STATISTICS & NUMERICAL METHODS FOR PLANT ENGINEERS AGE-214 By S. O. Duffuaa Systems Engineering Department

STATISTICS & NUMERICAL METHODSFOR PLANT ENGINEERS

AGE-214

ByBy

S. O. DuffuaaS. O. Duffuaa

Systems Engineering Systems Engineering DepartmentDepartment

Salih DuffuaaSalih Duffuaa

•Dr. Duffuaa is a Professor of Industrial and Systems Engineering at the Department of Systems Engineering at King Fahd University of Petroleum and Minerals, Dhahran, Saudi Arabia. He received his PhD in Operations Research from the University of Texas at Austin, USA. His research interests are in the areas of Operations research, Optimization, quality control, process improvement and maintenance engineering and

management .

He teaches course in the areas of Statistics, Quality control, Production and inventory control, Maintenance and reliability engineering and Operations Management. He consulted to industry on maintenance , quality control and facility planning. He authored a book on maintenance planning and control published by John Wiley and Sons and edited a book on maintenance optimization and control. He is the Editor of the Journal of Quality in Maintenance Engineering, published by Emerald in the United Kingdom.

King Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsDepartment of Systems EngineeringDepartment of Systems Engineering

Statistics for Plant Engineers and Statistics for Plant Engineers and Lab ScientistsLab Scientists

Oct, 18-22, 2008 Oct, 18-22, 2008

A closed short course for Saudi Aramco Employees

On

DayMorning Session

7:30 – 11:45 A.MAfternoon session

12:45 to 3:45 PMDr. Salih Duffuaa, Introduction to probability ,frequency & Probability distributions, mean and variance.

Dr. Mohammad Haboubi, The normal distribution, the central limit theorem and sampling distributions. .Dr. Hesham Al-Fares: , Point and interval estimation, statistical significance tests.

Dr. Mohammad Al Salamah, Simple regression, residual analysisDr. Shokri Selim:Multiple regression, adequacy of a regression model, Applications

Saturday

Sunday

Monday

Tuesday

Wednesday

Course OutcomesCourse Outcomes

• Apply probability concepts and laws to solve basic lab problems.

• Summarize and present data in meaningful ways.• Compute probabilities from probability

distributions.• Construct confidence intervals for sample data.• Test statistical hypothesis.• Construct regression models and use them in

various applications such as equipment calibration and prediction.

Day 1 Module ObjectivesDay 1 Module Objectives

• Concept and definition of probability.

• Axioms of probability

• Laws of probability


• Data Summary– Measures of central tendency.

• Mean X-bar and Median M

– Measures of variability – Range R, Variance S2, Standard deviation S

and coefficient of variation (CoV).

• Frequency Distribution.• Distributions• Expected value


• Random variables.

• Mass and distribution functions

• Expected value

Examples of a Random ExperimentExamples of a Random Experiment

• Measuring a current in a wire.

• Number of samples analyzed per day .

• Time to do a task. Time to analyze a sample.

• Yearly rain fall in Dhahran

Examples of a Random ExperimentExamples of a Random Experiment

• Throwing a coin

• Number of accidents on campus per month.

• Students must generate at least 5 examples.

Random ExperimentsRandom Experiments • Every time the experiment is repeated a

different out come results.

• The set of all possible outcomes is call Sample Space denoted by S.

• In the experiment of throwing the coin the sample space S = { H, T}.

Random ExperimentsRandom Experiments

• In the experiment on the number of defective parts in three parts the sample space S = { 0, 1, 2, 3}.

• Number of weekly traffic accidents on KFUPM campus.

EventEvent

• An event E is a subset of the sample space.

