fcm2063 - printed notes

5/27/2018 FCM2063 - Printed Notes

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Probability and Statistics

For Scientist and Technologist

A Course in Probability and Statistics

By

Radzuan Razali

Afza Shafie


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Contents:

Part One: Probabili ty and Distri bution

Chapter 1

1. The concept of probability

1.1Introduction1.2Sample Space, probability of events, counting rule1.3Conditional probability1.4Multiplication rule1.5Bayes theorem

Chapter 2:

2. Discrete random variable and probability distribution

2.1Introduction2.2Discrete random variable2.3Discrete probability distribution2.4Special functions for discrete probability distribution

Chapter 3:

3. Continuous random variable and probability distribution

3.1Introduction3.2Continuous random variable3.3Continuous probability distribution3.4Special functions for continuous probability distribution

Part Two: Descriptive Statistics

Chapter 4:

4. Data display and summary of data

4.1 Introduction4.2 The definition and the difference between sample and population4.3 Graphical display of data: Stem and leaf, and Box-plot4.4 The mean, variance and standard deviation of the data

Chapter 5:

5. Statistical process control: X-bar and R-charts.


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5.1 Introduction5.2 Statistical process control: X-bar and R-charts

Part Three: I nferential Statistics

Chapter 6:

6. Hypothesis testing for single population

6.1Introduction6.2Test about a sample mean for large sample, population variance is known6.3TheP-valuefor the test and confidence interval for mean6.4Test about sample mean for small sample, population variance is unknown6.5TheP-valuefor the test and confidence interval for mean6.6Test about proportion

6.7TheP-valuefor the test and confidence interval for proportion6.8Confidence interval for proportion

6.9Test about variance6.10 TheP-valuefor the test and confidence interval for variance

Chapter 7:

7. Simple linear regression model

8.1 Introduction8.2 Least squares estimator

8.3 The coefficient of determination8.4 Confidence intervals and significance tests

Part Four: Design of Experiments

Chapter 8:

8. The design and analysis of experiments

10.1 Introduction10.2 Two-Factorial design experiment

Appendix

Table 1: The normal - Z distributionTable 2: The Students t-distribution

Table 3: The chi-squared, distributionTable 4: The F-distribution


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Preface

This book provides an introduction to probability and statistics, with particular emphasis on

applications in applied sciences, technology and engineering. Typically introductory texts onengineering statistics spend a great deal of time on basic probability ideas for the first severalchapters. In fact, basic probabilities can easily fill up a standard introductory course. Becauseengineering students often have only one probability/statistics course, the material needs to bereorganized in order to allow for coverage of statistical methodology.

This book will be divided to four parts; part 1 is related to the basic concept of probability andthe distributions as in the Chapter 1 till Chapter 4. Chapter 1, we give a brief introduction to thebasic concept of the probability. In Chapter 2, we introduce the definition of discrete randomvariables and the probability distributions. The continuous and their probability distributions willbe discussed in Chapter 3.

Part 2 is covering on descriptive statistics as in the Chapter 4 and Chapter 5. In Chapter 4, weintroduce the types of how to display data and summary of the data. Meanwhile, the randomsample, central limit theorem, normal approximation and statistical process control will bediscussed in the Chapter 5.

The engineering students also must be given some experience on how to do a basic data analysis.The inferential statistics are important as a part of statistical methods to do the data analysis. PartIII is covering the inferential statistics such as in the Chapter 6 till Chapter 9. In Chapter 6, weintroduce the hypothesis testing for single population and the hypothesis testing for twopopulations will be discussed in Chapter 7. While in Chapter 8 and Chapter 9, we introduce the

simple and multiple linear regressions, respectively.

Finally, in part IV, we also want the students have some experience on the real application inengineering. The related topics such as a factorial design and design of experiment will bediscussed in the Chapter 10.

Afza Shafie

Radzuan Razali

April 2010.


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Chapter 1

1. Basic concept of probabilityLearning objectives:

At the end of this chapter, student should be able to:

Define and construct sample space of an experiment.

Define random events, identify types of events, apply Venn Diagram and laws Tofind event set including intersection, union and complement.

Identify mutually exclusive and exhaustive events. To apply Bayes theorem to find the conditional probability of an event when the

event is partitioned into several mutually exclusive and exhaustive subsets.

1.1 Introduction

Probability theory refers to the study of randomness and uncertainty. Probability formsthe basis knowledge which we can make inferences about a population based on thedistribution and its providemethods for quantifying the chances or likelihood associatedwith various outcomes. Probability helps to explain a lot of everyday occurrences and weactually discuss it frequently.

Probability also has been used everyday in engineering and technology. For example: theprobability of a good part being produce, the reliability of a new machine (reliabilities areactually probabilities) etc.

An engineer wants to be fairly certain that the percentage of good rods is at least 90%;otherwise he will shut down the process for recalibration. How certain that he has at least90% of the 1000 rods are good?

What is the different between probability and inferential statistics? Probability isinvolving properties of the population under study which are assumed known and

questions regarding a sample taken from the population are posed and answered. While,inferential statistics is involved a characteristics of a sample which are available to theexperimenter and this information enables experimenter to draw conclusions about thepopulations.


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1.1.1 Definition:

Some definitions or terms in basic probability must be known and well understand.Among the definitions are:

Random Process is a situation in which possible results are known but actual resultscannot be predicted with certainty in advance.

Outcome is related toeach possible result for a random process

Experiment is a process by which an observation or measurement is obtained (yieldoutcomes)

1,2 Sample Space, probability of events, counting rule

In the process of collecting data before analysis and interpretation being done, the methodof how to model the random experiment is crucial. The terms related to it such as samplespace and an event are important.

1.2.1 Sample Space:

Sample space denoted by S, is the set of all possible outcomes of an experiment.Event is any collection (subset) of outcomes contained in the sample space S.

An event is called simple if it consists of exactly one outcome and called compoundevent if it consists of more than one outcome. Mean while the null event is an event withno outcomes. This is actually impossible event or empty set.

Example 1.1:

Experiment of roll a die:

The sample space is: S = {1, 2, 3, 4, 5, 6}

The simple events (or outcomes) are:

E1: observe No. 1 = {1} E2= {2} E3 = {3}

E4= {4} E5= {5} E6= {6}

The compound events are:A : observe an odd number = {1, 3, 5}B: observe a number greater than or equal to 4 = {4, 5, 6}


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Example 1.2:

Toss a coin for three times and observed the number of heads. The sample space is,

S = {0, 1, 2, 3}

The sample space for the lifetime of a machine (in days) is,

S = {t | t 0 }= [0, )

The sample space for the number of calls at a telephone exchange during a specific timeinterval is,

S = {0, 1,.}

The knowledge in set theory is important to understand the basic of probability. Theunion of eventsA andBdenoted byA U Band read A orB is the event consisting of alloutcomes that are either inAor inBor in both events.

The intersection ofAandBdenoted byA Band read A andB, is the event consistingof all outcomes that are in bothAandB.

The complement of event A, denoted by AC, is the event of all outcomes in the sample

space Sthat are not contained in eventA.

If two events A and B have no outcomes in common they are said to be mutuallyexclusive or disjoint events. This means that if one of the events occurs the other cannot.

All these events can be visualized in term of Venn diagram:

1.2.2 Probability of Events

An event is a subset of all of the possible outcomes of an experiment. The probability of

event is to assign for each event, sayE, a number,P(E),called the probability ofEwhichwill give a precise measure of the chance that Ewill occur. The probability of an eventE,is defined as the ratio of the number of outcome favorable to the event, ndivided by thetotal number of all possible outcomes,N. That isP(E) = n/N.

For example, in the experiment tossing a die repeatedly, in the long run, what would weexpect that the probability of even number will occurs,P(E=2 or 4 or 6)?


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In this experiment, an event is even number will occur three times, so n=3.The totalpossible outcomes is six, so N=6. Hence the probability of even number will occur is,P(E=2 or 4 or 6)=3/6=0.5

Condition of Probability

A probability denoted byPis a rule (or function) which assigns a number between 0 and1 to each event and must satisfies:

0 P(E) 1 for any event E

P( ) = 0 , P(S) = 1,

If A1, A2, is an infinite collection of mutuall y exclusiveevents, then

The probability of the complementof any event A is given as

For example, ifP(rain tomorrow) = 0.6 then P(no rain tomorrow) = 0.4

Other notations for complement forA isAcor

Example 1.3:

An oil-prospecting firm plans to drill two exploratory wells. Past evidence is used toassess the possible outcomes listed in the following table:

Event Description Probability

A

BC

Neither well produces oil nor gas

Exactly one well produces oil or gas.Both wells produce oil or gas

0.85

0.120.03

Find and give description.

1 2 1 2( ...) ( ) ( ) ...P A A P A P A

( ') 1 ( )P A P A

)()(),( CBPandCBPBAP


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Solution:

EventsA, Band C are mutually exclusive because the occurrence of one eventprecludes the occurrence of either of the other two.

P(A or B) = P(A) + P(B) = 0.97 (probability at most one well produces oil orgas)

P(B or C) = P(B) + P(C)= 0.15(probability at least one well produces gas or oil

P(B) = 1 P(B) = 0.88(probability both wells not produce or both produce oil or gas)

1.2.3 General Addition Law

LetA andBbe two events defined in a sample space S.

If two eventsA andBare mutually exclusive, then

Thus

This can be expanded to consider more than twomutually exclusive events.

