the following figure that when =1, the Weibull distribution has
the same shape as the exponential distribution, i.e. it is the exponential distribution (this can
be seen also by setting = 1 in the above Weibull cdf and comparing the result to the cdf for
the exponential distribution). The Weibull distribution is therefore simply a generalization of
the exponential distribution.
1.5
1.4
1.3
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1 0
lambda = 1, beta = 0.5
lambda = 1, beta = 2
lambda = 1, beta = 3
1
2
3
4
5
6
7
x
Example: Pitting corrosion.
A study of pitting corrosion was carried out on buried cast iron oil and gas pipelines.
The study found that the maximum pit depth data followed a Weibull distribution with  =
0.45 and  = 1.8.
The average maximum pit depth is therefore
Mean 
1 
1 
1
Γ 1   
0.889  1.98mm
0.45  1.8  0.45
with a variance of
2
 1 
1 
1
  2

Variance  2  1     1    
         0.452
2

2
1  
 


 1     1     1.29

 1.8  
  1.8 

The probability that a randomly selected dug up pipe, on inspection, has a maximum pit
depth of 2 mm or less is
F(x) = F(2) = P(X ≤ 2) = F(x)  1  e(0.45x)  0.5628
1.8
vi.
The Normal Distribution
Among all the distributions used in applied research, the single most important is the
normal distribution. There are two fundamental reasons for this. First, the behavior of many
physical phenomena can be modeled well using this distribution. Secondly, and perhaps more
importantly, the behavior of averages can, when using large samples, be modeled well using
this distribution This latter fact leads to the development of many tests for important
hypothesis (see unit 4 for more on this).
The shorthand for saying X is normally distributed is
X ~ N()
with parameters and . In fact, if a random variable is normally distributed then its mean
value equals and 2 is its variance (as these are population parameters,  is the population
mean and  the population standard deviation). The pdf is
f(x) 
d F(x)

dx
1
2 2
 1  x   2 
exp  
 
 2    
Unfortunately there is no analytical solution to the integral of f(x) and so there is no
closed form expression for F(x). The integral must be evaluated numerically and this creates
some minor complications. Using numerical integration procedures it has been proved that:
1.
2.
3.
Approximately 68% of all the possible values that X can take on are within the numeric
range ± 1
Approximately 95% of all the possible values that X can take on are within the numeric
range ± 2
Virtually all (approximately 99.7%) the possible values that X can take on are within
the numeric range ± 3
Further, the pdf is bell shaped and symmetric around the peak point. The peak point is
located at the mean value , and ± 1represents the points of inflection for the pdf so that
 essentially determines how wide the distribution is. For this reason  is sometimes called
the location parameter and  the shape parameter. All this is illustrated in the screen shot
below.
Any normally distributed random variable can be converted into a standard normal
random variable, Z. This is important because of the inability to integrate f(x). Because any
normal variable can be converted into a standard normal variable, it is only required to
numerically integrate various areas under this single standard normal distribution and tabulate
the results for future use.
In fact the expression in round brackets in the formula for f(x) above is a standard
normal variable, which is usually given the symbol Z
Z
X 

So if X is a random variable following a normal distribution, its re scaled value, Z,
follows the standard normal distribution, whose pdf is
f(z) 
 1 2
exp  z  
 2

2
1
2
The following screenshot shows the numerically calculated integral for f(z), i.e. the
values for F(z) corresponding to various values for z. A table of F(z) is shown below and can
be found at the back of any decent textbook on statistics. This table is usually referred to as
the Z Table. Alternatively, F(z) can be “looked up” in Excel using the =NormsDist(z)
function. The Z Table can be read the other way around, i.e. finding the z value associated
with a given value for F(z) in Excel, using the =NormsInv(z) function. Notice the F(z) values
shown in the Z table below are the probabilities of observing the shown Z value or less.
It follows from the last two screenshots that a normally distributed variable and its
standardized equivalent are quite specifically related. When a normally distributed variable,
X, takes on a value equal to its mean, the corresponding standardized value is zero. So a
standardized normal variable has a mean of zero. When a normally distributed variable takes
on a value equal to its mean plus one standard deviation, the corresponding standardized
value is unity. So a standardized normal variable has a standard deviation of one.
All of this is best illustrated using an example.
Example: Concrete blocks.
A company manufactures concrete blocks that are used for construction purposes.
Suppose that the weight of all the individual concrete blocks produced by this company are
normally distributed with a mean value of  = 11 kg and a standard deviation of  = 0.3 kg.
What percentage of concrete blocks produced by this company weighs 10.5 kg or less?
The standardized value for 10.5 kg is Z = (10.5 – 11)/0.3 = -1.67. From the Z table above, the
probability of observing this Z value or less is 0.0475, or 4.75% of manufactured blocks will
have a weight of 10.5 kg or less.
Z values
0.6
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
0.5
f(x)
0.4
0.3
0.2
=Normsdist(-1.67)
= 0.0475
0.1
0
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Concrete block weights (kg)
vii.
Transformations of the normal distribution
If a random variable X has a normal distribution, so that Z has a standard normal
distribution, then its square has a chi square distribution with parameter v = 1. v is referred to
as the degrees of freedom. More generally if there v normally distributed variables that are
independent of each other, the squares of their standardized values once added together
follow a chi square distribution with v degrees of freedom. That is
2v = {(Z1)2 + (Z2)2 + …. + (Zn)2}
has a pdf given by
f(  v2 ) 
1
 /2
2
( / 2)
y ( / 2)1 exp(   v2 / 2)
As the chi square distribution is essentially a transformation of the standard normal
distribution, it is not surprising to realize that again there is not closed form expression for F(
 v2 ). Again values for F(  v2 ) have been numerically calculated and tabulated for various
values of v. Values from this table can be looked up in Excel using the function =ChiDist(  v2
,).
.