• Example of Events in the experiment of the number of defective in a sample of 3 parts are:

• E1 = { 0}, E2 = { 0,1}, E3 = { 1, 2}

Example of EventsExample of Events

• A sample of polycarbonate plastic is analyzed for scratch resistance and shock resistance. The results from 49 samples are:

• Shock resistance H L

H 40 4Scratch Resistance L 2 3

Let A denote the event a sample has high shock resistance and B denote the event a sample has high scratch resistance. Determine the the number of samples in AB, AB and A`

Solution of ExampleSolution of Example

• IAI = 42, IBI = 44

• IABI = 40

• IABI = 46

• A = 7 , B = 5

ExerciseExercise

• Refer to the event example and answer the following:

• Find the number in AB

• Find the number of elements in AB

• Find the number of elements in AB

Listing of Sample SpacesListing of Sample Spaces

• Tree Diagrams

• Experience

•

Listing of Sample SpacesListing of Sample Spaces

• The experiment of throwing a coin twice

H

T1

H

T

H

T

S = { HH, HT, TH, TT}

Example on Listing Sample SpacesExample on Listing Sample Spaces

• Draw the tree diagram for finding the sample space for the number of defect item in a sample of size three taken from a production line producing chips.

Types of Sample SpacesTypes of Sample Spaces

• A sample space is discrete if it consists of a finite ( or countable infinite ) set of outcomes. Examples are:

• S = { H, T}, S = { 1, 2, 3, …}

• Sample space is continuous if contains an interval (finite or infinite): { T: 0 ≤ T ≤ 60}.

• Students should give more examples

Notation

•P - denotes a probability

•A, B, ... - denote a specific event

•P (A) - denotes the probability of an event occurring

Concepts and Definition of Concepts and Definition of ProbabilityProbability

Four definitions of probability:

• Classical or a priori probability

• Statistical or a posteriori probability

• Subjective probability (used in Bayesian methods).

• Mathematical probability

CLASSICAL OR A PRIORI CLASSICAL OR A PRIORI PROBABILITYPROBABILITY

P(A) = # of ways A can occur (# favorable cases)

Total number of possible cases (# of totalpossible cases)

STATISTICAL OR A POSTERIORISTATISTICAL OR A POSTERIORIPROBABILITYPROBABILITY

# of successes

Number of trialsPr (A) =

In the limit as # of trials

Infinity

SUBJECTIVE PROBABILITYSUBJECTIVE PROBABILITY

• A measure of the degree of belief.

• There is a 10% chance it will rain today.

• There is a 95% chance you can see the new moon tomorrow morning.

• Subjective probability is the basis for Bayesian methods.

Probability LimitsProbability Limits The probability of an impossible event is 0.

The probability of an event that is certain to occur is 1.

0 ≤ P(A) ≤ 1

Impossibleto occur

Probability LimitsProbability Limits The probability of an impossible event is 0.

The probability of an event that is certain to occur is 1.

0 ≤ P(A) ≤ 1

Impossibleto occur

Certainto occur

MATHEMATICAL PROBABILITYMATHEMATICAL PROBABILITY

• A measure of uncertainty ( or possibility) that satisfy the following conditions:

• 0 ≤ P(A) ≤ 1

• P(S) = 1

• Pr (A U B) = Pr (A) + Pr (B) If A Π B = Ǿ

Possible Values for ProbabilitiesPossible Values for Probabilities

Certain

Likely

50-50 Chance

Unlikely

Impossible

1

0.5

0

Probability of an EventProbability of an Event

• For discrete a sample space, the probability of an event denoted as P(E) equals the sum of the probabilities of the outcomes in E.

• Example: S = { 1, 2, 3, 4, 5} each outcome is equally likely. E is even numbers within S. E = { 2, 4}, P(E) = 2/5.

Axioms of ProbabilityAxioms of Probability

• If S is the sample space and E is any event then the axioms of probability are:

1. P(S) = 1

2. 0 P(E) 1

3. If E1 and E2 are event such that E1 E2 = ,

then,

P(E1 E2) = P(E1 ) + P(E2)

mutually exclusivemutually exclusive

• Events A and B are mutually exclusive

if they cannot occur simultaneously.

DefinitionDefinition

Total Area = 1

P(A) P(B)

P(A and B)

Overlapping Events

DefinitionDefinition

Total Area = 1 Total Area = 1

P(A) P(B) P(A) P(B)

P(A and B)

Nonoverlapping EventsOverlapping Events

P(A)P(A)

Complementary Events

The complement of event A, denoted by A, consists of all outcomes in

which event A does not occur.

