Chapter 2 Probability, Random Variables and Probability Distributions (1)

29
Student Lecture Notes 1 Chapter 2 Chapter 2 Probability, Random Probability, Random Variables and Probability Variables and Probability Distributions Distributions 1 Learning Objectives Learning Objectives 1. 1. Differences between the Two Types of Random Differences between the Two Types of Random V i bl V i bl Variables Variables 2. 2. Discrete Random Variables Discrete Random Variables 1. 1. Describe Discrete Random Variables Describe Discrete Random Variables 2. 2. Compute the Expected Value & Variance of Discrete Compute the Expected Value & Variance of Discrete Random Variables Random Variables 3. 3. Continuous Random Variables Continuous Random Variables 2 1. 1. Describe Normal Random Variables Describe Normal Random Variables 2. 2. Introduce the Normal Distribution Introduce the Normal Distribution 3. 3. Calculate Probabilities for Continuous Random Variables Calculate Probabilities for Continuous Random Variables 4. 4. Assessing Normality Assessing Normality

Transcript of Chapter 2 Probability, Random Variables and Probability Distributions (1)

Page 1: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 1

Chapter 2 Chapter 2 Probability, Random Probability, Random

Variables and Probability Variables and Probability DistributionsDistributions

11

Learning ObjectivesLearning Objectives

1.1. Differences between the Two Types of Random Differences between the Two Types of Random V i blV i blVariablesVariables

2.2. Discrete Random VariablesDiscrete Random Variables1.1. Describe Discrete Random VariablesDescribe Discrete Random Variables2.2. Compute the Expected Value & Variance of Discrete Compute the Expected Value & Variance of Discrete

Random VariablesRandom Variables

3.3. Continuous Random VariablesContinuous Random Variables

22

1.1. Describe Normal Random VariablesDescribe Normal Random Variables2.2. Introduce the Normal DistributionIntroduce the Normal Distribution3.3. Calculate Probabilities for Continuous Random VariablesCalculate Probabilities for Continuous Random Variables

4.4. Assessing NormalityAssessing Normality

Page 2: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 2

Random VariablesRandom Variables

•• A variable defined by the probabilitiesA variable defined by the probabilities•• A variable defined by the probabilities A variable defined by the probabilities of each possible value in the of each possible value in the population.population.

33

Data TypesData Types

Data

Numerical Qualitative

44

Discrete Continuous

Page 3: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 3

Types of Types of Random VariablesRandom Variables

Discrete Random VariableDiscrete Random VariableDiscrete Random Variable Discrete Random Variable Whole Number (0, 1, 2, 3 etc.)Whole Number (0, 1, 2, 3 etc.)Countable, Finite Number of ValuesCountable, Finite Number of Values

Jump from one value to the next and cannot take any Jump from one value to the next and cannot take any values in between.values in between.

Continuous Random VariablesContinuous Random VariablesWhole or Fractional NumberWhole or Fractional Number

55

Whole or Fractional NumberWhole or Fractional NumberObtained by MeasuringObtained by MeasuringInfinite Number of Values in IntervalInfinite Number of Values in Interval

Too Many to List Like Discrete VariableToo Many to List Like Discrete Variable

Discrete Random Discrete Random Variable ExamplesVariable Examples

ExperimentExperiment RandomRandom PossiblePossibleExperimentExperiment RandomRandomVariableVariable

PossiblePossibleValuesValues

Children of One Children of One Gender in FamilyGender in Family

# Girls# Girls 0, 1, 2, ..., 10?0, 1, 2, ..., 10?

Answer 33 QuestionsAnswer 33 Questions # Correct# Correct 0 1 2 330 1 2 33

66

Answer 33 QuestionsAnswer 33 Questions # Correct# Correct 0, 1, 2, ..., 330, 1, 2, ..., 33

Count Cars at TollCount Cars at TollBetween 11:00 & 1:00Between 11:00 & 1:00

# Cars# CarsArrivingArriving

0, 1, 2, ..., 0, 1, 2, ..., ∞∞

Page 4: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 4

Discrete Discrete Probability DistributionProbability Distribution

11 List of All possible [List of All possible [xx pp((xx)] pairs)] pairs1.1. List of All possible [List of All possible [xx, , pp((xx)] pairs)] pairsxx = Value of Random Variable (Outcome)= Value of Random Variable (Outcome)pp((xx) = Probability Associated with Value) = Probability Associated with Value

2.2. Mutually Exclusive (No Overlap)Mutually Exclusive (No Overlap)33 Collectively Exhaustive (Nothing Left Out)Collectively Exhaustive (Nothing Left Out)

77

3.3. Collectively Exhaustive (Nothing Left Out)Collectively Exhaustive (Nothing Left Out)4. 4. 0 0 ≤≤ pp((xx) ) ≤≤ 115. 5. ΣΣ pp((xx) = 1) = 1

Marilyn says: It may sound strange, but more Marilyn says: It may sound strange, but more families of 4 children have 3 of one gender and one of families of 4 children have 3 of one gender and one of the other than any other combination. Explain this.the other than any other combination. Explain this.

