chap 02 probability.ppt

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© 2008 Prentice-Hall, Inc. 2 – 6

Introduction

! Life is uncertain, we are not surewhat the future will bring

! Risk and probability is a part ofour daily lives

! Probability is a numericalstatement about the likelihood

that an event will occur


Fundamental Concepts

1. The probability, P , of any event orstate of nature occurring is greaterthan or equal to 0 and less than orequal to 1. That is:

0 P (event) 1

2. The sum of the simple probabilitiesfor all possible outcomes of anactivity must equal 1



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Diversey Paint Example

! Demand for white latex paint at Diversey Paintand Supply has always been either 0, 1, 2, 3, or 4gallons per day

! Over the past 200 days, the owner has observedthe following frequencies of demand

QUANTITYDEMANDED

NUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200 Total 1.00 (= 200/200)


Diversey Paint Example! Demand for white latex paint at Diversey Paint

and Supply has always been either 0, 1, 2, 3, or 4gallons per day

! Over the past 200 days, the owner has observedthe following frequencies of demand

QUANTITYDEMANDED

NUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)

3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200 Total 1.00 (= 200/200)

Notice the individual probabilitiesare all between 0 and 1

0 ! P (event) ! 1

And the total of all eventprobabilities equals 1

" P (event) = 1.00



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Determining objective probability

! Relative frequency! Typically based on historical data

Types of Probability

P (event) =Number of occurrences of the event

Total number of trials or outcomes

! Classical or logical method! Logically determine probabilities without

trials

P (head) =1

2

Number of ways of getting a head

Number of possible outcomes (head or tail)


Types of Probability

Subjective probability is based onthe experience and judgment of theperson making the estimate

! Opinion polls

! Judgment of experts

! Delphi method

! Other methods



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Mutually Exclusive Events

Events are said to be mutuallyexclusive if only one of the events canoccur on any one trial

! Tossing a coin will resultin either a head or a tail

! Rolling a die will result inonly one of six possibleoutcomes


Collectively Exhaustive Events

Events are said to be collectivelyexhaustive if the list of outcomesincludes every possible outcome! Both heads and

tails as possibleoutcomes ofcoin flips

! All six possibleoutcomesof the rollof a die

OUTCOMEOF ROLL

PROBABILITY

1 1 /62 1 /63 1 /

64 1 /65 1 /66 1 /6

Total 1



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Drawing a Card

Draw one card from a deck of 52 playing cards

P (drawing a 7) = 4 /52 =1 /13

P (drawing a heart) = 13 /52 =1 /4

! These two events are not mutually exclusivesince a 7 of hearts can be drawn

! These two events are not collectivelyexhaustive since there are other cards in thedeck besides 7s and hearts


Table of Differences

DRAWSMUTUALLYEXCLUSIVE

COLLECTIVELYEXHAUSTIVE

1. Draws a spade and a club Yes No

2. Draw a face card and anumber card

Yes Yes

3. Draw an ace and a 3 Yes No

4. Draw a club and a nonclub Yes Yes

5. Draw a 5 and a diamond No No

6. Draw a red card and adiamond

No No



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Adding Mutually Exclusive Events

We often want to know whether one or asecond event will occur

! When two events are mutuallyexclusive, the law of addition is –

P (event A or event B) = P (event A) + P (event B)

P (spade or club) = P (spade) + P (club)= 13 /52 +

13 /52

= 26 /52 =1 /2 = 0.50 = 50%


Adding Not Mutually Exclusive Events

P (event A or event B) = P (event A) + P (event B)

– P (event A and event B both occurring)

P ( A or B) = P ( A) + P ( B) – P ( A and B) P (five or diamond) = P (five) + P (diamond) – P (five and diamond)

= 4 /52 +13 /52 –

1 /52

= 16 /52 =4 /13

The equation must be modified to accountfor double counting

! The probability is reduced bysubtracting the chance of both eventsoccurring together



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Venn Diagrams

P ( A) P ( B)

Events that are mutually

exclusive

P ( A or B) = P ( A) + P ( B)

Figure 2.1

Events that are not

mutually exclusive

P ( A or B) = P ( A) + P ( B) – P ( A and B)

Figure 2.2

P ( A) P ( B)

P ( A and B)


Statistically Independent Events

Events may be either independent ordependent! For independent events, the occurrence

of one event has no effect on theprobability of occurrence of the secondevent



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Which Sets of Events Are Independent?