P(A)(read “not A”)

Rules for Complementary Rules for Complementary EventsEvents

P(A) + P(A)= 1


P(A) + P(A) = 1

= 1 – P(A)

P(A)


P(A) + P(A) = 1

= 1 – P(A)

P(A) = 1 – P(A)

P(A)

Venn Diagram for the Venn Diagram for the Complement of EventComplement of Event A A

Total Area = 1

P(A)

P(A) = 1 – P(A)

Probability of ‘At Least OneProbability of ‘At Least One’’

‘At least one’ is equivalent to one or more.

The complement of getting at least one item of a particular type is that you get no items of that type.

Probability of ‘At Least Probability of ‘At Least OneOne’’

• If P(A) = P (getting at least one), then

• P(A) = 1 – P(A)

• where P(A) is P (getting none)

• Any event combining 2 or more events Compound Even Notation

• P(A or B) = P (event A occurs or event B

occurs or they both occur)

• P(A and B) = P (event A occurs and event B occurs)

Definitions

Addition RuleAddition Rule

• P(A or B) = P (event A occurs or event

B occurs or they both occur).• P(AB) = P(A) + P(B) – P( AB)

• If AB) = , then,

• P(AB) = P(A) + P(B)

Addition RuleAddition Rule

• A B

P(AB) = P(A) + P(B) – P( AB)

Addition Rule: ExampleAddition Rule: Example

• Let S = { 1, 2, 3, 4, 5, 6, 7, 8,9,10)

• A = { 2,3,4,5,6}, B = {4, 5,6,7,9,10}

• AB = { 2,3,4,5,6,7,9,10}

• P(A) = 5/10 =0.5 P(B) = 6/10 = 0.6

• P(AB) = 0.8

• P(AB) = 0.5 + 0.6 – 0.3 = 0.8

ProblemProblem

• Let assume A, B and C are events from the sample space S. P(A) = 0.4. P(B) = 0.5,

P(C) = 0.3, P(BC) = 0.1, P(AC) = 0.2, A and B are mutually exclusive. Compute the following:

(i) P(A'), (ii) P(AυC), (iii) P[(AυB)C)

(iv) P(A υB υC), (v) P(B'UC‘)

Note A' means A compliment.

Conditional ProbabilityConditional Probability

• Conditional Probability Concept

• P(A B) = P(A B)/ P(B) for P(B) > 0

• Give Examples

• Solve problems

Example on Conditional ProbabilityExample on Conditional Probability

• Let S = { 1, 2, 3, 4, 5, 6, 7, 8,9,10)

• A = { 2,3,4,5,6}, B = {4, 5,6,7,9,10}

• P(A|B) = P(AB)/P(B) = 0.3/0.6 = 0.5

• This is as if we consider B our sample space and see how many elements from A in B. This will make P(B) = 1

Conditional ProbabilityConditional Probability

Dependent Events P(A and B) = P(A) • P(B|A)

Conditional ProbabilityConditional Probability Dependent Events P(A and B) = P(A) • P(B|A)Formal P(B|A) =IntuitiveThe conditional probability of B given A can be found by assuming the event A has occurred and, operating under that assumption, calculating theprobability that event B will occur.

P(A and B)P(A)

Independent EventsIndependent Events

Independent EventsTwo events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other.

• P(A|B) = P(A)

Dependent EventsDependent Events

Dependent EventsIf A and B are not independent, they are said to be dependent.

Formal Multiplication RuleFormal Multiplication Rule

• P(A and B) = P(A) • P(B) if A and B are independent

• P(AB) = P(A) • P(B)

Figure 3-9 Applying the Multiplication Rule

P(A and B)

Multiplication Rule

AreA and B

independent?

P(A and B) = P(A) • P(B | A)

P(A and B) = P(A) • P(B)Yes

No

Generalization of Addition RulesGeneralization of Addition Rules

• Addition Rule

P(AB) = P(A) + P(B) – P( AB)

• If AB) = , then,

• P(AB) = P(A) + P(B)

• This rule can be generalized to k events

• If Ei Ej = , then

• P( E1 E2 … Ek) = P(E1) + P(E2) + … + P(EK)

Multiplication RuleMultiplication Rule

• P(A B) = P(AB) ) P(B) = P(BA) ) P(A)• Example:

The probability that an automobile battery subject to high engine compartment temperature suffer low charging is 0.7. The probability a battery is subject to high engine compartment temperature is 0.05.