Construct a sample space and look at the total number of Construct a sample space and look at the total number of ways each event can occur out of the total number ofways each event can occur out of the total number of

Sample SpaceSample Space

BBBBBBBBways each event can occur out of the total number of ways each event can occur out of the total number of combinations that can occur, and calculate frequencies.combinations that can occur, and calculate frequencies. GBBBGBBB

BGBBBGBB

BBGBBBGB

BBBGBBBG

GGBBGGBB

GBGBGBGB

GBBGGBBG

BGGBBGGB

P (girl) = 1/2 P (boy) = 1/2

so, P (BBBB) = ½ x ½ x ½ x ½ = 1/16

•• Are all 16 combinations equally likely? Is the sex Are all 16 combinations equally likely? Is the sex of each child independent of the other three?of each child independent of the other three?

88

BGGBBGGB

BGBGBGBG

BBGGBBGG

BGGGBGGG

GBGGGBGG

GGBGGGBG

GGGBGGGB

GGGGGGGG

•• If you have a family of four, what is the probability of…If you have a family of four, what is the probability of…

P(all girls or all boys) = P(all girls or all boys) = P (2 boys, 2 girls)=P (2 boys, 2 girls)=P(3 boys, 1 girl or 3 girls, 1 boy)=P(3 boys, 1 girl or 3 girls, 1 boy)=

8/16=4/8=1/2 8 ways to have 3 of 1 and 2 of the other.

6/16 = 3/8 six different ways to have 2 boys and 2 girls

2/16 = 1/8

Page 5: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 5

Assume the random variable X represents the number Assume the random variable X represents the number of girls in a family of 4 kids. (lower case x is a of girls in a family of 4 kids. (lower case x is a particular value of X, ie: x=3 girls in the family)particular value of X, ie: x=3 girls in the family)

Sample SpaceSample Space

BBBBBBBB

Random Variable XRandom Variable X

x=0x=0Number of Number of

Girls, xGirls, xProbability, Probability,

P(x)P(x)GBBBGBBB

BGBBBGBB

BBGBBBGB

BBBGBBBG

GGBBGGBB

GBGBGBGB

GBBGGBBG

BGGBBGGB

x=1x=1

x=1x=1

x=1x=1

x=1x=1

x=2x=2

x=2x=2

x=2x=2

x=2x=2

,, ( )( )00 1/161/1611 4/164/1622 6/166/1633 4/164/1644 1/161/16

TotalTotal 16/16=1 0016/16=1 00

99

What is the probability of exactly 3 girls in 4 kids?What is the probability of exactly 3 girls in 4 kids?

What is the probability of at least 3 girls in 4 kids?What is the probability of at least 3 girls in 4 kids?

P(X=3) = 4/16

BGGBBGGB

BGBGBGBG

BBGGBBGG

BGGGBGGG

GBGGGBGG

GGBGGGBG

GGGBGGGB

GGGGGGGG

x=2x=2

x=2x=2

x=3x=3

x=3x=3

x=3x=3

x=3x=3

x=4x=4

TotalTotal 16/16=1.0016/16=1.00

P(X≥3) = 5/16

Visualizing Discrete Visualizing Discrete Probability DistributionsProbability Distributions

ListingListing TableTable

Probability, P(x)

6/16

0 300.350.40

Number of Girls, xNumber of Girls, x Probability, P(x)Probability, P(x)

00 1/161/16

11 4/164/16

22 6/166/16

33 4/164/16

44 1/161/16

TotalTotal 16/16=1.0016/16=1.00

{(0,1/16), (1,.25), (2,3/8),(3,.25),(4,1/16) }{(0,1/16), (1,.25), (2,3/8),(3,.25),(4,1/16) }

GraphGraph

1010

1/16

4/16 4/16

1/16

0.000.050.100.150.200.250.30

0 1 2 3 4

Number of Girls, x

P(x) X is random and x is fixed. We X is random and x is fixed. We

can calculate the probability can calculate the probability that different values of X will that different values of X will occur and make a probability occur and make a probability distribution.distribution.

Page 6: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 6

Probability DistributionsProbability Distributions

Probability, P(x)

1/16

6/16

4/16 4/16

1/16

0 000.050.100.150.200.250.300.350.40

P(x)

1111

0.000 1 2 3 4

Number of Girls, x

Probability distributionsProbability distributions can be written as probability histograms.can be written as probability histograms.Cumulative probabilitiesCumulative probabilities: Adding up probabilities of a range of : Adding up probabilities of a range of values.values.