Example 1.4

One of the residential in Ipoh, 45% of all households subscribe to the Sinar Hariannewspaper published in a nearby city, 75% subscribe to the Utusan Malaysia, and 30% ofall households subscribe to both papers. Draw a Venn diagram for this problem.If a household is selected at random, what is the probability that it subscribes toa) At least one of the two newspapersb) Exactly one of the two newspapers

Solution:

a)A= event subscribe to Sinar Harian,B= event subscribe to Utusan Malaysia

P(A U B) = [ P(A) + P(B)P(A B)] = 0.45 + 0.750.30 = 0.9

b)P (exactly one) =P (A B) +P (A B) = 0.15 + 0.45 = 0.6

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

P(A B) 0

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


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The probability of an event Aequals the number of outcomes (sample points) containedinAdivided by the total number of possible outcomes. That is:

P(A) = n(A) / n(S)

Important condition: all outcomes are equally likely to occur. Inefficient when n(S) islarge.

1.2.4 Counting Rule:

Eliminates the need for listing each simple event and help to easily assigned probabilitiesto various events when the outcomes are equally likely. Especially helpful if the samplespace is quite large.

Product (Multiplication) Rule

If there are kelements ( or things) to choose and there are n1choices for the firstelement, n2for the second element, and so on to nkchoices for the k

th element,

then the number of possible ways of selecting them is only applies when elements aredifferent or the order of elements matters.

Example 1.5:

A chemical engineer wishes to conduct an experiment to determine how these fourfactors affect the quality of the coating. She is interested in comparing two charge levels,three density levels, four temperature levels, and five speed levels. How many

experimental conditions are possible?

Solution:

The possible experiment conditions are 2x3x4x5=120

Permutations and Combinations

Permutation is an orderedarrangement of kobjects taken from a set of ndistinct objects (k n ).

The number of waysof permutation of kobjects from ndistinct objects will be denotedby the symbolPk,n

)!(

!PP

kn

nnknk

,


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Example 1.6:

8 teaching assistants are available to grade an exam of four questions. Wish to select adifferent assistant to grade each question (only one assistant per question). How manypossible ways can the assistant are chosen for grading?

Solution:

The number of possible ways is

Combination

Combination: an unorderedsubset of kobjects taken from a set of ndistinct objects.

The number of waysof combination of kobjects from ndistinct objects is denoted by thesymbol Ck,n

Permutation vs. Combination

Permutations are larger in number than combinations: e.g., the three numbers (1,2, 3), (1,3,2) (2,3,1) , (3,1,2), (3,2,1) are all different permutations of the numbers 1, 2 and 3.However, they all represent the same combination of numbers.

Example 1.7:

Fifteen players compete in a tournament. In how many ways cana) rankings be assigned to the top five competitors?b) the best five competitors be randomly chosen?

Solution

The number of rankings that can be assigned to the top five competitors is

The number of ways that five competitors can be chosen is

168084 PPn

k

!!

!,

)( knk

n

k

nCC

k

nnk

,!!( )! !

k nk

nn PnCk k n k k

360,360155 PPn

k

003,3!10!5

!15

5

15

nkC


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1.3 Conditional Probability; Independent Events;

Sometimes it is useful to know the probability that an event will occur given that anotherevent occurred. Given two possible events, if we know that one event occurred then this

information can be applied in calculating the other events probability.

1.3.1 Conditional Probability

The conditional probability ofA, given that Bhas already occurred, is denoted as P (A |B)and defined as:

0)(provided,)(

)()(

Bp

Bp

BApBAp

The conditional probability of B, given thatAhas already occurred, is denoted as P ( B |

A)and defined as:

0)(provided,)(

)()(

Ap

Ap

BApABp

Example 1.8:

The Information Resource Center(IRC), UTP displays three types of books entitledScience (S), Engineering (E), and Technology (T). Reading habits of randomlyselected reader with respect to these types of books are

Read regularly S E T SE ST ET SETProbability 0.14 0.23 0.37 0.08 0.09 0.13 0.05

Find the following probabilities and interpret

a) P( S | E )b) P( S |E U T )c) P( S | reads at least one )d) P( S U E | T)Solution:


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3478.023.0

08.0

)(

)(

)(

Ep

ESp

ESp

2553.047.0

12.0

)(

)()(

TEp

TESpTESp

2857.049.0

14.0

)(

)(

)(

)(

)()oneleastatreads(

TESp

Sp

TESp

TESSp

TESSpSp

5946.037.022.0

)()()(

TpTESpTESp

1.3.2 Independent Events

The probability of both events occurring can be calculated by rearranging the terms in theexpression of conditional probability.

Two eventsAandBare called independent if the probability of eventAis not affected bythe occurrence of eventB, so and

Example 1.9:

In rolling a fair die, let eventA = {1, 3, 5}and eventB = {4, 5, 6}.Are eventsAandBindependent?

0.07

0.20

0.02

0.04

0.03

0.080.05

S E

T39.3x

)()|( APBAP

)()()( BPBAPBAP

)()()( BPAPBAP


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Solution:

P(A) = , P(B)=1/2 and

Since , soA andB are not independent events.

Whats the difference between mutually exclusiveand independentevents?Two events mutually exclusive (disjoint): both cannot happen when the experiment isperformed, soP( A| B) = 0, or vice versa

Two mutually exclusive events: P(A B) = 0and P( A U B) = P(A) + P(B)Mutually exclusive events must be dependent.Two events are independent:P( A U B) = P(A) + P(B)P(A B)

Example 1.10:

Toss a single die and observe the eventsA: a number less than 4

B: a number less than or equal to 2C: a number greater than 3

Are eventsAandBindependent? Are eventsAandBmutually exclusive?Are eventsAand Cindependent? Are eventsAand Cmutually exclusive?

Solution:

P(A) = , P(B) = 1/3, P(C ) = P( A | B) P(A),AandBdependent but not mutually exclusive.Aand Care dependent but mutually exclusive.

1.4 Bayes theorem

1.4.1 Multiplicative Law of Probability and Independence

For two events A and B,

Events A and B are independentif and only if

If eventsA1, .., Akare independentthen,

( ) ( | ). ( )P A B P A B P B

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

1 2 1 2( ... ) ( ) ( ) ( )k kP A A A P A P A P A

)()()( BPAPBAP

6/1)( BAP


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Multiplication rule is most useful when the experiment consists of several stages insuccession. The conditioning event, B, describes the outcome of the first stage andAtheoutcome of the second, so that P( A| B)conditioning on what occurs first will often be

known.

Example 1.11:

During a space shot, the primary computer system is backed up by two secondarysystems. They operate independently of one another, and each is 95% reliable.What is the probability that all three systems will be operable at the time of the launch?

SolutionLet,

A1: event main system is operableA2: event first backup is operableA3: event second backup is operable

GivenP(A1) = P(A2) = P(A3) = 0.95Since they operate independentlyP(A1 A2A3) = P(A1)P(A2) P(A3) = 0.857

1.4.2 The Law of Total Probability

Suppose B1, B2,, Bn are mutually exclusiveand exhaustivein S, then for any eventA

1.4.3 Bayes Theorem

Suppose B1, B2,, Bn are mutually exclusive and exhaustive (whose union is S). LetAbe an event such thatP(A)> 0. Then for any event Bj, j =1, 2, , n,

Example 1.12:

A store stocks bulbs for LCD projector from three suppliers. Suppliers A, B, and Csupply 10%, 20%, and 70% of the bulbs respectively. It has been determined thatcompanyAs bulbs are 1% defective while company Bs are 3% defective and company

1

( ) ( | ) ( )( | )

( ) ( | ) ( )

k k k

k n

i i

i

P A B P A B P BP B A

P A P A B P B

1 1

( ) ( ) ( | ) ( )n n

i i i

i i

P A P A B P A B P B


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Cs are 4% defective. If a bulb is selected at random and found to be defective, what isthe probability that it came from supplierB?

Solution:

LetDis a defective, then the probability that it came from supplierBis

||

| | |

P B P D BP B D

P A P D A P B P D B P C P D C

0.2 0.03

0.1 0.01 0.2 0.03 0.7 0.04

0.1714


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Exercise Chapter 1:

1. Each message in a digital communication system is classified as to whether it is receivedwithin the time specified by the system design. If 3 messages are classified, what is an

appropriate sample space for this experiment?

2. A digital scale is used that provide weights to the nearest gram. Let event A: a weightexceeds 11 grams, B: a weight is less than or equal to 15 grams, C: a weight is greater than orequal to 8 grams and less than 12 grams. What is the sample space for this experiment? andfind

(a) A UB (b) A (c) A B

(d) (A UC) (e) A B C (f) B C

3. Samples of building materials from three suppliers are classified for conformance to air-quality specifications. The results from 100 samples are summarized as follows:

Conforms

Yes No

Supplier

R 30 10

S 22 8

T 25 5

Let A denote the event that a sample is from supplier R, and B denote the event that a sample

conforms to the specifications. If sample is selected at random, determine the followingprobabilities:

(a) P(A) (b) P(B) (c) P(B)(d) P(AUB) (e) P(AB) (f) P(AUB)

(g) )( BAP (h) )( ABP

4. The compact discs from a certain supplier are analyzed for scratch and shock resistance. Theresults from 100 discs tested are summarized as follows:

ScratchResistance

High Low

ShockResistance

High 30 10

Medium 22 8

Low 25 5


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Let A denote the event that a disc has high shock resistance, and B denote the event that adisc has high scratch resistance. If sample is selected at random, determine the followingprobabilities:

(a) P(A) (b) P(B) (c) P(B)

(d) P(AUB) (e) P(AB) (f) P(AUB)(g) )( BAP (h) )( ABP

5. The reaction times ( in minutes) of a reactor for two batches are measured in an experiment.

(a) Define the sample space of the experiment.(b) Define event A where the reaction time of the first batch is less than 45 minutes and event

B is the reaction time of the second batch is greater than 75 minutes.(c) FindA UB, A B and A(d) Verify whether events A and B are mutually exclusive.