What is the probability a battery is subject to low charging current and high engine compartment temperature?


• Let A denote the event a battery suffers low charging current. Let B denote the event that a battery is subject to high engine compartment temperature. The probability the battery is subject to both low charging current and high engine compartment temperature is the intersection of A and B.

• P(A B) = P(AB) ) P(A) = 0.7 x 0.05 = 0.035

Example On Conditional and Example On Conditional and Multiplication ( Product) RuleMultiplication ( Product) Rule

• Consider a town that has a population of 900 persons, out of which 600 are males. The rest are females. A total of 600 are employed, out of which 500 are males. Let M denote male, F denote female and E employed and NE not employed. A person is picked at random. Find the following probabilities. P(M), P(E), P(EF), P(EF), P(E F).

Town Population ExampleTown Population Example

• Town Population Distribution

Employed

Unemployed

Male Female

500

100

100

200

600

300

600 300 900 Total

Total


• P(M) = 600/900 = 2/3

• P(E) = 600/900 = 2/3

• P(EF) = 100/300 = 1/3

• P(EF) = P(EF) P(F) = (1/3) x (1/3) = 1/9

• P(E F) = P(E) + P(F) – P(EF)

= 2/3 + 1/3 – 1/9 = 8/9

Statistical IndependenceStatistical Independence

• Two events are statistically independent if the knowledge about one occurring does not affect the probability of the other happing. Mathematically expressed as:

• P(AB) = P(A)

• P(A B) = P(A) P(B) Why ?

Example of IndependenceExample of Independence

• Let us consider the experiment of throwing the coin twice. Let B denote the event of having a head (H) in the first throw and A denote having a tale (T) in the second throw.

• P(AB) ) = ½ = P(A)• P(A B) = ½ x ½ = ¼ = P(A) P(B)• Therefore A and B are independent

Example of DependentExample of Dependent

• A daily production of manufactured parts contains 50 parts that do not meet specifications while 800 meets specification. Two parts are selected at random without replacement from the batch. Let A denote the event the first part is defective and B the event the second part is defective.

• Are A and B independent?• The answer is NO. Work it out before

you see the next slide

Example of DependentExample of Dependent

• P( BA ) = 49/849 why?

• P(B) = P(B A )P(A) + P(B A)P(A) = (49/849)(50/850) + (50/849)(800/850)

= 50/850

Therefore A and B are not independent.

Problem 1Problem 1

• A box contains 9 components of which 3 are defective.

• If one unit is drawn at random what is the probability it is defective?

• If two components are chosen at random– What is the probability both are defective?– what is the probability just one is

defective?

Problem 2Problem 2

• A company has two machines one is less reliable than the other. The better one has probability of 0.9 of working through out the week without repair. The probability for the other is 0.7.

• What is the probability both machines is working satisfactory through out the week?

• What is the probability at least one of them requires repair?

Total Probability RuleTotal Probability RuleMotivational ProblemMotivational Problem

• Aramco has three labs that perform the same oil analysis. Lab1 in Abqaiq, Lab2 in Ras Tanura and Lab3 is in Dhahran. 100 samples were analyzed in lab 1, 70 samples in lab2 and 30 samples in lab 3. The chance of error in lab1 is 5%, in lab2 is 10% and in lab3 is 8%.

What is the probability that a sample analyzed by Aramco is in error. If the analysis of a sample is found to be in error what is the chance the analysis is done in lab1.

Total Probability RuleTotal Probability Rule

• In a chip manufacturing process 20% of the chips produced are subjected to a high level of contamination. 0.1 of these chips causes product failure. The probability is 0.005 that a chip that is not subjected to high contamination levels during manufacturing causes a product failure.

• What is the probability that a product using one of these chips fails?

Total Probability RuleTotal Probability Rule

• Let B the event that a chip causes product failure. We can write B as part of B in A and part of B in A.

• B = (B A) (B A)• P(B) = P(B A) + P(B A)• P(B) = P(BA) ) P(A) + P(B A) ) P(A) • Graphically on next slide.

