1 1 Slide HJ Copyrights Chapter 5 Discrete Probability Distributions n Random Variables n Discrete Probability Distributions n Expected Value and Variance

1 1 Slide SlideHJ CopyrightsHJ Copyrights

Chapter 5Chapter 5 Discrete Probability Distributions Discrete Probability Distributions

Random VariablesRandom Variables Discrete Probability DistributionsDiscrete Probability Distributions Expected Value and VarianceExpected Value and Variance Binomial Probability DistributionBinomial Probability Distribution Poisson Probability DistributionPoisson Probability Distribution Hypergeometric Probability DistributionHypergeometric Probability Distribution

.10.10

.20.20

.30.30

.40.40

0 1 2 3 4 0 1 2 3 4


Introduction to Introduction to Probability DistributionsProbability Distributions

Random VariableRandom Variable

•Represents a possible numerical value Represents a possible numerical value from a random experimentfrom a random experiment

Random

Variables

Discrete Random Variable

ContinuousRandom Variable

Ch. 5 Ch. 6


Random VariablesRandom Variables

A A random variable random variable （随机变量）（随机变量） is a numerical is a numerical description of the outcome of an experiment.description of the outcome of an experiment.



Random VariablesRandom Variables

A A discrete random variablediscrete random variable （离散随机变（离散随机变量）量） may assume either a finite number of may assume either a finite number of values or an infinite sequence of values.values or an infinite sequence of values.

A A continuous random variablecontinuous random variable （连续随机变（连续随机变量）量） may assume any numerical value in an may assume any numerical value in an interval or collection of intervals.interval or collection of intervals.




Discrete Probability Distributions Discrete Probability Distributions （离散概率分布（离散概率分布））

The The probability distributionprobability distribution （概率分布）（概率分布） for a for a random variable describes how probabilities are random variable describes how probabilities are distributed over the values of the random distributed over the values of the random variable.variable.

The probability distribution is defined by a The probability distribution is defined by a probability functionprobability function （概率函数）（概率函数） , denoted by , denoted by ff((xx), which provides the probability for each ), which provides the probability for each value of the random variable.value of the random variable.

The required conditions for a discrete The required conditions for a discrete probability function are:probability function are:

ff((xx) ) >> 0 0

ff((xx) = 1) = 1 We can describe a discrete probability We can describe a discrete probability

distribution with a table, graph, or equation.distribution with a table, graph, or equation.


Using past data on TV sales (below left), a Using past data on TV sales (below left), a tabular representation of the probability tabular representation of the probability distribution for TV sales (below right) was distribution for TV sales (below right) was developed.developed.

NumberNumber

Units SoldUnits Sold of Daysof Days xx ff((xx))

00 80 80 0 0 .40 .40

11 50 50 1 1 .25 .25

22 40 40 2 2 .20 .20

33 10 10 3 3 .05 .05

44 2020 4 4 .10 .10

200200 1.00 1.00

Example: JSL AppliancesExample: JSL Appliances



Graphical Representation of the Probability Graphical Representation of the Probability DistributionDistribution

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.20.20

.30.30

.40.40

.50.50

0 1 2 3 40 1 2 3 4Values of Random Variable Values of Random Variable xx (TV sales) (TV sales)Values of Random Variable Values of Random Variable xx (TV sales) (TV sales)

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Discrete Uniform Probability Distribution Discrete Uniform Probability Distribution （离散均匀概率函数）（离散均匀概率函数）

The The discrete uniform probability distributiondiscrete uniform probability distribution is is the simplest example of a discrete probability the simplest example of a discrete probability distribution given by a formula.distribution given by a formula.

The The discrete uniform probability functiondiscrete uniform probability function is is

ff((xx) = 1/) = 1/nn

where:where:

nn = the number of values the = the number of values the randomrandom

variable may assumevariable may assume Note that the values of the random variable Note that the values of the random variable

are equally likely.are equally likely.