1. (a) Your education

(b) Your income level

2. (a) Draw a jack of hearts from a full 52-card deck

(b) Draw a jack of clubs from a full 52-card deck

3. (a) Chicago Cubs win the National League pennant

(b) Chicago Cubs win the World Series

4. (a) Snow in Santiago, Chile

(b) Rain in Tel Aviv, Israel

Dependent events

Dependentevents

Independentevents

Independent events


Three Types of Probabilities! Marginal (or simple) probability is just the

probability of an event occurring

P ( A)

! Joint probability is the probability of two or moreevents occurring and is the product of theirmarginal probabilities for independent events

P ( AB) = P ( A) x P ( B)

! Conditional probability is the probability of event

B given that event A has occurred P ( B | A) = P ( B)

! Or the probability of event A given that event Bhas occurred

P ( A | B) = P ( A)



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Joint Probability Example

The probability of tossing a 6 on the firstroll of the die and a 2 on the second roll

P (6 on first and 2 on second)

= P (tossing a 6) x P (tossing a 2)

= 1 /6 x1 /6 =

1 /36 = 0.028


Independent Events

1. A black ball drawn on first draw

P ( B) = 0.30 (a marginal probability )

2. Two green balls drawn

P (GG ) = P (G ) x P (G ) = 0.7 x 0.7 = 0.49(a joint probability for two independent events)

A bucket contains 3 black balls and 7 green balls

! We draw a ball from the bucket, replace it, anddraw a second ball



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Independent Events

3. A black ball drawn on second draw if the firstdraw is green

P ( B | G ) = P ( B) = 0.30(a conditional probability but equal to the marginal

because the two draws are independent events)

4. A green ball is drawn on the second if the first

draw was green P (G | G ) = P (G ) = 0.70

(a conditional probability as in event 3)

A bucket contains 3 black balls and 7 green balls

! We draw a ball from the bucket, replace it, anddraw a second ball


Statistically Dependent EventsThe marginal probability of an event occurring iscomputed the same

P ( A)

The formula for the joint probability of two events is

P ( AB) = P ( B | A) P ( A)

P ( A | B) = P ( AB)

P ( B)

Calculating conditional probabilities is slightly morecomplicated. The probability of event A given thatevent B has occurred is



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When Events Are Dependent

Assume that we have an urn containing 10 balls ofthe following descriptions

! 4 are white (W ) and lettered ( L)

! 2 are white (W ) and numbered ( N )

! 3 are yellow (Y ) and lettered ( L)

! 1 is yellow (Y ) and numbered ( N )

P (WL) = 4 /10 = 0.4 P (YL) = 3 /10 = 0.3

P (WN ) = 2 /10 = 0.2 P (YN ) = 1 /10 = 0.1 P (W ) = 6 /10 = 0.6 P ( L) = 7 /10 = 0.7

P (Y ) = 4 /10 = 0.4 P ( N ) = 3 /10 = 0.3



4 ballsWhite (W )

andLettered ( L)

2 ballsWhite (W )

andNumbered ( N )

3 balls Yellow (Y )

andLettered ( L)

1 ball Yellow (Y )and Numbered ( N )

Probability (WL) =4

10

Probability (YN ) =110

Probability (YL) =3

10

Probability (WN ) =2

10

Urn contains10 balls

Figure 2.3



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The conditional probability that the ball drawnis lettered, given that it is yellow, is

P ( L | Y ) = = = 0.75 P (YL)

P (Y )

0.3

0.4

Verify P (YL) using the joint probability formula

P (YL) = P ( L | Y ) x P (Y ) = (0.75)(0.4) = 0.3


Joint Probabilities

for Dependent Events

P ( MT ) = P (T | M ) x P ( M ) = (0.70)(0.40) = 0.28

If the stock market reaches 12,500 point by January,there is a 70% probability that Tubeless Electronicswill go up

! There is a 40% chance the stock market willreach 12,500

! Let M represent the event of the stock marketreaching 12,500 and let T be the event thatTubeless goes up in value



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Sample Problem 1

! The nose cone on a space vehicle mustseparate before the satellite can be placedinto orbit and put into use. Explosive boltsare fired in order to allow the nose coneto be jettisoned. If four such bolts areused, each having a probability of 0.98 offiring correctly, what is the probability thatall four will function properly? Assume

that the bolts are not wired in series andhence are independent of each other.