6. When a die is rolled and a coin is tossed, use a tree diagram to describe the set of possibleoutcomes and find the probability that the die shows an odd number and the coin shows ahead.

7. A bag contains 3 black and 4 while balls. Two balls are drawn at random one at a timewithout replacement.

(i) What is the probability that a second ball drawn is black?(ii) What is the conditional probability that first ball drawn is black if the second ball is

known to be black?

8. An oil-prospecting firm plans to drill two exploratory wells. Past evidence is used to assessthe possible outcomes listed in the following table:

Find and give description for

9. In a residential suburb, 60% of all households subscribe to the metro newspaper published ina nearby city, 80% subscribe to the local paper, and 50% of all households subscribe to bothpapers. Draw a Venn diagram for this problem. If a household is selected at random, what isthe probability that it subscribes to(a) at least one of the two newspapers(b) exactly one of the two newspapers

Event Description Probability

ABC

Neither well produces oil or gasExactly one well produces oil or gas

Both wells produce oil or gas

0.800.180.02

)'()(),( BPandCBPBAP


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10.In a student organization election, we want to elect one president from five candidates, onevice president from six candidates, and one secretary from three candidates. How manypossible outcomes?

11.Suppose each student is assigned a 5 digit number. How many different numbers can becreated?

12.A chemical engineer wishes to conduct an experiment to determine how these four factorsaffect the quality of the coating. She is interested in comparing three charge levels, fivedensity levels, four temperature levels, and three speed levels. How many experimentalconditions are possible?

13.A menu has five appetizers, three soup, seven main course, six salad dressings and eightdesserts. In how many ways can(a)a full meal be chosen?

(b)a meal be chosen if either and appetizer or a soup is ordered, but not both?

14.Ten teaching assistants are available to grade a test of four questions. Wish to select adifferent assistant to grade each question (only one assistant per question). How manypossible ways can the assistant be chosen for grading?

15.Participant samples 8 products and is asked to pick the best, the second best, and the thirdbest. How many possible ways?

16.Suppose that in the taste test, each participant samples eight products and is asked to selectthe three best products. What is the number of possible outcomes?

17.A contractor has 8 suppliers from which to purchase electrical supplies. He will select 3 ofthese at random and ask each supplier to submit a project bid. In how many ways can theselection of bidders be made?

18.Twenty players compete in a tournament. In how many ways can(a) rankings be assigned to the top five competitors?(b)the best five competitors be randomly chosen?

19.Three balls are selected at random without replacement from the jar below. Find theprobability that one ball is red and two are black.

20.A university warehouse has received shipment of 25 printers, of which 10 are laser printersand 15 are inkjet models. If 6 of these 25 are selected at random by a technician, what is theprobability that exactly 3 of those selected are laser printers?

21.There are 17 broken light bulbs in a box of 100 light bulbs. A random sample of 3 light bulbsis chosen without replacement.(a)How many ways are there to choose the sample?


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(b)How many samples contain no broken light bulbs?(c)What is the probability that the sample contains no broken light bulbs?(d)How many ways to choose a sample that contains exactly 1 broken light bulb?(e)What is the probability that the sample contains no more than 1 broken light bulb?

22.An agricultural research establishment grows vegetables and grades each one as either goodor bad for taste, good or bad for its size, and good or bad for its appearance. Overall, 78% ofthe vegetables have a good taste. However, only 69% of the vegetables have both a goodtaste and a good size. Also, 5% of the vegetables have a good taste and a good appearance,but a bad size. Finally, 84% of the vegetables have either a good size or a good appearance.(a) if a vegetable has a good taste, what is the probability that it also has a good size?(b) if a vegetable has a bad size and a bad appearance, what is the probability that it has a

good taste?

23.A local library displays three types of books entitled Science (S), Arts (A), andNovels (N). Reading habits of randomly selected reader with respect to these types of

books are

Read regularly S A N SA SN AN SANProbability 0.14 0.23 0.37 0.08 0.09 0.13 0.05

Find the following probabilities and interpret(a)P( S | A )(b)P( S | A U N )(c)P( S | reads at least one )(d)P( S U A | N)

24.A batch of 500 containers for frozen orange juice contains 5 that are defective. Two areselected at random, without replacement, from the batch. Let A and B denote that the firstand second selected is defective respective(a)Are A and B independent events?(b)If the sampling were done with replacement, would A and B be independent?

25.Everyday (Mon to Fri) a batch of components sent by a first supplier arrives at certaininspection facility. Two days a week, a batch also arrives from a second supplier. Eightypercent of all batches from supplier 1 pass inspection, and 90% batches of supplier 2 passinspection. On a randomly selected day, what is the probability that two batches passinspection?

26.The probability is 1% that an electrical connector that is kept dry fails during the warrantyperiod of a portable computer. If the connector is ever wet, the probability of a failure duringthe warranty period is 5%. If 90% of the connectors are kept dry and 10% are wet, whatproportion of connectors fail during the warranty period?

27.Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects(88%). Mechanical defects are related to loose keys (27%) or improper assembly (73%).


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Electrical connect defects are caused by defective wires (35%), improper connections (13%)or poorly welded wires (52%). Find the probability that a failure is due to(a) loose keys(b)improperly connected or poorly welded wires.

28.During a space shot, the primary computer system is backed up by two secondary systems.They operate independently of one another, and each is 90% reliable. What is the probabilitythat all three systems will be operable at the time of the launch?

29.A store stocks light bulbs from three suppliers. SuppliersA,B, and Csupply 10%, 20%, and70% of the bulbs respectively. It has been determined that company As bulbs are 1%defective while companyBs are 3% defective and company Cs are 4% defective. If a bulbis selected at random and found to be defective, what is the probability that it came fromsupplierB?

30.A particular city has three airports. Airport A handles 50% of all airline traffic, while airports

B and C handle 30% and 20%, respectively. The rates of losing a baggage in airport A, B andC are 0.3, 0.15 and 0.14 respectively. If a passenger arrives in the city and losses a baggage,what is the probability that the passenger arrives at airport A?

31.A company rated 75% of its employees as satisfactory and 25% unsatisfactory. Of thesatisfactory ones 80% had experience, of the unsatisfactory only 40%. If a person withexperience is hired, what is the probability that (s)he will be satisfactory?

32.In a certain assembly plant, three machines, B1, B2, B3, make 30%, 45% and 25%,respectively, of the products. It is known from past experience that 2%,3% and 2% of theproducts made by each machine, respectively, are defective. Now, suppose that a finished

product is randomly selected.(a)What is the probability that it is defective?(b)If a product was chosen randomly and found to be defective, what is the probability that

it was produced by machine B3?

33.Three machines A, B and C produce identical items of their respective output 5%, 4% and3% of the items are faulty. On a certain day A has produced 25%, B has produced 30% andC has produced 45% of the total output. An item selected at random is found to be faulty.What are the chances that it was produced by C?

34.Suppose that a test for Influenza A, H1N1 disease has a very high success rate: if a tested

patient has the disease, the test accurately reports this, a positive, 99% of the time, and if atested patient does not have the disease, the test accurately reports that, a negative, 95% of

the time. Suppose also, however, that only 0.1% of the population have that disease.(a)What is the probability that the test returns a positive result?(b)If the patient has a positive, what is the probability that he has the disease?(c)What is the probability of a false positive?


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35.An insurance company charges younger drivers a higher premium than it does older driversbecause younger drivers as a group tend to have more accidents. The company has 3 agegroups: Group A includes those less than 25 years old, have a 22% of all its policyholders.Group B includes those 25-39 years old, have a 43% of all its policyholders, Group Cincludes those 40 years old and older, have 35% of all its policyholders. Company records

show that in any given one-year period, 11% of its Group A policyholders have an accident.The percentages for groups B and C are 3% and 2%, respectively.(a)What is the probability that the companys policyholders are expected to have an accident

during the next 12 months?(b)Suppose Mr. Chong has just had a car accident. If he is one of the companys

policyholders, what is the probability that he is under 25?


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Chapter 2

2. Discrete random variable and discrete probability distributionsLearning objectives:


Define the random variables

Differentiate between discrete and continuous random variables

Define the discrete probability distributions

Know the special functions for discrete probability distribution

2.1 Introduction

A random variable is a rule that assigns a number to each outcome of an experiment.These numbers are called the measured valuesof the random variable. The capital letterslikeX, YandZis used to denote a random variable and the small letters like x, y and z todenote the measured values.

Example 2.1:

Select a soccer player; the random variable Yis the number of goals the player has

scored during the season.

The measured values of Yare 0, 1, 2, 3,

The test marks for 100 engineering students; the random variableZis the average numberof goals scored by the students.

The values ofZare 65.4, 67.8, 70.5, 77.3,

There are two types of random variables called a discrete random variable and acontinuous random variable.

2.2 Discrete random variableThe measured values for a discrete random variable are finite or countable. The valuesare in terms of integer value. The number of students in this class is the example of adiscrete random variable.