The The expected valueexpected value （数学期望）（数学期望） , or mean, of a , or mean, of a random variable is a measure of its central random variable is a measure of its central location.location.

EE((xx) = ) = = = xfxf((xx))

The The variancevariance （方差）（方差） summarizes the summarizes the variability in the values of a random variable.variability in the values of a random variable.

Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx))

The The standard deviationstandard deviation, , , is defined as the , is defined as the positive square root of the variance.positive square root of the variance.

Expected Value and VarianceExpected Value and Variance



Expected Value of a Discrete Random VariableExpected Value of a Discrete Random Variable

xx ff((xx)) xfxf((xx))

00 .40 .40 .00 .00

11 .25 .25 .25 .25

22 .20 .20 .40 .40

33 .05 .05 .15 .15

44 .10 .10 .40.40

EE((xx) = 1.20) = 1.20

The expected number of TV sets sold in a day The expected number of TV sets sold in a day is 1.2is 1.2


Variance and Standard DeviationVariance and Standard Deviation

of a Discrete Random Variableof a Discrete Random Variable

xx x - x - ( (x - x - ))22 ff((xx)) ((xx - - ))22ff((xx))

00 -1.2-1.2 1.44 1.44 .40.40 .576 .57611 -0.2-0.2 0.04 0.04 .25.25 .010 .01022 0.8 0.8 0.64 0.64 .20.20 .128 .12833 1.8 1.8 3.24 3.24 .05.05 .162 .16244 2.8 2.8 7.84 7.84 .10.10 .784 .784

1.660 = 1.660 =

The variance of daily sales is 1.66 TV sets The variance of daily sales is 1.66 TV sets squaredsquared.. The standard deviation of sales is 1.2884 TV The standard deviation of sales is 1.2884 TV sets.sets.



Binomial Probability DistributionBinomial Probability Distribution （二项分（二项分布）布）

Properties of a Binomial ExperimentProperties of a Binomial Experiment

1.1. The experiment consists of a sequence of The experiment consists of a sequence of nn identical trials.identical trials.

2.2. Two outcomes, Two outcomes, successsuccess and and failurefailure, are , are possible on each trial. possible on each trial.

3.3. The probability of a success, denoted by The probability of a success, denoted by pp, , does not change from trial to trial.does not change from trial to trial.

4.4. The trials are independent.The trials are independent.StationarityStationarityAssumptionAssumption


Binomial Probability DistributionBinomial Probability Distribution

Our interest is in the Our interest is in the number of successesnumber of successes occurring in the occurring in the nn trials. trials.

We let We let xx denote the number of successes denote the number of successes occurring in the occurring in the nn trials. trials.



Number of Experimental Outcomes Number of Experimental Outcomes Providing Exactly Providing Exactly xx Successes in Successes in nn Trials Trials

where:where: nn! = ! = nn((nn – 1)( – 1)(nn – 2) . . . (2)(1) – 2) . . . (2)(1)

0! = 10! = 1

!!( )!

n nx x n x

!!( )!

n nx x n x



Probability of a Particular Sequence of Probability of a Particular Sequence of Trial Outcomes with x Successes in Trial Outcomes with x Successes in nn Trials Trials

( )(1 )x n xp p ( )(1 )x n xp p



Binomial Probability FunctionBinomial Probability Function

where:where:

ff((xx) = the probability of ) = the probability of xx successes in successes in nn trialstrials

nn = the number of trials = the number of trials

pp = the probability of success on any = the probability of success on any one trialone trial

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1


Example: Evans ElectronicsExample: Evans Electronics


Evans is concerned about a low retention Evans is concerned about a low retention rate for employees. On the basis of past rate for employees. On the basis of past experience, management has seen a turnover experience, management has seen a turnover of 10% of the hourly employees annually. of 10% of the hourly employees annually. Thus, for any hourly employees chosen at Thus, for any hourly employees chosen at random, management estimates a probability random, management estimates a probability of 0.1 that the person will not be with the of 0.1 that the person will not be with the company next year.company next year.

