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8/12/2019 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
© 2008 Prentice-Hall, Inc. 2 – 7
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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© 2008 Prentice-Hall, Inc. 2 – 8
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)
© 2008 Prentice-Hall, Inc. 2 – 9
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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© 2008 Prentice-Hall, Inc. 2 – 10
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)
© 2008 Prentice-Hall, Inc. 2 – 11
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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© 2008 Prentice-Hall, Inc. 2 – 12
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
© 2008 Prentice-Hall, Inc. 2 – 13
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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© 2008 Prentice-Hall, Inc. 2 – 14
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
© 2008 Prentice-Hall, Inc. 2 – 15
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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© 2008 Prentice-Hall, Inc. 2 – 16
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%
© 2008 Prentice-Hall, Inc. 2 – 17
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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© 2008 Prentice-Hall, Inc. 2 – 18
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)
© 2008 Prentice-Hall, Inc. 2 – 19
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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© 2008 Prentice-Hall, Inc. 2 – 20
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
© 2008 Prentice-Hall, Inc. 2 – 21
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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© 2008 Prentice-Hall, Inc. 2 – 22
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
© 2008 Prentice-Hall, Inc. 2 – 23
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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© 2008 Prentice-Hall, Inc. 2 – 24
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
© 2008 Prentice-Hall, Inc. 2 – 25
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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© 2008 Prentice-Hall, Inc. 2 – 26
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
© 2008 Prentice-Hall, Inc. 2 – 27
When Events Are Dependent
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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© 2008 Prentice-Hall, Inc. 2 – 28
When Events Are Dependent
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
© 2008 Prentice-Hall, Inc. 2 – 29
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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© 2008 Prentice-Hall, Inc. 2 – 32
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.
© 2008 Prentice-Hall, Inc. 2 – 33
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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© 2008 Prentice-Hall, Inc. 2 – 34
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?
© 2008 Prentice-Hall, Inc. 2 – 35
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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© 2008 Prentice-Hall, Inc. 2 – 46
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
© 2008 Prentice-Hall, Inc. 2 – 47
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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© 2008 Prentice-Hall, Inc. 2 – 48
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
© 2008 Prentice-Hall, Inc. 2 – 49
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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© 2008 Prentice-Hall, Inc. 2 – 50
Probability Distribution of a
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
© 2008 Prentice-Hall, Inc. 2 – 51
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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© 2008 Prentice-Hall, Inc. 2 – 52
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
© 2008 Prentice-Hall, Inc. 2 – 53
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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© 2008 Prentice-Hall, Inc. 2 – 54
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
© 2008 Prentice-Hall, Inc. 2 – 55
Variance of a
Discrete Probability Distribution
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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© 2008 Prentice-Hall, Inc. 2 – 56
Variance of a
Discrete Probability Distribution
A related measure of dispersion is thestandard deviation
2!Variance!
where
!
= square root
= standard deviation
© 2008 Prentice-Hall, Inc. 2 – 57
Variance of a
Discrete Probability Distribution
A related measure of dispersion is thestandard deviation
2!Variance!
where
!
= square root
= standard deviation
For the textbook questionVariance!
141291 ..
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© 2008 Prentice-Hall, Inc. 2 – 58
Probability Distribution of a
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 )
© 2008 Prentice-Hall, Inc. 2 – 59
Probability Distribution of a
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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© 2008 Prentice-Hall, Inc. 2 – 60
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
© 2008 Prentice-Hall, Inc. 2 – 61
The Binomial Distribution
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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© 2008 Prentice-Hall, Inc. 2 – 62
The Binomial Distribution
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
© 2008 Prentice-Hall, Inc. 2 – 63
The Binomial Distribution
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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© 2008 Prentice-Hall, Inc. 2 – 64
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%
© 2008 Prentice-Hall, Inc. 2 – 65
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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© 2008 Prentice-Hall, Inc. 2 – 66
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
© 2008 Prentice-Hall, Inc. 2 – 67
Solving Problems with
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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© 2008 Prentice-Hall, Inc. 2 – 68
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
Solving Problems with
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 ....
© 2008 Prentice-Hall, Inc. 2 – 69
Solving Problems with
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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© 2008 Prentice-Hall, Inc. 2 – 70
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,
© 2008 Prentice-Hall, Inc. 2 – 71
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
! 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
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The Normal Distribution
| | |
40 ! = 50 60
| | |
! = 40 50 60
Smaller !, same
| | |
40 50 ! = 60
Larger !, same
Figure 2.8
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!
The Normal Distribution
Figure 2.9
Same !, smaller
Same !, larger
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The Normal Distribution
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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The Normal Distribution
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 )
© 2008 Prentice-Hall, Inc. 2 – 77
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
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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?
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Haynes Construction Company
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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Haynes Construction Company
Haynes builds apartment buildings
! Total construction time follows a normaldistribution
! For triplexes, ! = 100 days and = 20 days
! Completion in 75 days or less will earn abonus of $5,000
! What is the probability of getting thebonus?
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Haynes Construction Company
! 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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Haynes Construction Company
! 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
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Haynes Construction Company
! = 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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Haynes Construction Company
! = 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
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Haynes Construction Company
Haynes builds apartment buildings
! Total construction time follows a normaldistribution
! For triplexes, ! = 100 days and = 20 days
! 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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Haynes Construction Company
! 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
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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 *