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2.3 Continuous random variableThe measured values for continuous random variables are in terms of real number in therange. It can be any values within the range. The weight of students in this class is theexample of a continuous random variable.

Example 2.2:

Identify whether the random variable below is discrete or continuous random variable.

(i) The number of female students in the class.(ii) The number of telephone calls.(iii) The time between two accidents.(iv) The number of cracks in a certain length of road.(v) The height of the athletes participated in the Asian Game.(vi) The volume of water in the tank.

(vii) A score on the statistics final examination.(viii) The number of cars on the road at a certain period of time.

Solution:

(i) discrete(ii) discrete(iii) continuous(iv) discrete(v) continuous(vi) continuous

(vii) continuous(viii) discrete

2.4Discrete probability distribution

If arandom variableis a discrete variable, itsprobability distributionis called adiscrete probability distribution or probability mass function, pmf.

Suppose the experiment is flipping a coin two times. This simple experimentcan have

four possible outcomes or sample space: HH, HT, TH, and TT. Now, let the random

variableX

represent the number of Heads that result from this experiment. The random

variableXcan only take on the values 0, 1,or 2, so it is a discrete random variable.
http://stattrek.com/Help/Glossary.aspx?Target=Random_variablehttp://stattrek.com/Help/Glossary.aspx?Target=Random_variablehttp://stattrek.com/Help/Glossary.aspx?Target=Random_variablehttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Statistical_experimenthttp://stattrek.com/Help/Glossary.aspx?Target=Statistical_experimenthttp://stattrek.com/Help/Glossary.aspx?Target=Statistical_experimenthttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Random_variable


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The probability distribution for this experiment appears as below.

Number ofheads, X

Probability function, P(X)

0 1/4

1 1/2

2 1/4

The above table represents a discrete probability distribution and the probability function,

P(X=xi)is called probability mass function (pmf) of Xbecause it relates each value of a

discrete random variable with its probability of occurrence.

2.4.1 The properties of the probability mass function (pmf)The pmf, P(X=xi)of a discrete random variable Xmust satisfied two conditions;

(i) 1)(0 ixXP

(ii) 1)( ix

ixXP

Given pmf, the probability of Xoccurs can be calculated. For example the probability atmost one occurs is )1()0()1( XPXPXP

Example 2.3:

Two balls are drawn at random in succession without replacement from an urn containing4 red balls and 6 black balls. Find the probabilities of all the possible outcomes.

Solution:

Let Xdenote the number of red balls in the outcome.

Possibleoutcomes

RR RB BR BB

X 2 1 1 0

Here,x1= 2,x2= 1,x3= 1,x4= 0


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Now, the probability of getting 2 red balls when we draw out the balls one at a time is:

Probability of first ball being red = 4/10

Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out

of a total of 9 balls left.) So:

Likewise, for the probability of red first is 4/10 followed by black is 6/9 (because thereare 6 black balls still in the urn and 9 balls all together). So:

Similarly for black then red:

Finally, for 2 black balls:

So the probability distribution is:

X 2 1 0P(X=x) 2/15 8/15 5/15

Example 2.4:

Given the probability distribution,

X 0 1 2 3 4 5

P(X=x) 1/10 1/5 k 1/5 3/10 1/10

Find the value of kthat makes P(X=x)a valid pmf of X.

Solution:

For P(X=x) is truly pmf of X, its must satisfy 1)( ix

ixXP . Hence,


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10/1110/110/35/15/110/1)( kkxXPix

i.

2.4.2 The cumulative distribution function (cdf)

The cdf of a discrete random variable Xis defined by,

)()()( xx

ii

xXPxXPxF

Example 2.5:

Given pmf,

X 2 1 0

P(X=x) 2/15 8/15 5/15

Find the cdf of X.

Solution:

For x< 0, 0)()( xXPxF

For 15/5)0()0()(,10 xPXPxFx

For 15/1315/815/5)1()0()1()(,21 xPxPXPxFx

For

115/214/815/5

)2()1()0()2()(,2

xPxPxPXPxFx

So the cdf of Xis:

2,1

21,15/13

10,15/5

0,0

)(x

x

x

x

xF


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2.4.3 The mean and the variance of X

Given the pmf of X, P(X=x), all the parameters of X such as the mean, the variance andthe standard deviation can be determined by using the expectation definition.

The mean of Xis defined by,

)()(1

n

iii xXPxXE

The variance of Xis defined by,

2

1

2222 )())(()()(

n

iii xXPxXEXEXVar

The standard deviation, is a square root of the variance.

Example 2.6:

Let Xis a random variable with pmf,

X 2 1 0

P(X=x) 2/15 8/15 5/15

Find the mean, the variance and the standard deviation of X.

Solution:

The mean of X is:

8.015/12)15/5(0)15/8(1)15/2(2)()(1

n

iii xXPxXE

The variance of X is:

0.4270.64-16/15

)8.0()15/5(0)15/8(1)15/2(2

)())(()()(

2222

2

1

2222

n

iii xXPxXEXEXVar


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Standard deviation is = 0.653

2.5Special functions for discrete probability distribution

There are many special discrete probability distributions such as Bernoulli distribution,Binomial distribution and Poisson distribution.

2.5.1 Bernoulli distribution

The experiment conducted with only two possible outcomes. In an experiment of tossinga fair coin for 1 time and Xis the number of head. There are only two possible outcomes,X =0 or X=1with probability distribution:

Possibleoutcomes

Head Tail

X 1 0P(X=x) 1/2 1/2

2.5.2 Binomial distribution

If the Bernoulli experiment conducted for n times, and the random variable X is thenumber of success, then the probability distribution of Xis called Binomial distributionwith pmf,

,.......3,2,1,0,)(

xqpx

n

xXP

xnx

where pis the probability of success and q=1-p.

By using the definition, it can be shown that, if X is a Binomial distribution, then themean of Xis E(X) = npand the variance of X, is Var(X) = npq.

Example 2.7:

In the experiment of tossing a fair coin for 10 times, and Xis the number of head.(i) What is the pmf of X?.(ii) Find the probability the head will appear exactly 5 times.(iii) What is the probability no head?(iv) Find the mean and the variance ofX.

Solution:

(i) The probability mass function of Xis given by:


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,.......3,2,1,0,)5.0()5.0(10

)( 10

x

xxXP xx

(ii) 246.0)5.0()5.0(!5!5

!10)5.0()5.0(

5

10)5( 5555

XP

(iii) 00097.0)5.0()5.0(!10!0

!10)5.0()5.0(

0

10)0(

100100

XP

(i i i) The mean of Xis np = 10(0.5)=5The variance of Xis npq=10(0.5)(0.5)=2.5

2.5.3 Poisson distribution

Another important discrete distribution is a Poisson distribution. The random variable Xis the number of occurrences in the interval of interest. The example of Poissondistribution is the number of accidents within a certain period of times. If Xis a randomvariable with a Poisson distribution then the pmf of Xis given by,

,.......3,2,1,0,!

)(

xx

exXP

x

where is the mean of Xfor the interval of interest.

By using the definition, it can be shown that, if X is a Binomial distribution, then the

mean of Xis E(X) = and the variance of X, is Var(X) = .Example 2.8:

Anne's answering machine receives about 6 telephone calls between 8 a.m. and 10 a.m.What is the probability that Anne receives more than 1 call in the next 15 minutes?

Solution:

Let X= the number of calls Anne receives in 15 minutes. (The interval of interest is 15minutes. The random variable X takes on the values 0, 1, 2,.. If Anne receives, on theaverage, 6 telephone calls in 2 hours, then Anne will receives 1/8 = 0.75 calls in 15

minutes, on the average. So, it means that is 0.75.

Hence, the probability that Anne receives more than 1 call in the next 15 minutes is,


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174.0)354.0472.0(1!1

)75.0(

!0

)75.0(1

)]1()0([1)1(1)1(

175.0075.0

ee

XPxPXPXP

Exercise 2

1. Identify each of the random variables as continuous or discrete random variable.

(a) The number of atoms(b) The number of fish in a pond(c) The home team score in a football game(d) The voltage on a power line(e) A score on the mathematic final exam

(f) The volume of gas in the tank(g) The number of cars at the petrol station(h) The number of accidents in Ipoh(i) The number of cakes left in the pantry(j) The height of civil engineering students in UTP

2. Let,

x 0 1 2 3

0.15 0.25 k 0.35

(i) Find the value of k that result in a valid probability distribution.(ii)Find the expected number ofXand the standard deviation ofX.(iii)What is the probability thatXgreater than or equal to 1?

3. At UTP, the business students run an investment club. Each semester they create investmentportfolios in multiples of RM1,000 each. Records from the past several years show thefollowing probabilities of profits (rounded to the nearest RM50). In the table below, x =profit per RM1, 000 andP(x) is the probability of earning that profit.

x 0 50 100 150 200

0.15 0.35 k 0.2 0.05

(a)Determine the value of kthat results in a valid probability distribution.(b)The profit per RM1, 000 is a random variable. Is it discrete or continuous? Explain.(c)Find the expected value of the profit in a $1,000 portfolio.(d)Find the standard deviation of the profit.(e)What is the probability of a profit of $150 or more in a RM1, 000 portfolios?


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(b)What is the probability that none of the student pilots successfully lands the plane usingthe simulator?

(c)What is the probability that exactly eight of the student pilots successfully land the planeusing the simulator?