Graphical RepresentationGraphical Representation

A AB

A

General Form of Total Probability RuleGeneral Form of Total Probability Rule

• Assume E1, E2, … Ek are mutually exclusive and exhaustive events. Then

• P(B) = P(B E1) + P(B E2) + …+ P(B Ek) )

= P(B E1) P(E1) + P(B E2) P(E2) + …+ P(B Ek) P(Ek)

Bayes RuleBayes Rule

• P(A B) = P(AB) ) P(B) = P(BA) ) P(A)

• Implies

• P(AB) ) = P(BA) ) P(A)/ P(B) , P(B) > 0

• OR Refer to the slide on about the general total probability rule, we get

• P(Ei B) = P(Ei B)/ P(B) = P(B Ei )P(Ei)/ P(B)

• = P(B Ei )P(Ei)/ P(B E1) P(E1) + P(B E2) P(E2) + …+ P(B Ek) P(Ek)

Example on Bayes TheoremExample on Bayes Theorem

• Refer to the example about the chip production. If you know a chip caused failure what is the chance that the chip is subjected to a high level of contamination when its produced.

• We want P(A B)• P(A B) = P(B A) P(A)/ P(B) = (.1)(.2)/0.024

= 5/6 = 0.833

What is the probability of the chip is not subjected to a high level of contamination when produced ?

Answer in two ways.

Aramco Example SolutionAramco Example Solution

•

KFUPM Example on Bayes TheoremKFUPM Example on Bayes Theorem• KFUPM students when driving to building 24 th

use two roads. The main road that passes in front of gate 1 and the second road that passes in front of gate 2. The students use the main road 80% of the time because it is shorter. The radar is on 60% of the time on the main road and 30% of the time on the other road. The students are always speeding. Find the chance a student will be caught speeding. If you know student is caught speeding what is the probability he is coming to building 24 by the main road. Answer the same question for the other road.

Solution of KFUPM ExampleSolution of KFUPM Example

Data Table Compressive Strength of 80 Aluminum Data Table Compressive Strength of 80 Aluminum Lithium AlloyLithium Alloy

105 221 183 186 121 181 180 143 97 154 153 174 120 168 167 141 245 228 174 199 181 158 176 110 163 131 154 115 160 208 158 133 207 180 190 193 194 133 156 123 134 178 76 167 184 135 229 146 218 157 101 171 165 172 158 169 199 151 142 163 145 171 148 158

160 175 149 87 160 237 150 135196 201 200 176 150 170 118 149

How to Summarize The Data in How to Summarize The Data in The Table AboveThe Table Above

• Point summary

• Tabular format

• Graphical format

Point SummaryPoint Summary

• Measures of Central Tendency

• Measures of Variation

• Why are we interested in these type of measures?

Measures of Central Tendency or Measures of Central Tendency or LocationLocation

• Central tendency measures

– Mean xi/n

– Median --- Middle value

– Mode --- Most frequent value

Percentiles: Measure of LocationPercentiles: Measure of Location

• Pth percentile of the data is a value where at least P% of the data takes on this value or less and at least (1-P)% of the data takes on this value or more.

• Median is 50th percentile. ( Q2)

• First quartile Q1 is the 25th percentile.

• Third quartile Q3 is the 75th percentile.

Measures of VariabilityMeasures of Variability

• Range = Max xi - Min xi

• Variance = V = (xi – x )2/ n-1

• Standard deviation = S S = Square root (V)

Tabular and Graphical SummaryTabular and Graphical Summary

• Frequency distribution table.

• Histogram

• Cumulative frequency plot.

Steps for Constructing Frequency Steps for Constructing Frequency TableTable

• Determine number of classes/intervals.– Between 5 and 20– Close to Square root of n : number of data points.

• Count how many data points in each

class. This is the frequency.• The relative frequency is the frequency divided

by n.• The cumulative frequency is the sum of the

frequency up to certain level.

Class Interval

(psi)

Tally

Frequency

Relative FrequencyCumulative

Relative Frequency

70 ≤ x < 90 || 2 0.0250 0.0250

90 ≤ x < 110 ||| 3 0.0375 0.0625

110 ≤ x < 130 |||| | 6 0.0750 0.1375

130 ≤ x < 150 |||| |||| |||| 14 0.1750 0.3125

150 ≤ x < 170 |||| |||| |||| |||| || 22 0.2750 0.5875

170 ≤ x <1 90 |||| |||| |||| || 17 0.2125 0.8000

190 ≤ x < 210 |||| |||| 10 0.1250 0.9250

210 ≤ x < 230 |||| 4 0.0500 0.9750

230 ≤ x < 250 || 2 0.0250 1.0000

HistogramHistogram

• Plot class versus frequency or relative frequency to get the histogram.

