Introduction to Probaility[1]

7/29/2019 Introduction to Probaility[1]

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1

Statistics

Alan D. Smith


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2An Experiment ...


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3An Experiment

100 Pennies

Random Toss ...

How many heads do you expect?

How many ways could they land?

What is the probability of getting all heads?What is the probability of getting 50 heads &

50 tails?


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4Analysis

Consider 4 pennies

Consider 100 pennies !!!


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5

Who Cares about Pennies?

Consider a situation where a manufacturer is

concerned about market share. Historically,

10% of the population uses the product. Amarket survey of 1,000 people reveals that 120

people in the survey use the product. Has

market share increased? Should we expandproduction? How certain are you?


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6

Discrete Probabil i ty

Distributions

Chapter 5


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7TO DEFINE THE TERMS PROBABILITY

DISTRIBUTIONAND RANDOM VARIABLE.

TO DISTINGUISH BETWEEN A DISCRETEAND

CONTINUOUSPROBABILITY DISTRIBUTION.

TO CALCULATE THE MEAN, VARIANCE, ANDSTANDARD DEVIATIONOF A DISCRETE

PROBABILITY DISTRIBUTION.

TO DESCRIBE THE CHARACTERISTICS OF THE

BINOMIAL DISTRIBUTION.

TO INTRODUCE THE POISSON DISTRIBUTION

WEEKS GOALS


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8RANDOM VARIABLES Definition:Arandom var iableis a numerical

value determined by the outcome of anexperiment. (A quantity resul ting from a random

exper iment that, by chance, can assume dif ferent

values). EXAMPLE :Consider a random experiment in

which a coin is tossed three times. Let Xbe the

number of heads. Let H represent the outcome

of a Head and T the outcome of a Tail.


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9EXAMPLE (Con tinued)

TTT

TTH

THT

THH

HTT

HTH

HHT

HHH

0

1

1

2

1

2

2

3Sample

SpaceNumber of

Heads = X


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10PROBABIL ITY DISTRIBUTIONS Definition:Aprobabil i ty distr ibutionis a listing

of all the outcomes of an experiment and theirassociated probabilities.


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11Disc rete Distr ibut ion


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12CHARACTERISTICS OF APROBABIL ITY DISTRIBUTION

The probability of an outcome must always bebetween 0 and 1.

EXAMPLE:P(0 head) = 0.125, P(1 head) = 0.375,

etc. in the coin tossing experiment. The sum of the probabilities of all mutually

exclusive outcomes is always 1.

P(0) + P(1) + P(2) + P(3) = 1


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13DISCRETE RANDOM VARIABLE Definition:Adiscrete random var iableis a

variable that can assume only certain clearlyseparated values resulting from a count of some

item of interest.

EXAMPLE:Let X be the number of heads whena coin is tossed 3 times. Here the values for X

are: 0, 1, 2 and 3.

EXAMPLE:Let X be the number of start-upbusinesses that fail given a starting set of 100.

The values for X are: 0, 1, , 99, 100.


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14CONTINUOUS RANDOM VARIABLE Definition:Acontinuous random var iableis a

variable that can assume one of an infinitelylarge number of values (within certain

limitations).

EXAMPLE:(a) Average GPA for all Students(b) Income

(c) ROI


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15THE MEAN OF A DISCRETEPROBABILITY DISTRIBUTION

Themeanreports the central locationof the data.

Themeanis the long-run averagevalue of the

random variable.

Themeanof a probability distribution is alsoreferred to as its expected value, E(X).

Themeanis aweighted average.


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16MEAN o f a PROBABILITYDISTRIBUTION (con tinued)

The mean of a discrete probability distributionis computed by the formula:

wherem(Greek letter , mu) represents the meanandP(X)is the probabil i ty of the var ious outcomes

X.

m E X X P X( ) [ ( )]


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17The Variance (Standard Deviat ion )of a Disc rete Probabi l i ty Distr ibu t ion

Thevariancemeasures the amount ofspread orvariationof a distribution.

Thevarianceof a discrete distribution is denoted

by the Greek letters2(sigma squared). Thestandard deviationis obtained by taking the

square rootofs2.


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18The Variance (Standard Deviat ion)-

(cont inued)

The varianceof a discrete probabilitydistributionis computed from the formula:

Thestandard deviationis computed froms (s 2)1/2(sigma).

( )s m22

X P X( )


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19 U-Rent-It is a car rental

agency specializing inrenting cars to families

who need an additional

car for a short period of

time. Pat Padgett,

President, has studied

her records for the last

20 weeks and reportsthe following number of

cars rented per week.

# o f C a r s

r e n t e d

W e e k s

1 0 5

1 1 6

1 2 7

1 3 2

EXAMPLE


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20Does the above data qualify as a probability

distribution?Convert the number of cars rented per week to a

probability distribution. This is shown on the

next slide.

EXAMPLE (Con t inued)


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21EXAMPLE (cont inued)


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22Compute the mean number of cars rented per

week. The meanm = E(X)= S[XP(X)]=Compute the variance of the number of cars

rented per week.

The variance s2 S[(X- m)2P(X)]=

EXAMPLE (cont inued)

CALCULATIONS FOR IN TABLE


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23CALCULATIONS FOR m IN TABLE

FORM


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24CALCULATIONS FORs2i n TABLEFORM

N u m b e r

o f C a r s

R e n t e d , X

P r o b .