Sample Problem 2

! Of the repair jobs that Bennie’sMachine Shop receives, 20 percentare welding jobs and 80 percent aremachining jobs.1. What is the probability that the next

three jobs to come in will be welding jobs?

2. What is the probability that two of thenext three jobs to come in will bemachining jobs?



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Sample Problem 3

! The Adams Tool Company uses an aptitude test toscreen applicants for machinist positions. The testis a standard one which has been used in similarapplications with other companies. Past recordsshow that 70 percent of all persons taking the testscore at least 700. Further, it has beendemonstrated that all persons who have scored 700or more have a 60 percent probability of performingsatisfactorily on the job, whereas those who scoreless than 700 only have a 30 percent probability of

satisfactory performance.1. What is the probability that a person can achieve at least

700 on the test and also perform satisfactorily on the job?

2. What is the probability that a person will score less than700 and also perform satisfactorily?


Sample 4 and 5

(Group Seatwork)

4 A class contains 30students. Ten arefemale (F) and UScitizens (U); 12 aremale (M) and UScitizens; 6 are femaleand non-US citizens(U); 2 are male andnon-US citizens. Aname is randomlyselected from theclass roster and it isfemale. What is theprobability that thestudent is a UScitizen?

5. Your professor tellsyou that if you scorean 85 or better onyour midterm exam,then you have a 90%chance of getting anA for the course. Youthink you only have a50% chance of

scoring 85 or better.Find the probabilitythat both your scoreis 85 and better andyou receive an A inthe course?



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Random Variables

Discrete random variables can assume onlya finite or limited set of values

Continuous random variables can assumeany one of an infinite set of values

A random variable assigns a real number toevery possible outcome or event in anexperiment

X = number of refrigerators sold during the day


Random Variables – Numbers

EXPERIMENT OUTCOMERANDOM

VARIABLES

RANGE OFRANDOM

VARIABLES

Stock 50Christmas trees

Number of Christmastrees sold

X 0, 1, 2,#, 50

Inspect 600items

Number of acceptableitems

Y 0, 1, 2,#, 600

Send out 5,000sales letters

Number of peopleresponding to theletters

Z 0, 1, 2,#, 5,000

Build an

apartmentbuilding

Percent of building

completed after 4months

R 0 ! R ! 100

Test the lifetimeof a lightbulb(minutes)

Length of time thebulb lasts up to 80,000minutes

S 0 ! S ! 80,000

Table 2.4



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Random Variables – Not Numbers

EXPERIMENT OUTCOMERANDOM

VARIABLES

RANGE OFRANDOM

VARIABLES

Studentsrespond to aquestionnaire

Strongly agree (SA)Agree (A)Neutral (N)Disagree (D)Strongly disagree (SD)

5 if SA4 if A..

X = 3 if N.. 2 if D.. 1 if SD

1, 2, 3, 4, 5

One machineis inspected

Defective Y=Not defective

0 if defective

1 if not defective

0, 1

Consumersrespond tohow they likea product

GoodAveragePoor

3 if good#. Z = 2 if average

1 if poor #..

1, 2, 3

Table 2.5


Probability Distribution of a

Discrete Random Variable

Dr. Shannon asked studentsto respond to the statement,

“The textbook was wellwritten and helped meacquire the necessaryinformation.”