8. SupposeXhas a Poisson distribution a mean of 7Determine the following.(a)P(X = 0);(b)P(X = 5);(c)P(X < 3); and(d) )4( XP .

9. At the Mc Donald drive-thru window of food establishment, it was found that during slowerperiods of the day, vehicles visited at the rate of 15 per hour. Determine the probability that(a)no vehicles visiting the drive-thru within a ten-minute interval during one of these slow

periods;(b)only 3 vehicles visiting the drive-thru within a ten-minute interval during one of these

slow periods; and(c)at least three vehicles visiting the drive-thru within a ten-minute interval during one of

these slow periods.

10.The number of cracks in a section of PLUS highway that are significant enough to requirerepair is assumed to follow a Poisson distribution with a mean of two cracks per kilometer.Determine the probability that(a) there are no cracks at all in 2km of highway;(b) at least one crack in 500meter of highway; and(c) there are exactly 3 cracks in 0.5km of highway.


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Chapter 3

3. Continuous probability distributionsLearning objectives:


Define the continuous probability distributions

Know the special functions for continuous probability distribution

3.1 Introduction

If the outcomes of the experiment conducted are continuous random variables, itsprobability distribution is called a continuous probability distribution or probability

density function, pdf.

3.1.1 The properties of the probability density function (pdf)

The pdf, f(x)of a continuous random variable Xmust satisfied two conditions;

(j) 1)(0 xf

(ii)1)(

dxxf

Given pdf, the probability of Xoccurs can be calculated. For example the probability atmost one occurs is

1)()1( dxxfXP

Example 3.1:

Let Xbe continuous random variable with pdf given by,
http://stattrek.com/Help/Glossary.aspx?Target=Random_variablehttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Probability_distributionhttp://stattrek.com/Help/Glossary.aspx?Target=Random_variable


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elsewhere,0

20,)(

2 xkxxf

Find the value of kthat makes f(x)a valid pdf of X.

Solution:

To be a valid pdf, f(x)must satistify,

1)(

dxxf

So,

8

31

3

8

0

2

3)(

32

0

2

kkx

kdxkxdxxf

3.1.2 The cumulative distribution function (cdf)

The cdf of a continuous random variable Xis defined by,

xdxxfxXPxF x

,)()()(

Example 3.2:


elsewhere,0

10,3)(

2 xxxf

Find,

(i ) P(X < 0.5), P( 0.5 < X < 0.75)(ii) The cdf of X.Solution:

(i)

8

1)5.0(

0

5.03)()5.0( 33

5.0

0

25.0 xdxxdxxfXP


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64

19)5.0()75.0(

5.0

75.03)()75.05.0(

33

375.0

5.0

275.0

5.0

xdxxdxxfXP


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(ii) The cdf of X,

x

dxxfxXPxF 0)()()(0,For x

x x

xdxxdxdxxfxXPxFx0

0

3230)()()(,10For

xx

dxdxxdxdxxfxXPxFx1

0 1

0

2 1030)()()(,1For

So the cdf of Xis:

1,1

10,

0,0

)( 3

x

xx

x

xF

3.1.3 The mean and the variance of X

Given the pdf of X, f(x), all the parameters of Xsuch as the mean, the varianceand the standard deviation can be determined by using the expectation definition.

The mean of Xis defined by,

dxxxfXE )()(

The variance of Xis defined by,

2222 )())(()()(

dxxxfXEXEXVar

The standard deviation, is a square root of the variance.

Example 3.3:



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elsewhere,0

20,8

3

)(

2 xxxf

Find the mean and the variance of X.

Solution:

The mean of Xis

5.12

3

0

2

32

3

8

3)()(

43

2

0

xdxxdxxxfXE

The variance of Xis,

15.020

3

4

9

5

12

4

9

0

2

40

3

4

9

8

3

)5.1(8

3)()(

52

0

4

22

0

22222

xdxx

dxxxdxxfxXVar

3.2Special functions for continuous probability distribution

There are many special continuous probability distributions such as Uniformdistribution, Exponential distribution, Gamma distribution and Normaldistribution.

3.2.1 Uniform distribution

The random variable Xis a uniform distribution, and then the pdf of X is givenby,

elsewhere,0

,1

)(bxa

abxf

By using the definition, it can be shown that, if Xis a uniform distribution, then

the mean of Xis E(X) = and the variance of X, is Var(X) = .


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3.2.2 Exponential distribution

The random variable X is an exponential distribution, and then the pdf of X isgiven by,

elsewhere,0

0,)(

xexf

x

By using the definition, it can be shown that, if Xis a uniform distribution, then

the mean of Xis E(X) = and the variance of X, is Var(X) =1/.Example 3.4:

Let Xis the number of individuals failing in a large group and has a exponentialdistribution. If we assume that the mean of X under a certain situation is 10, whatis the probability that more than 20 will fail at the same time?

Solution:

The random variable Xhas pdf,

elsewhere,0

0,)(

xexf

x

where is the 1/mean of X. So, = 1/10=0.1

Hence,

135.020

)1.0()20(21.0

20

1.0

eedxeXP xx

3.2.3 Normal distribution

The random variable Xis a Normal distribution, and then the pdf of Xis given by,

xexfx

,2

1)(

2

2

)(

By using the definition, it can be shown that, if X is a random variable with

normal distribution, then the mean of X is E(X) = and the variance of X, isVar(X) = . If Xis a random variable with normal distribution, then Xis alwaysbe written as,


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Exercise 3

1. Suppose that X is a continuous random variable having the probability density

function

(a)Find the value of constant k(b)Find P(-0.5 1)(c)P(0 < X < 2)

(d)the mean ofX(e) the variance ofX.

4. LetXbe a continuous random variable with pdf given by

elsewhere

forkf

,0

1x1x)x(

2


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elsewhere,0

30,)(

2 xkxxf

Find(a) the value of constant k(b)the cdf, F(x)(c)P(X >1)(d)the mean ofX(e) the variance ofX.

5. Find the cumulative probability distribution ofX given that the density function is

Find(a) the value of constant k(b)the cdf, F(x)(c)P(0.25 < X < 0.5)(d)the mean ofX(e) the variance ofX.

6. Suppose a random variable,X has a uniform distribution with a = 5 and b = 9. Find(a)P(5.5 60)(c)P(50


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(b)will be between 50 to 100 seconds?; and(c)at most 150 seconds?

10.The mean weight of 500 UTP students is 68kg and the variance is 72.25kg. Find theprobability of students who weight

(a)between 65kg and 72kg(b)more than 70kg

11.An average LCD Projector bulb manufactured by the ABC Corporation lasts 300 dayswith variance of 2500days. By assuming that the bulb life is normally distributed,what is the probability that the bulb will last(a)at most 365 days?(b)between 250days and 350days?(c)at least 400days?

12.The line width of a tool used for semiconductor manufacturing is assumed to be

normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05micrometer.(a)What is the probability that a line width is greater than 0.62 micrometer?(b)What is the probability that a line width is between 0.47 and 0.63 micrometer?(c)The line width of 90% of samples is below what value?

oooOOOooo


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Chapter 4

4. Data display and summary of dataLearning objectives:


Explain the different between population and sample

Find the sample mean, sample variance and sample standard deviation

Plot data using stem and leaf display

Construct the Box-Plot

4.1 Introduction

The major use of inferential statistics is to use information from a sampleto infer

something about a population. A population is a collection of data whoseproperties are analyzed. The population is the complete collection to be studied; itcontains all subjects of interest. A sample is a part of the population of interest, asub-collection selected from a population. A parameter is a numericalmeasurement that describes a characteristic of a population, while a statistic is anumerical measurement that describes a characteristic of a sample. In general, wewill use a statistic to infer something about a parameter.

4.2Mean and variance

The mean is the sum of all numbers in the list divided by the total numbers in the

list. If the given list is Statistical Population then the mean is called Population

Mean and the given list is a Statistical Sample, then the mean is called Sample

mean. The mean has an expected value of , known as the population mean. The

sample mean makes a good estimator of the population mean, as its expected value

which is as the same as the population mean.


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Often, since the population variance is an unknown parameter, it is estimated by

the mean sum of squares, which changes the distribution of the sample mean from

a normal distribution to a Student's t distribution with n1 degrees of freedom.

The mean and the variance of population and sample mean and sample variance

can be expressed as follows. By using the following equations we can identify thedifference.

Population Mean and Variance are defined as:

2

1

21 )(1

VarianceMean

N

ii

N

ii

xN

N

x

where Nis the size of the Population.

Sample Mean and sample variance are defined as:

2

1

21 )(1

1VarianceMean xx

ns

n

x

xN

ii

n

ii

where nis the sample size

Example 4.1:

Given the sample data as 55, 68, 90, 42, 89, 70. Find the sample mean and thesample variance of this data.

Solution:


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6.353]6

)414(30334[

5

1

)(

1

1)(

1

1

is,varianceThe

696

414

6

708942906855

is,MeanThe

2

1

2

1

22

1

2

1

n

x

xn

xxn

s

n

x

x

n

iin

ii

n

ii

n

ii

4.3 A Stem and Leaf plot

Data can be shown in a variety of ways including graphs, charts and tables. AStem and Leaf plot is a type of graph that is similar to a histogram but showsmore information. The Stem-and-Leaf plot summarizes the shape of a set of data(the distribution) and provides extra detail regarding individual values.