P ( X )( X - m ) ( X - m ) 2 ( X - m ) 2 P ( X )

1 0 0 . 2 5 1 0 - 1 1 . 3 1 . 6 9 0 . 4 2 2 51 1 0 . 3 0 1 1 - 1 1 . 3 0 . 0 9 0 . 0 2 7 0

1 2 0 . 3 5 1 2 - 1 1 . 3 0 . 4 9 0 . 1 7 1 5

1 3 0 . 1 0 1 3 - 1 1 . 3 2 . 8 9 0 . 2 8 9 0

s 2 = 0 . 9 1 3 5m = 1 1 . 3 s = 0 . 9 5 5 8 .


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25B INOMIAL PROBAB IL ITY DISTRIBUTION Thebinomial distributionhas the following

characteristics:An outcome of an experiment is classified into

one of two mutually exclusive categories -

success or failure. The data collected are the results of counts.

The probability of a success stays the same for

each trial. So does the probability of failure. The trials are independent, meaning that the

outcome of one trial does not affect the outcome

of any other trial.

BINOMIAL PROBABILITY


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26BINOMIAL PROBABILITYDISTRIBUTION (con tinued)

To construct a binomial distr ibution, we let

nbe the number of trials.

rbe the number of observed successes.

pbe the probability of success on each trial. qbe the probability of failure, found by q =1-p.

4! = 4*3*2*1

0! = 1! = 1

P rn

r n r prqn r( )!

!( )!


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27 The Department of Labor Statistics for the state

of Kentucky reports that 20% of the workforcein Treble County is unemployed. A sample of 14

workers is obtained from the county. Compute

the following probabilities (Binomial): three are unemployed. (n= 14, p= 0.2).

P(r= 3) = ?

EXAMPLE


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28P(r = 3)USING THE FORMULAGiven : (r= 3, n= 14, p= 0.2, q= 0.8)

Recall

so,

P(r= 3) =

=BINOMDIST(3, 14, 0.2, FALSE) via EXCEL

P r

n

r n r p

r

q

n r

( )

!

!( )!


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29 at least one of the sampled workers is

unemployed.P(r 1) = at most two of the sampled workers are

unemployed.

P(r 2) = three or more are unemployed.

P(r 3) =



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30Return toWho Cares about Pennies?

Consider a situation where a manufacturer is

concerned about market share. Historically,

10% of the population uses the product. Amarket survey of 1,000 people reveals that 120

people in the survey use the product. Has

market share increased? Should we expandproduction? How certain are you?

BINOMIAL PROBABILITY


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31BINOMIAL PROBABILITYDISTRIBUTION

To construct a binomial distr ibution, we let

nbe the number of trials. (1000)

rbe the number of observed successes. (120 ...)

pbe the probability of success on each trial. (0.1)

qbe the probability of failure, 1-p. (0.90)

P(120) + P(121) + + P(1000) = 0.017 =

1 - P(r


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32A company manufactures ball bearings to be

used on mountain bikes. It is known that 5% ofthe diameters of the bearings will be outside the

accepted limit (defective). If 6 bearings are

selected at random, what is the probability that:

Exactlyzerowil l be defective? Exactly one?

Exactly two?Exactly three?Exactly four?

Exactly five?Exactly six?

(Note: Binomial n = 6, p = 0.05)

EXAMPLE


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33Note that the binomial conditions are met:

There is a constant probability of success (0.05).

There is a fixed number of trials (6).

The trials are independent. (Why?)

There are only two possible outcomes (a bearing

is either defective or nondefective).

EXAMPLE (continued)

3


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34Binomial Probability Distribution for n = 6and p = 0.05

EXAMPLE (continued)

How would you use this table?

35MEAN & VARIANCE OF THE BINOMIAL


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35MEAN & VARIANCE OF THE B INOMIALDISTRIBUTION

Themeanis given by

The varianceis given by

mnp

s2 1 np p( )

36EXAMPLE


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36EXAMPLE For the previous example regarding defective

bearings, recall that: p= 0.05 and n= 6.

Hence, m = np = s2= np(1 - p) =

37

POISSON PROBABILITY


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37

The binomial distribution of probabilities willbecome more and more skewed to the right as

the probability of success become smaller.

The limiting form of the binomial distribution

where the probability of success pis very small

andnis large is called the Poisson probabil i ty

distribution.

ThePoissondistribution can be described

mathematically using the formula:

POISSON PROBABILITY

DISTRIBUTION

38POISSON PROBABIL ITY


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38

m(mu)is the arithmetic mean number of occurrences(successes) in a particular interval of time.

e is the constant 2.71828 (base of the Naperian

logarithmic system)

x number of occurrences (successes)

P(X)is the required probability.

=POISSON(x, m, FALSE) via EXCEL

POISSON PROBABIL ITY

DISTRIBUTION

P X ex

x( )

! m

m

39

EXAMPLE


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39 The Maumee Urgent Care facility specializes in

caring for minor injuries, colds and flu. For theevening hours of 6 - 10 PM the mean number of

arr ivals is 4.0 per hour. Assume the arrivals

follow the Poisson distribution. Compute thefollowing probabilities.

What is the probability ofexactly 4arrivals in an

hour?

P(4) = 0.1954.

EXAMPLE

P Xe

x

x( )

!

m m

40


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40

What is the probability of less than 4 arrivals in

an hour?

What is the probability of at least 4 arrivals in

an hour?


P Xe

x

x

( ) !

m m

Documents

Introduction to Probaility[1]