Selecting the right probability distributionis important

! For discrete random variables aprobability is assigned to each event

5. Strongly agree4. Agree

3. Neutral2. Disagree1. Strongly disagree



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Discrete Random Variable

OUTCOMERANDOMVARIABLE ( X )

NUMBERRESPONDING

PROBABILITY P ( X )

Strongly agree 5 10 0.1 = 10/100

Agree 4 20 0.2 = 20/100

Neutral 3 30 0.3 = 30/100

Disagree 2 30 0.3 = 30/100

Strongly disagree 1 10 0.1 = 10/100

Total 100 1.0 = 100/100

Distribution follows all three rules1.Events are mutually exclusive and collectively exhaustive

2.Individual probability values are between 0 and 1

3.Total of all probability values equals 1

Table 2.6


Probability Distribution for

Dr. Shannon s Class

P ( X )

0.4 –

0.3 –

0.2 –

0.1 –

0 – | | | | | |1 2 3 4 5

XFigure 2.5



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Probability Distribution for

Dr. Shannon

s Class

P ( X )

0.4 –

0.3 –

0.2 –

0.1 –

0 – | | | | | |1 2 3 4 5

XFigure 2.5

Central tendency ofthe distribution isthe mean orexpected value

Amount ofvariability or spreadis the variance


Expected Value of a Discrete

Probability Distribution

=

=n

i

i i X P X X E

1

)(...)( 2211 nn X P X X P X X P X

The expected value is a measure of the centraltendency of the distribution and is a weightedaverage of the values of the random variable

where

i X

)(i

X P

n

i 1

)( X E

= random variable’s possible values

= probability of each of the random variable’spossible values

= summation sign indicating we are adding all n possible values

= expected value or mean of the random sample



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Variance of a

Discrete Probability Distribution

For a discrete probability distribution thevariance can be computed by

)()]([ n

i

i i X P X E X

1

22Variance!

where

i X

)( X E

)(i

X P

= random variable’s possible values

= expected value of the random variable

= difference between each value of the randomvariable and the expected mean

= probability of each possible value of therandom sample

)]([ X E X i


Variance of a


For Dr. Shannon’s class

)()]([variance5

1

2

i

i i X P X E X

).().().().(variance 2092410925 22

).().().().( 3092230923 22

).().( 10921 2

291

36102430003024204410

.

.....



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Variance of a


A related measure of dispersion is thestandard deviation

2!Variance!

where

!

= square root

= standard deviation


Variance of a


A related measure of dispersion is thestandard deviation

2!Variance!

where

!

= square root

= standard deviation

For the textbook questionVariance!

141291 ..



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Continuous Random VariableSince random variables can take on an infinitenumber of values, the fundamental rules forcontinuous random variables must be modified

! The sum of the probability values must stillequal 1

! But the probability of each value of therandom variable must equal 0 or the sumwould be infinitely large

The probability distribution is defined by a

continuous mathematical function called theprobability density function or just the probabilityfunction

! Represented by f ( X )



Continuous Random Variable

P r o b a b i l i t y

| | | | | | |

5.06 5.10 5.14 5.18 5.22 5.26 5.30

Weight (grams)

Figure 2.6



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The Binomial Distribution

! Many business experiments can becharacterized by the Bernoulli process

! The Bernoulli process is described by thebinomial probability distribution

1. Each trial has only two possible outcomes

2. The probability stays the same from one trialto the next

3. The trials are statistically independent

4. The number of trials is a positive integer



The binomial distribution is used to find theprobability of a specific number of successesout of n trials

We need to know

n = number of trials

p = the probability of success on anysingle trial

We letr = number of successes

q = 1 – p = the probability of a failure



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The binomial formula is

r nr q p

r nr

nnr

)!(!

!trialsinsuccessesof yProbabilit

The symbol ! means factorial, and

n! = n(n – 1)(n – 2)#(1)

For example

4! = (4)(3)(2)(1) = 24By definition

1! = 1 and 0! = 1



NUMBER OFHEADS (r ) Probability = (0.5)r (0.5)5 – r

5!r !(5 – r )!