The data is arranged byplace value.The digits in the largest place are referred toas the stem and the digits in the smallest place are referred to as the leaf (leaves).The leaves are always displayed to the left of the stem. Stem and Leaf plots aregreat organizers for large amounts of information. It provides an at a glance toolfor specific information in large sets of data, otherwise one would have a long ofmarks to sift through and analyze. The totals of data, median and mode are alsocan be determined by Stem and Leaf plots. They are usually used when there arelarge amounts of numbers or data to analyze. Series of scores on sports teams,series of temperatures or rainfall over a period of time, series of classroom testscores are examples of when Stem and Leaf plots could be used.

Example 4.2:

The following data is the temperatures for August in Malaysia.

77 80 82 68 65 59 6155 50 62 61 70 69 6465 70 62 65 65 75 76
http://math.about.com/cs/baseten/index.htmhttp://math.about.com/cs/baseten/index.htmhttp://math.about.com/cs/baseten/index.htm


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85 80 82 83 79 79 7180 77 89

Use the Stem and Leaf plot to determine the mode and the median for thetemperatures.

Solution:

First step should be to place the numbers in order from smallest to the largest.

The mode is 65 and the median is 70.

4.4 A Box plot

Indescriptive statistics,a box plot or (also known as a box-and-whisker diagram)is an excellent visual summary of many important aspects of a data distributionthrough their five-number summaries: the smallest observation (sampleminimum), lower quartile (Q1), median (Q2), upper quartile (Q3), and largest

observation (sample maximum). A box plot may also indicate which observations,if any, might be consideredoutliers.Box plot can be drawn either horizontally orvertically.

4.4.1 Construct a box plot

Step 1: Place the numbers in order from smallest to the largest.

Step 2: Find the median,Q2, thelower quartile, Q2and the upper quartile, Q3of agiven set of data.

Step 3: Find the interquartile range(IQR). The IQR is the difference between theupper quartile and the lower quartile.

Step 4: Start to draw the Box-plot either horizontally or vertically.

Temperatures

Tens Ones

5 0 5 9

6 1 1 2 2 4 5 5 5 5 8 9

7 0 0 1 5 6 7 7 9 9

8 0 0 0 2 2 3 5 9
http://en.wikipedia.org/wiki/Descriptive_statisticshttp://en.wikipedia.org/wiki/Five-number_summaryhttp://en.wikipedia.org/wiki/Sample_minimumhttp://en.wikipedia.org/wiki/Sample_minimumhttp://en.wikipedia.org/wiki/Quartilehttp://en.wikipedia.org/wiki/Medianhttp://en.wikipedia.org/wiki/Quartilehttp://en.wikipedia.org/wiki/Sample_maximumhttp://en.wikipedia.org/wiki/Outlierhttp://en.wikipedia.org/wiki/Outlierhttp://en.wikipedia.org/wiki/Sample_maximumhttp://en.wikipedia.org/wiki/Quartilehttp://en.wikipedia.org/wiki/Medianhttp://en.wikipedia.org/wiki/Quartilehttp://en.wikipedia.org/wiki/Sample_minimumhttp://en.wikipedia.org/wiki/Sample_minimumhttp://en.wikipedia.org/wiki/Five-number_summaryhttp://en.wikipedia.org/wiki/Descriptive_statistics


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Step 5: Calculate the 1.5IQR and determine the range of 1.5IQR from upperquartile and the lower quartile. The value(s) that place outside of the1.5IQR range called the outlier(s). The value(s) that place outside of the3IQR range called the extreme outlier(s).

Example 4.3

Suppose that thirty UTP students live in Village 2. These are the following ages:

18, 20, 21, 26, 24, 19, 25, 20, 22, 21,

19, 24, 25, 28, 24, 20, 26, 20, 35, 17,

18, 24, 20, 21, 22, 27, 25, 28, 27, 24.

Step 1: Place the numbers in order from smallest to the largest.

17, 18, 18, 19, 19, 20, 20, 20, 20, 20,

21, 21, 21, 22, 22, 24, 24, 24, 24, 24,

25, 25, 25, 25, 26, 26, 27, 27, 28, 35.

Step 2: Find the median,Q2, the lower quartile, Q2and the upper quartile, Q3of agiven set of data.

The median, Q2= (X15+ X16)/2 = (22+24)/2=23

The position of Q1= (0.25) (n+1) = 0.25(31) = 7.75

So the lower quartile, Q1is X7+ 075(X8-X7) =20 + 0.75(20-20)=20

The position of Q3= (0.75) (n+1) = 0.75(31) = 23.25

So the upper quartile, Q3is X23+ 0.25(X24-X23) =25+0.25(25-25) = 25

Step 3: The interquartile range(IQR) = Q3- Q1= 2520 = 5


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The 1.5IQR = 7.5 and 3IQR = 15

Step 4: Start to draw the Box-plot either horizontally or vertically.

outlier

17 28 o35

Q1=20 Q2=23 Q3=25

12.5.< ------------..IQR=5 ---------------->.32.5

Exercise 4:

1. Find the mean, median and mode for the following observations:

6.5 7.8 4.6 3.7 6.5 9.2 12.1 6.5 3.7 10.8

2. Find the mean, median and mode for the following observations:

2.3 3.6 2.6 2.8 3.2 3.6 4.3 5.2 6.9 2.8 3.6

3. Seven oxide thickness measurements of wafers are studied to assess quality in asemiconductor manufacturing process. The data (in angstroms) are: 1264, 1280, 1301,

1300, 1292, 1307, and 1275. Calculate the sample average, variance and standarddeviation.

4. The following data are direct solar intensity measurements (watts/m2) on different

days at a location in southern Spain: 562, 869, 708, 775, 775, 704, 809, 856, 655, 806,878, 909, 918, 558, 768, 870, 918, 940, 946, 661, 820, 898, 935, 952, 957, 693, 835,905, 939, 955, 960, 498, 653, 730, 753. Calculate the sample mean, variance andsample standard deviation.


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5. Find the mean, variance and standard deviation of the following samples of marks forthe probability and statistics final examination.

84.981.9 80.8 79.4 78.2 76.5

75.073.8 72.7 72.6 71.4 70.969.368.6 67.5 66.8 65.2 64.4

59.558.3 58.5 57.6 56.9 55.248.248.0 47.8 46.5 45.9 44.638.337.4 36.8 36.5 35.6 34.938.4

6. Find the mean, variance and standard deviation of the following samples of marks forthe engineering drawing course.

98.4 98.1 98.0 97.8 96.4 95.2 94.3 92.6 91.8 90.589.6 88.7 87.3 86.8 85.7 84.2. 83.7 82.8 80.5 80.8

79.7 78.2 77.4 77.4 76.8 75.9 74.2 73.9 72.6 71.469.8 68.6 67.5 66.8 65.2 64.4 63.7 62.8 61.4 60.759.2 58.3 58.5 57.6 56.9 55.2 54.7 53.9 52.9 51.259.6 48.0 47.8 46.5 45.9 44.6 43.8 42.7 41.8 40.639.8 37.4 36.8 36.5 35.6 34.9 33.2 33.8 32.7 31.6

7. The shear strengths of 100 spot welds in a titanium alloy follow. Construct a stem-and-leaf diagram for the weld strength data and comment on any important featuresthat you notice.

5408 5431 5475 5442 5376 5388 5459 5422 5416 5435

5420 5429 5401 5446 5487 5416 5382 5357 5388 5457

5407 5469 5416 5377 5454 5375 5409 5459 5445 5429

5463 5408 5481 5453 5422 5354 5421 5406 5444 5466

5399 5391 5477 5447 5329 5473 5423 5441 5412 5384

5445 5436 5454 5453 5428 5418 5465 5427 5421 5396

5381 5425 5388 5388 5378 5481 5387 5440 5482 5406

5401 5411 5399 5431 5440 5413 5406 5342 5452 5420

5458 5485 5431 5416 5431 5390 5399 5435 5387 5462

5383 5401 5407 5385 5440 5422 5448 5366 5430 5418

(a)Construct a stem-and-leaf display for these data.(b)Find the median, the quartiles, and the 5th and 95th percentiles.

8. The data that follow represent the yield on 90 consecutive batches of ceramicsubstrate to which a metal coating has been applied by a vapor-deposition process.


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94.1 87.3 94.1 92.4 84.6 85.4

93.2 84.1 92.1 90.6 83.6 86.6

90.6 90.1 96.4 89.1 85.4 91.7

91.4 95.2 88.2 88.8 89.7 87.5

88.2 86.1 86.4 86.4 87.6 84.2

86.1 94.3 85.0 85.1 85.1 85.1

95.1 93.2 84.9 84.0 89.6 90.5

90.0 86.7 78.3 93.7 90.0 95.6

92.4 83.0 89.6 87.7 90.1 88.3

87.3 95.3 90.3 90.6 94.3 84.1

86.6 94.1 93.1 89.4 97.3 83.7

91.2 97.8 94.6 88.6 96.8 82.9

86.1 93.1 96.3 84.1 94.4 87.3

90.4 86.4 94.7 82.6 96.1 86.4

89.1 87.6 91.1 83.1 98.0 84.5

(a)Construct a cumulative frequency plot and histogram for the yield(b)Construct a stem-and-leaf display for these data.(c)Find the median, the quartiles, and the 5th and 95th percentiles for the yield

9. The average age of the football players on each team of the premier league as follows.

29.4 29.8 29.4 31.8 32.7 34.0

28.5 27.9 30.9 29.3 28.8 28.6

29.1 31.0 30.7 30.3 29.7 31.0

28.4 28.9 27.7 28.7 30.5 29.8

26.6 27.9 27.9 29.9 29.3 28.1

(a)Construct a cumulative frequency plot and histogram for the yield(b) Construct a stem-and-leaf display for these data.(c) Find the median, the quartiles, and the 5th and 95th percentiles for the yield

10.The following cold start ignition time of an automobile engine obtained for a testvehicle are as follows:

1.75 1.92 2.62 2.35 3.09 3.15 2.53 1.91

(a)Calculate the sample median, the quartiles and the IQR(b)Construct a box plot of the data.