Choosing 3 hourly employees at random, Choosing 3 hourly employees at random, what is the probability that 1 of them will leave what is the probability that 1 of them will leave the company this year?the company this year?



Using the Binomial Probability FunctionUsing the Binomial Probability Function

LetLet: p: p = .10, = .10, nn = 3, = 3, xx = 1 = 1

= (3)(0.1)(0.81)= (3)(0.1)(0.81)

= .243 = .243

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2



Using the Tables of Binomial ProbabilitiesUsing the Tables of Binomial Probabilities

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250


Using a Tree DiagramUsing a Tree Diagram


1st Worker 1st Worker 2nd Worker2nd Worker 3rd Worker3rd Worker xx Probab.Probab.

Leaves (.1)Leaves (.1)

Stays (.9)Stays (.9)

33

22

00

22

22



S (.9)S (.9)



S (.9)S (.9)

S (.9)S (.9)

S (.9)S (.9)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1)

L (.1)L (.1) .0010.0010

.0090.0090

.0090.0090

.7290.7290

.0090.0090

11

11

11

.0810.0810

.0810.0810

.0810.0810



SD( ) ( )x np p 1SD( ) ( )x np p 1

Expected ValueExpected Value

EE((xx) = ) = = = npnp

VarianceVariance

Var(Var(xx) = ) = 22 = = npnp(1 - (1 - pp))

Standard DeviationStandard Deviation




• Expected ValueExpected Value

EE((xx) = ) = = 3(.1) = .3 employees out = 3(.1) = .3 employees out of 3of 3

• VarianceVariance

Var(x) = Var(x) = 22 = 3(.1)(.9) = .27 = 3(.1)(.9) = .27

• Standard DeviationStandard Deviationemployees 52.)9)(.1(.3)(SD x employees 52.)9)(.1(.3)(SD x






Poisson Probability DistributionPoisson Probability Distribution （泊松概率分布）（泊松概率分布）

A discrete random variable following this A discrete random variable following this distribution is often useful in estimating the distribution is often useful in estimating the number of occurrences over a number of occurrences over a specified specified interval of time or spaceinterval of time or space..

It is a discrete random variable that may It is a discrete random variable that may assume an assume an infinite sequence of valuesinfinite sequence of values (x = 0, (x = 0, 1, 2, . . . ).1, 2, . . . ).

ExamplesExamples: :

• the number of knotholes in 14 linear feet of the number of knotholes in 14 linear feet of pine boardpine board

• the number of vehicles arriving at a toll the number of vehicles arriving at a toll booth in one hourbooth in one hour


Poisson Probability DistributionPoisson Probability Distribution

Properties of a Poisson ExperimentProperties of a Poisson Experiment

• The probability of an occurrence is the same The probability of an occurrence is the same for any two intervals of equal length.for any two intervals of equal length.

• The occurrence or nonoccurrence in any The occurrence or nonoccurrence in any interval is independent of the occurrence or interval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.


Poisson Probability DistributionPoisson Probability Distribution

Poisson Probability FunctionPoisson Probability Function

where:where:

f(x) f(x) = probability of = probability of xx occurrences in an occurrences in an intervalinterval

= = mean numbermean number of occurrences in an of occurrences in an intervalinterval

ee = 2.71828 = 2.71828

f xex

x( )

!

f x

ex

x( )

!


Example: Mercy HospitalExample: Mercy Hospital

Using the Poisson Probability FunctionUsing the Poisson Probability Function

Patients arrive at the emergency room of Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a probability of 4 arrivals in 30 minutes on a weekend evening?weekend evening?

= 6/hour = 3/half-hour, = 6/hour = 3/half-hour, xx = 4 = 44 33 (2.71828)

(4) .16804!

f

4 33 (2.71828)

(4) .16804!

f


x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

Example: Mercy HospitalExample: Mercy Hospital

Using the Tables of Poisson ProbabilitiesUsing the Tables of Poisson Probabilities







Hypergeometric Probability DistributionHypergeometric Probability Distribution（超几何概率分布）（超几何概率分布）

The The hypergeometric distributionhypergeometric distribution is closely is closely related to the binomial distribution.related to the binomial distribution.