• Plot the classes versus cumulative frequency to get the cumulative frequency plot.

70 90 110 130 150 170 190 210 230 250

Compressive Strength (psi )

Fre

quen

cy

25 20 15 10 5 0

Histogram of Compressive Strength Data

0

10

20

30

40

50

60

70

80

90

1

Strength

Cum

ulat

ive

Fre

quen

cy

100 150 200 250

Cumulative Frequency Plot of Compressive Strength Data

Concept of DistributionConcept of Distribution

Histogram ShapesHistogram Shapes• While the number of shapes that a histogram ca

take is unlimited, certain shapes appear often then others.

• Drawing a line that connects the edges of the bars in a Histogram forms a curve. We can make certain inferences about the data from the shape of the curve.

Distribution

Random Experiment and random Random Experiment and random variablesvariables

• Throwing a coin

• S = { H, T}.

• Define a mapping X: { H, T} R

• X(H) = 1 and X(T) = 0, Also the probability of 1 and 0 are the same as for H and T.

• Then we call X a random variable.

Random Experiment and random Random Experiment and random variablesvariables

• In the experiment on the number of defective parts in three parts the sample space S = { 0, 1, 2, 3}

• Find P(0), P(1), P(2) and P(3)

• P(0) = 1/8, P(1) = 3/8, P(2) = 3/8

and P(3) = 1/8

Probability Mass FunctionProbability Mass Function

• X o 1 2 3

f(x)1/8 3/8 1/83/8

Properties of f(x)

f(x) 0

f(x) = 1

Give many examples in class

Probability Mass FunctionProbability Mass Function

• Build the probability mass functions for the following random variables:– Number of traffic accidents per month on

campus.– Class grade distribution– Number of “ F” in SE 205 class per semester– Number of students that register for SE 205

every semester.

Cumulative Distribution FunctionCumulative Distribution Function

• It is a function that provide the cumulative probability up to a point for a random variable (r.v). Defined as follows for a discrete r.v:

• P( X x) = F(x) = f(t) t x

Cumulative Distribution Function Cumulative Distribution Function (CDF)(CDF)

• Example of a cumulative distribution function

• F(x) = 0 x -2 = 0.2 -2 x 0 = 0.7 0 x 2 = 1.0 2 x What is the density function for the

above F(x). Note you need to subtract

Probability Mass Function Corresponding Probability Mass Function Corresponding to Previous CDFto Previous CDF

• X -2 0 2

f(x)0.2 0.5 0.3

The above density function is the one corresponding to the previous CDF is

Mean /Expected Value of a Discrete Mean /Expected Value of a Discrete Random Variable (r.v)Random Variable (r.v)

• The mean of a discrete r.v denoted as E(X) also called the expected value is given as:

• E(X) = μ = x f(x) • The expected value provides a good idea a

bout the center of the r.v.• compute the mean of the r.v in previous

slide:• E(X) = (-2) (0.2) + (0) (0.5) + (2)(0.3) = 0.2

x

Variance of A Random VariableVariance of A Random Variable

• The variance is a measure of variability.• What is variability?• The variance is defined as:• V(X) =σ2 = E(X-μ)2 = (x-μ)2f(x) • Compute the variance of the r.v in the slide before the

previous one.

• σ2 = (-2-0.2)2 (0.2) + (0-0.2)2(0.5) + (2-0.2)2(0.3)

Expected Value of a Function of a r.vExpected Value of a Function of a r.v

• Let X be a r.v with p.m.f f(x) and let h(X) is a function of X. Then the expected value of h(X) is given as:

• E(H(X)) = h(x) f(x)

• Compute the expected value of h(X) = X2 - X for the r.v in the previous slides.

.

x

Problem 3Problem 3

• Compute the expected value and the variance for the random variable on slide 94 page 47.

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STATISTICS & NUMERICAL METHODS FOR PLANT ENGINEERS AGE-214 By S. O. Duffuaa Systems Engineering Department