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11.The following data are the joint temperatures of the O-rings (F) for each test firing oractual launch of the space shuttle rocket motor (fromPresidential Commission on theSpace Shuttle Challenger Accident,Vol. 1, pp. 129131): 84, 49, 61, 40, 83, 67, 45,66, 70, 69, 80, 58, 68, 60, 67, 72, 73, 70, 57, 63, 70, 78, 52, 67, 53, 67, 75, 61, 70, 81,76, 79, 75, 76, 58, 31.

(a)Compute the sample mean and sample standard deviation;(b)Calculate the median, the quartiles and the IQR;(c)Construct a box plot of the data and comment on the possible presence of outliers.

12.Ipoh Pantai Hospital compiles data on the length of stay by patients in short-termhospitals. A random sample of 28 patients yielded the following data on length ofstay, in days.

3 6 15 7 3 55 14 4 12 18 9 6 12

5

10 13 7 1 23 96 8 11 9 4 21 10

(a)Compute the sample mean and sample standard deviation;(b)Calculate the median, the quartiles and the IQR;(c)Construct a box plot of the data and comment on the possible presence of outliers.

oooOOOooo

Chapter 5

5. Random sample, central limit theorem and Normal Approximation;Statistical process control

Learning objectives:


Define the random sample and sample mean

Use the Central Limit Theorem to define the sample mean distribution

Define the Normal Approximation to Binomial and Poisson distribution

Construct the X-bar chart and R chart in statistical process control

5.1Random sample and sample mean


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In statistical terms, a random sample is a set of independent random variables X1,X2, , Xn that have been drawn from a population in such a way that eachrandom variable was selected has the same distribution and has the same chanceof being selected.

Sample mean is the average of the sample. If we have n observation in onesample, the sample mean is the total of the observation divide by the number ofsample size, n.

5.2Central Limit Theorem and sample mean distribution

Central limit theorem says that if the sample size is large, and a random sample isa set of independent random variables X1, X2, , Xn has a normal distribution

with mean, and variance, then the sample mean, X is also normallydistributed with mean, and variance, n. That is

)/,(~ 2 nNX

Example 5.1:

At chemical engineering department, Universiti Teknologi PETRONAS, the meanage of the students is 20.6 years old, and the variance is 20 years. A randomsample of 80 students is drawn from 250 students. What is the probability that theaverage age of these students is greater than 22 years old?

Solution:

0808.09192.01)4.1(1

)4.1(1)4.1()

25.0

6.2022()22(So,

)25.0,6.20(~Hence,

25.080

20)(and6.20)(ofmeanthe,80For

20)(ofvariancetheand6.20)(ofmeanThe

2

ZPZPZPXP

NX

nXVXEXn

XVXXEX

5.3Normal Approximation

The binomial and Poisson distributions are discrete random variables, whereas thenormal distribution is continuous. We need to take this into account when we are


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using the normal distribution to approximate a binomial or Poisson using acontinuity correction.

The continuity correction, 5.0 for probability of X is depend on the inequalitysign, ,,, . For example P(X < a) = P(X - 0.5 < a - 0.5) and for

)5.05.0()( aXPaXP

5.3.1 Normal approximation to Binomial

The Central Limit Theorem says that as n increases, the binomial distributionwith n trials and probability p of success gets closer and closer to a normaldistribution. That is, the binomial probability of any event gets closer and closerto the normal probability of the same event.

The normal distribution is a good approximation to Binomial when n issufficiency large and pis not too close to0or 1. How largenneeds to be depends

on the value of p. It is better to be conservative and limit the use of the normaldistribution as an approximation to the binomial when np > 5and n(1 - p) > 5.

That is, if we have a random variable X ~ Bin(n , p) and nis large and pis smallsuch that np > 5, than X can be calculated approximately using the Normaldistribution. It means that the random variable Xwill be normally distributed with

mean = np and variance, )1(2 pnp i.e X ~ N( ).

Example 5.2:

Suppose in experiment of tossing a fair coin for 50 times. What is the probabilityof getting between 9 and 11 heads?

Solution:

Let X be the random variable representing the number of heads thrown.

X ~ Bin (50, 0.5)

Since nis large and np > 5, then we can use normal approximation to find theprobability. It mean that now, X is normally distributed with mean np =25andvariance 12.5. i.e X ~ N (25, 12.5). Hence,


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7665.00934.08599.0)32.1()08.1(

)8.0.132.1(5.12

255.11

5.12

255.8

5.12

)25)5.011(

5.12

)25)5.09()119(

ZPZP

ZPXP

5.3.2 Normal approximation to Poisson

Thenormal distribution can also be used to approximate thePoisson distribution

for large values of (the mean of the Poisson distribution).That is, if we have a random variable X ~ Poisson () and is large than Xcan becalculated approximately using the Normal distribution. It means that the random

variable Xwill be normally distributed with mean = and variance, 2 i.eX ~ N ( )Example 5.3:

A car hire firm has 20 cars to hire. The number of demands for a car is hired per

day is a Poisson distribution with mean of 3. Calculate the probability that at most

ten cars will be hired in one day.

Solution:

Let a random variable Xdenotes the number of demands for a car.

The given mean value is 3. By the Poisson distribution

20.....,....,4,3,2,1,0,!

)(

xx

exXP

x

Since is large, then the probability can be calculated using a normalapproximation with mean and variance is also i.eX ~ N( Hence,

994.0)5.2()5.2(

33)5.010(

33)5.0()10(

ZP

XPXP

5.4Statistical process control
http://www.mathsrevision.net/alevel/pages.php?page=74http://www.mathsrevision.net/alevel/pages.php?page=75http://www.mathsrevision.net/alevel/pages.php?page=75http://www.mathsrevision.net/alevel/pages.php?page=74


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Statistical process control (SPC), is a powerful tools that implement the conceptof prevention as a shift from the traditional quality by inspection/correction. SPCis a technique that employs statistical tools for controlling and improvingprocesses. It is an important ingredient in continuous process improvement (CPI)strategies. It uses simple statistical means to control, monitor, and improve

processes.

Among the most commonly used tools of SPC:

histograms cause-and-effect diagrams Pareto diagrams control charts scatter or correlation diagrams run charts process flow diagrams

The most important SPC tool is called control charts. That is a graphicalrepresentations of process performance over time concerned with how (orwhether) processes vary at different intervals and identifying nonrandom orassignable causes of variation. The control charts are also providing a powerfulanalytical tool for monitoring process variability and other changes in process

mean. There are two common charts use in the SPC. The X - chart and R-chart.

The X and Range, R Charts are a set of control charts for variables data (data thatis both quantitative and continuous in measurement, such as a measured

dimension or time). The X - chart monitors the process location over time, based

on the average of a series of observations, called a subgroup. While the R-chartmonitors the variation between observations in the subgroup over time.

The X - chart or R- chart are used when you can rationally collect measurementsin groups (subgroups) of between two and ten observations. The charts' x-axes aretime based, so that the charts show a history of the process. The data is time-ordered; that is, entered in the sequence from which it was generated.

5.4.1 How to construct X - chart and R-chart

In order to construct the chart, the sample mean, the average of the sub-group andthe limits must be calculated.

The sample mean is calculated from a set of ndata values as

n

iix

nx

1

1.

The average of the subgroups data is calculated as

m

j

n

iijx

mnx

1 1

1


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where n is the subgroup size and mis the total number of subgroups included inthe analysis.

Thisx is a centre line of the chart and is called the estimate process mean.

The average range is calculated as

m

iir

mr

1

1 , whereris range between the largest

and the smallest value in each subgroup.

The upper and lower limits for the X - chart are calculated by using the formula

rAx 2 where A2can be find from the process control chart table.

While the upper for the R-chart is calculated by using the formula rD4 and for

lower limit using the formula rD3 where D4and D3can be find from the processcontrol chart table.

After the centre line and limits are calculated, and then the chart can be

constructed by plotting the observations of sample number versus x for X -chart

and the sample number versus rfor R-chart.

Example 5.4:

A component part for a jet aircraft engine is manufactured by an investmentcasting process. The vane opening on this casting is an important functionalparameter of the part.

We will illustrate the use of X and R control charts to assess the statisticalstability of this process. The table presents 20 samples of five parts each. Thevalues given in the table have been coded by using the last three digits of thedimension; that is, 31.6 should be 0.50316 inch.

Sample Number x1 x2 x3 x4 x5 X r1 33 29 31 32 33 31.6 4

2 33 31 35 37 31 33.4 6

3 35 37 33 34 36 35.0 44 30 31 33 34 33 32.2 4

5 33 34 35 33 34 33.8 2

6 38 37 39 40 38 38.4 3

7 30 31 32 34 31 31.6 4

8 29 39 38 39 39 36.8 10

9 28 33 35 36 43 35.0 15

10 38 33 32 35 32 34.0 6


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11 28 30 28 32 31 29.8 4

12 31 35 35 35 34 34.0 4

13 27 32 34 35 37 33.0 10

14 33 33 35 37 36 34.8 4

15 35 37 32 35 39 35.6 7

16 33 33 27 31 30 30.8 617 35 34 34 30 32 33.0 5

18 32 33 30 30 33 31.6 3

19 25 27 34 27 28 28.2 9

20 35 35 36 33 30 33.8 6

(a) Construct X andR control charts.(b) After the process is in control, estimate the process mean and standard

deviation.