0 0.03125 = (0.5)0(0.5)5 – 0

1 0.15625 = (0.5)1(0.5)5 – 1

2 0.31250 = (0.5)2(0.5)5 – 2

3 0.31250 = (0.5)3(0.5)5 – 3

4 0.15625 = (0.5)4(0.5)5 – 4

5 0.03125 = (0.5)5(0.5)5 – 5

5!0!(5 – 0)!

5!1!(5 – 1)!

5!2!(5 – 2)!

5!3!(5 – 3)!

5!4!(5 – 4)!

5!5!(5 – 5)!

Table 2.7



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Solving Problems with the

Binomial Formula

We want to find the probability of 4 heads in 5 tosses

n = 5, r = 4, p = 0.5, and q = 1 – 0.5 = 0.5

Thus454

5050454

5trials5insuccesses4

..

)!(!

!)( P

156250500625011234

12345.).)(.(

)!)()()((

))()()((

Or about 16%


Solving Problems with the

Binomial Formula

P r o b a b i l i t y

P ( r )

| | | | | | |1 2 3 4 5 6

Values of r (number of successes)

0.4 –

0.3 –

0.2 –

0.1 –

0 –

Figure 2.7



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Solving Problems with

Binomial Tables

MSA Electronics is experimenting with themanufacture of a new transistor

! Every hour a sample of 5 transistors is taken

! The probability of one transistor beingdefective is 0.15

What is the probability of finding 3, 4, or 5 defective?

n = 5, p = 0.15, and r = 3, 4, or 5So

We could use the formula to solve this problem,but using the table is easier



Binomial Tables P

n r 0.05 0.10 0.15

5 0 0.7738 0.5905 0.4437

1 0.2036 0.3281 0.3915

2 0.0214 0.0729 0.1382

3 0.0011 0.0081 0.0244

4 0.0000 0.0005 0.0022

5 0.0000 0.0000 0.0001

Table 2.8 (partial)

We find the three probabilities in the tablefor n = 5, p = 0.15, and r = 3, 4, and 5 andadd them together



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Table 2.8 (partial)

We find the three probabilities in the tablefor n = 5, p = 0.15, and r = 3, 4, and 5 andadd them together


Binomial Tables

P

n r 0.05 0.10 0.15

5 0 0.7738 0.5905 0.4437

1 0.2036 0.3281 0.3915

2 0.0214 0.0729 0.1382

3 0.0011 0.0081 0.0244

4 0.0000 0.0005 0.0022

5 0.0000 0.0000 0.0001

)()()()( 543defectsmoreor 3 P P P P

02670000100022002440 ....



Binomial Tables It is easy to find the expected value (or mean) andvariance of a binomial distribution

Expected value (mean) = np

Variance = np(1 – p)

For the MSA example

6375085015051Variance7501505valueExpected

.).)(.()(.).( pnp

np



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The Normal Distribution

The normal distribution is the most popularand useful continuous probabilitydistribution

! The formula for the probability densityfunction is rather complex

2

2

2

2

1

)(

)(

x

e X

! The normal distribution is specifiedcompletely when we know the mean, !,and the standard deviation,


Sample Problem 1 (Binomial)

! A candidate for public office hasclaimed that 60% of voters will votefor her. If 5 registered voters weresampled, what is the probability thatexactly 3 would say they favor thiscandidate.



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! The normal distribution is symmetrical,with the midpoint representing the mean

! Shifting the mean does not change theshape of the distribution

! Values on the X axis are measured in thenumber of standard deviations away fromthe mean

! As the standard deviation becomes larger,the curve flattens

! As the standard deviation becomessmaller, the curve becomes steeper



| | |

40 ! = 50 60

| | |

! = 40 50 60

Smaller !, same

| | |

40 50 ! = 60

Larger !, same

Figure 2.8



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!