The key differences are:The key differences are:

• the trials are not independentthe trials are not independent

• probability of success changes from trial to probability of success changes from trial to trialtrial


Hypergeometric Probability DistributionHypergeometric Probability Distribution

n

N

xn

rN

x

r

xf )(

n

N

xn

rN

x

r

xf )(

Hypergeometric Probability FunctionHypergeometric Probability Function

for 0 for 0 << xx << rr

where: where: ff((xx) = probability of ) = probability of xx successes in successes in nn trialstrials

nn = number of trials = number of trials NN = number of elements in the = number of elements in the

populationpopulation rr = number of elements in the = number of elements in the

populationpopulation labeled successlabeled success



Hypergeometric Probability FunctionHypergeometric Probability Function

• is the number of ways a sample of size is the number of ways a sample of size nn can be selected from a population of size can be selected from a population of size NN..

• is the number of ways is the number of ways xx successes can be successes can be selected from a total of selected from a total of rr successes in the successes in the population.population.

• is the number of ways is the number of ways nn – – xx failures can failures can be selected from a total of be selected from a total of NN – – rr failures in the failures in the population.population.

N

n

N

n

r

x

r

x

N r

n x

N r

n x


　例：在一个口袋中装有　例：在一个口袋中装有 3030 个球，其中有个球，其中有 1010 个红球个红球，其余为白球，这些球除颜色外完全相同，其余为白球，这些球除颜色外完全相同 .. 游戏者一游戏者一次从中摸出次从中摸出 55 个球个球 .. 摸到摸到 44 个红球就中一等奖，那么个红球就中一等奖，那么获一等奖的概率是多少？获一等奖的概率是多少？

解：由题意可见此问题归结为超几何分布模型。解：由题意可见此问题归结为超几何分布模型。　其中　其中 N = 30. r = 10. n = 5. N = 30. r = 10. n = 5. P(P( 一等奖一等奖 ) = P(X=4 or 5) = P(X=4) + P(X=5) ) = P(X=4 or 5) = P(X=4) + P(X=5) 　　　　　　 P(X=4) = C(4,10)*C(1,20)/C(5,30) P(X=4) = C(4,10)*C(1,20)/C(5,30) 　　　　 P(X=5) = C(5,10)*C(0,20)/C(5,30) P(X=5) = C(5,10)*C(0,20)/C(5,30) 　　　　 P(P( 一等奖一等奖 ) = 106/3393 ) = 106/3393


Example: NevereadyExample: Neveready


Bob Neveready has removed two dead Bob Neveready has removed two dead batteries from a flashlight and inadvertently batteries from a flashlight and inadvertently mingled them with the two good batteries he mingled them with the two good batteries he intended as replacements. The four batteries intended as replacements. The four batteries look identical.look identical.

Bob now randomly selects two of the Bob now randomly selects two of the four batteries. What is the probability he four batteries. What is the probability he selects the two good batteries?selects the two good batteries?


Example: NevereadyExample: Neveready


where:where: xx = 2 = number of = 2 = number of goodgood batteries selected batteries selected

nn = 2 = number of batteries selected = 2 = number of batteries selected NN = 4 = number of batteries in total = 4 = number of batteries in total rr = 2 = number of = 2 = number of goodgood batteries in total batteries in total

167.6

1

!2!2

!4

!2!0

!2

!0!2

!2

2

4

0

2

2

2

)(

n

N

xn

rN

x

r

xf 167.6

1

!2!2

!4

!2!0

!2

!0!2

!2

2

4

0

2

2

2

)(

n

N

xn

rN

x

r

xf


End of Chapter 5End of Chapter 5

Documents

1 1 Slide HJ Copyrights Chapter 5 Discrete Probability Distributions n Random Variables n Discrete Probability Distributions n Expected Value and Variance