Exercise 5

1. Suppose X1, X2, , X20 is a sample from normal distribution N (2) with = 5, 2 = 4. Find(a)Expectation and Variance of(b)Distribution of

2. Given that Xis normally distributed with mean 50 and standard deviation 4, computethe following for n=25.

(a)Mean and variance of X

(b) )49( XP

(c) )52( XP

(d) )5.5149( XP

3. Given that Xis normally distributed with mean 20 and standard deviation 2, computethe following for n=40.

X

X


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(a)Mean and variance of X

(b) )19( XP

(c) )22( XP

(d) )5.2119( XP

4. LetXdenote the number of flaws in a 1 in length of copper wire. The pmf ofXisgiven in the following table

X=x 0 1 2 3

P(X=x) 0.48 0.39 0.12 0.01

100 wires are sampled from this population. What is the probability that the averagenumber of flaws per wire in this sample is less than 0.5?

5. At a large university, the mean age of the students is 22.3 years, and the standard

deviation is 4 years. A random sample of 64 students is drawn. What is theprobability that the average age of these students is greater than 23 years?

6. Assuming an equal chance of a new baby being a boy or a girl, what is the probabilitythat 60 or more out of the next 100 births at Pantai Hospital will be girls?

7. If 10% of UTP students are international students, what is the probability that fewerthan 100 in a random sample of 818 students are coming from overseas?

8. Suppose that a sample of n = 1,600 tires of the same type are obtained at random froman ongoing production process in which 8% of all such tires produced are defective.

What is the probability that in such a sample 150 or fewer tires will be defective?

9. For overseas flights, an airline has three different choices on its dessert menuicecream, apple pie, and chocolate cake. Based on past experience the airline feels thateach dessert is equally likely to be chosen.(a) If a random sample of four passengers is selected, what is the probability that at

least two will choose ice cream for dessert?(b)If a random sample of 21 passengers is selected, what is the approximate

probability that at least two will choose ice cream for dessert?

10.Suppose that at a certain automobile plant, the number of work stoppage is a Poisson

distribution with an average per day due to equipment problems during the productionprocess is 12.0.What is the approximate probability of having 15 or fewer workstoppages due to equipment problems on any given day?

11.The number of cars arriving per minute at a toll booth on a particular bridge isPoisson distributed with a mean of 2.5.What is the probability that in any givenminute


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(a)no cars arrive?(b)not more than two cars arrive?

If the expected number of cars arriving at the toll booth per ten-minute interval is25.0, what is the approximateprobability that in any given ten-minute period

(c)not more than 20 cars arrive?(d)between 20 and 30 cars arrive?

12.A component part for a jet aircraft engine is manufactured by an investment castingprocess. The vane opening on this casting is an important functional parameter of the

part. We will illustrate the use of X and Rcontrol charts to assess the statisticalstability of this process. The table presents 20 samples of five parts each. The valuesgiven in the table have been coded by using the last three digits of the dimension; that

is, 31.6 should be 0.50316 inch.

Sample Number x1 x2 x3 x4 x5 X r1 33 29 31 32 33 31.6 4

2 33 31 35 37 31 33.4 6

3 35 37 33 34 36 35.0 4

4 30 31 33 34 33 32.2 4

5 33 34 35 33 34 33.8 2

6 38 37 39 40 38 38.4 3

7 30 31 32 34 31 31.6 4

8 29 39 38 39 39 36.8 10

9 28 33 35 36 43 35.0 15

10 38 33 32 35 32 34.0 611 28 30 28 32 31 29.8 4

12 31 35 35 35 34 34.0 4

13 27 32 34 35 37 33.0 10

14 33 33 35 37 36 34.8 4

15 35 37 32 35 39 35.6 7

16 33 33 27 31 30 30.8 6

17 35 34 34 30 32 33.0 5


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18 32 33 30 30 33 31.6 3

19 25 27 34 27 28 28.2 9

20 35 35 36 33 30 33.8 6

(a)Construct X andR control charts.

(b)After the process is in control, estimate the process mean and standard deviation.

13.The overall length of a skew used in a knee replacement device is monitored usingand R charts. The following table gives the length for 20 samples of size 4.(Measurements are coded from 2.00 mm; that is, 15 is 2.15 mm.)

Observation Observation

Sample 1 2 3 4 Sample 1 2 3 4

1 16 18 15 13 11 14 14 15 13

2 16 15 17 16 12 15 13 15 16

3 15 16 20 16 13 13 17 16 15

4 14 16 14 12 14 11 14 14 21

5 14 15 13 16 15 14 15 14 13

6 16 14 16 15 16 18 15 16 14

7 16 16 14 15 17 14 16 19 16

8 17 13 17 16 18 16 14 13 19

9 15 11 13 16 19 17 19 17 13

10 15 18 14 13 20 12 15 12 17

(a)Using all the data, find trial control limits for andRcharts, construct the chart,and plot the data.

(b)Use the trial control limits from part (a) to identify out-of-control points. Ifnecessary, revise your control limits, assuming that any samples that plot outsidethe control limits can be eliminated.

(c)Assuming that the process is in control, estimate the process mean and processstandard deviation.

14.The thickness of a printed circuit board (PCB) is an important quality parameter. Dataon board thickness (in cm) are given below for 25 samples of three boards each.

Sample 1 2 3 Sample 1 2 3

1 0.0629 0.0636 0.0640 14 0.0645 0.0640 0.0631

2 0.0630 0.0631 0.0622 15 0.0619 0.0644 0.0632

3 0.0628 0.0631 0.0633 16 0.0631 0.0627 0.0630


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Sample 1 2 3 Sample 1 2 3

4 0.0634 0.0630 0.0631 17 0.0616 0.0623 0.0631

5 0.0619 0.0628 0.0630 18 0.0630 0.0630 0.0626

6 0.0613 0.0629 0.0634 19 0.0636 0.0631 0.0629

7 0.0630 0.0639 0.0625 20 0.0640 0.0635 0.0629

8 0.0628 0.0627 0.0622 21 0.0628 0.0625 0.0616

9 0.0623 0.0626 0.0633 22 0.0615 0.0625 0.0619

10 0.0631 0.0631 0.0633 23 0.0630 0.0632 0.0630

11 0.0635 0.0630 0.0638 24 0.0635 0.0629 0.0635

12 0.0623 0.0630 0.0630 25 0.0623 0.0629 0.0630

13 0.0635 0.0631 0.0630

(a)Using all the data, find trial control limits for andRcharts, construct the chart,

and plot the data.(b)Use the trial control limits from part (a) to identify out-of-control points. If

necessary, revise your control limits, assuming that any samples that plot outsidethe control limits can be eliminated.

(c) Assuming that the process is in control, estimate the process mean and processstandard deviation.

oooOOOooo

Chapter 6

6. Hypothesis TestingOne populationLearning objectives:


Explain the concept of hypothesis testing

Understand the procedure or steps to perform the test


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Do a testing about the mean when the population variance is known and isunknown

Do a testing about the proportion

Perform the testing about the variance

6.1 Introduction

There are two types of statistical inferences: estimation of population parametersand hypothesis testing. Hypothesis testing is one of the most important tools ofapplication of statistics to real life problems. Most often, decisions are required tobe made concerning populations on the basis of sample information. Statisticaltests are used in arriving at these decisions.

Statistical hypotheses are based on the concept of proof by contradiction. For

example, say, we test the mean () of a population to see if an experiment has

caused an increase or decrease in . We do this by proof of contradiction by

formulating a null hypothesis against alternative hypothesis.

6.1.1 Null Hypothesis:

It is a hypothesis which states that there is no difference between the proceduresand is denoted by H0. For the above example the corresponding H0would be thatthere has been no increase or decrease in the mean. Always the null hypothesis istested, i.e., we want to either accept or reject the null hypothesis because we haveinformation only for the null hypothesis.

6.1.2 Alternative Hypothesis:

It is a hypothesis which states that there is a difference between the proceduresand is denoted by H1.

In hypothesis testing there will be a correct decision or false decision would bemade on the null hypothesis as summarize in this table.

Suppose Accept H0as true Reject H0as false

H0is trueCorrect decision. Probability:1 -

Type I error. Probability:

H0is false Type II error. Probability: Correct decision.Probability: 1


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We will make a correct decision if we accept H0when H0is true or we will rejectH0when actually H0is false.

The risk of rejecting the null hypothesis when we should not reject it is calledtype I error with probability .It means that we make a false decision because we

reject H0when actually H0is true with probability .

While when we accept H0 but actually H0 is not true, then we make a wrongdecision and this decision is called type II error is. The probability of type II error

is . We cannot determine (beta) with the statistical tools you learn in thiscourse.

The probability type I error, is called the level of significance and (1- )100%is called the confidence level of the test and (1 ) is called the "power" of thetest.

6.1.3 Types of test

In hypothesis testing there are three types of test on any parameters of interestcalled two tailed (sided) test, upper tailed test and lower tailed test such as in thetable below.

Type NullHypothesis,

fcm2063 - printed notes

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