Figure 2.9

Same !, smaller

Same !, larger



Figure 2.10

68%16% 16%

–1 +1 a ! b

95.4%2.3% 2.3%

–2 +2 a ! b

99.7%0.15% 0.15%

–3 +3 a ! b



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If IQs in the United States were normally distributedwith ! = 100 and = 15, then

1. 68% of the population would have IQsbetween 85 and 115 points (±1 )

2. 95.4% of the people have IQs between 70 and130 (±2 )

3. 99.7% of the population have IQs in therange from 55 to 145 points (±3 )

4. Only 16% of the people have IQs greater than115 points (from the first graph, the area tothe right of +1 )


Using the Standard Normal TableStep 1

Convert the normal distribution into a standardnormal distribution

! A standard normal distribution has a meanof 0 and a standard deviation of 1

! The new standard random variable is Z

X Z

where X = value of the random variable we want to measure

! = mean of the distribution

= standard deviation of the distribution

Z = number of standard deviations from X to the mean, !



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Using the Standard Normal Table

For example, ! = 100, = 15, and we want to findthe probability that X is less than 130

15

100130 X Z

devstd215

30

| | | | | | |

55 70 85 100 115 130 145

| | | | | | |

–3 –2 –1 0 1 2 3

X = IQ

X Z

! = 100= 15 P ( X < 130)

Figure 2.11


Using the Standard Normal TableStep 2

Look up the probability from a table of normalcurve areas

! Use Appendix A or Table 2.9 (portion below)

! The column on the left has Z values

! The row at the top has second decimalplaces for the Z values

AREA UNDER THE NORMAL CURVE

Z 0.00 0.01 0.02 0.03 1.8 0.96407 0.96485 0.96562 0.96638

1.9 0.97128 0.97193 0.97257 0.97320

2.0 0.97725 0.97784 0.97831 0.97882

2.1 0.98214 0.98257 0.98300 0.98341

2.2 0.98610 0.98645 0.98679 0.98713

Table 2.9

P ( X < 130)= ( Z < 2.00)

= 97.7%



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Haynes Construction Company

Haynes builds apartment buildings

! Total construction time follows a normaldistribution

! For triplexes, ! = 100 days and = 20 days

! Contract calls for completion in 125 days

! Late completion will incur a severe penaltyfee

! What is the probability of completing in 125days?



From Appendix A, for Z = 1.25 the area is 0.89435

! There is about an 89% probability Hayneswill not violate the contract

20

100125 X Z

25120

25.

! = 100 days X = 125 days

= 20 days Figure 2.12



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! Completion in 75 days or less will earn abonus of $5,000

! What is the probability of getting thebonus?



! But Appendix A has only positive Z values, theprobability we are looking for is in the negative tail

20

10075 X Z

25120

25.

Figure 2.12 !

= 100 days X = 75 days

P(X < 75 days)Area ofInterest



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! Because the curve is symmetrical, we can lookat the probability in the positive tail for thesame distance away from the mean

20

10075 X Z

25120

25.

! = 100 days X = 125 days

P(X > 125 days)Area ofInterest



! = 100 days X = 125 days

! We know the probabilitycompleting in125 days is 0.89435

! So the probabilitycompleting in morethan 125 days is1 – 0.89435 = 0.10565



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! = 100 days X = 75 days

! The probability of completing in less than75 days is 0.10565 or about 11%

! Going back tothe left tail of thedistribution

! The probabilitycompleting in morethan 125 days is1 – 0.89435 = 0.10565






! What is the probability of completingbetween 110 and 125 days?

! We know the probability of completing in 125days, P ( X < 125) = 0.89435

! We have to complete the probability ofcompleting in 110 days and find the areabetween those two events



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! From Appendix A, for Z = 0.5 the area is 0.69146

! P (110 < X < 125) = 0.89435 – 0.69146 = 0.20289or about 20%

20

100110 X Z

5020

10.

Figure 2.14

! = 100days

125days

= 20 days

110days


Problem 1 (Normal Distr.)

! The length of the rods coming out of our newcutting machine can be said to approximate anormal distribution with a mean of 10 inches anda standard deviation of 0.2 inch. Find theprobability that a rod selected randomly will havea length:a. Of less than 10.0 inches *

b. Between 10.0 and 10.4 inches *

c. Between 10.0 and 10.1 inches *

d. Between 10.1 and 10.4 inchese. Between 9.9 and 9.6 inches *

f. Between 9.9 and 10.4 inches *

g. Between 9.886 and 10.406 inches *

chap 02 probability.ppt

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