Washington State Population Washington State Population Survey and Random VariablesSurvey and Random Variables

A telephone survey of A telephone survey of h h ld th h th h ld th h t

number of number of telephones,xtelephones,x P(x)P(x)households throughout households throughout Washington State.Washington State.But some households don’t have But some households don’t have phones.phones.

00 0.035000.03500

11 0.705530.70553

22 0.217690.21769

33 0.029660.02966

44 0.007750.00775

55 0.003320.00332

66 0.000880.00088

77 0 000020 00002

0.71

0 50

0.60

0.70

1212

77 0.000020.00002

88 0.000000.00000

99 0.000150.00015

TotalTotal 1.000001.000000.04

0.22

0.03 0.01 0.000.00

0.10

0.20

0.30

0.40

0.50

0 1 2 3 4 5 6 7 8 9

Number of Telephone Lines (x)

P(x

)

Page 7: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 7

Probabilities about Telephone in Probabilities about Telephone in Washington StateWashington State

•• What is the probability that a household will have noWhat is the probability that a household will have noWhat is the probability that a household will have no What is the probability that a household will have no telephone?telephone?

•• What is the probability that a household will have 2 or What is the probability that a household will have 2 or more telephone lines?more telephone lines?

•• What is the probability that a household will have 2 to 4 What is the probability that a household will have 2 to 4 phone lines?phone lines?Wh t i th b bilit h h ld ill h hWh t i th b bilit h h ld ill h h

1313

•• What is the probability a household will have no phone What is the probability a household will have no phone lines or more than 4 phone lines?lines or more than 4 phone lines?

•• Who do you think is in that 3.5% of the population? Who do you think is in that 3.5% of the population? •• What are the implications of this for the quality of the What are the implications of this for the quality of the

survey?survey?

Probability Histogram of Probability Histogram of Telephone Lines, 1998Telephone Lines, 1998

0.710.70

0.220.20

0.30

0.40

0.50

0.60

P(x)

1414

0.04 0.03 0.01 0.000.00

0.10

0 1 2 3 4 5 6 7 8 9

Number of Telephone Lines (x)

Page 8: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 8

1.1. Expected ValueExpected ValueMean of Probability DistributionMean of Probability Distribution

Summary MeasuresSummary Measures

Mean of Probability DistributionMean of Probability DistributionWeighted Average of All Possible ValuesWeighted Average of All Possible Valuesμμ = = EE((XX)) = = ΣΣxx pp((xx))

2.2. VarianceVarianceWeighted Average Squared Deviation about Weighted Average Squared Deviation about MeanMean

mumu

1515

Mean Mean σσ22 = = V(X)= EV(X)= E[ ([ (xx − μ)− μ)22 ] = Σ ] = Σ ((xx − μ)− μ)22 pp((xx))σσ22 = = V(X)=E(XV(X)=E(X22)) −−[[E(XE(X))]]22

3.3. Standard DeviationStandard Deviationσσ ==√σσ22 = = SD(X)SD(X)

Sigma Sigma --squaredsquared

What is the average number of telephones in What is the average number of telephones in Washington Households and how much does size vary Washington Households and how much does size vary from the average?from the average?

# of# ofPhonesPhones

Approach 1: Variance Approach 1: Variance Approach 2: Approach 2: VarianceVariance

FF P( )P( ) P( )P( ) (( )) (( ))22 (( ))22 ( )( ) 22 22P( )P( )xx FrequencyFrequency P(x)P(x) xP(x)xP(x) (x(x--μμ)) (x(x-- μμ))22 (x(x--μμ))22P(x)P(x) xx22 xx22P(x)P(x)00 198,286 198,286 0.040.04 0.000.00 --1.31.3 1.651.65 0.060.06 00 0.000.0011 4,142,030 4,142,030 0.710.71 0.710.71 --0.30.3 0.080.08 0.060.06 11 0.710.7122 1,278,026 1,278,026 0.220.22 0.440.44 0.70.7 0.510.51 0.110.11 44 0.870.8733 174,110 174,110 0.030.03 0.090.09 1.71.7 2.942.94 0.090.09 99 0.270.2744 45,499 45,499 0.010.01 0.030.03 2.72.7 7.387.38 0.060.06 1616 0.120.1255 19,473 19,473 0.000.00 0.020.02 3.73.7 13.8113.81 0.050.05 2525 0.080.08

1616

66 5,170 5,170 0.000.00 0.010.01 4.74.7 22.2422.24 0.020.02 3636 0.030.0377 118 118 0.000.00 0.000.00 5.75.7 32.6732.67 0.000.00 4949 0.000.0088 -- 0.000.00 0.000.00 6.76.7 45.1045.10 0.000.00 6464 0.000.0099 897 897 0.000.00 0.000.00 7.77.7 59.5359.53 0.010.01 8181 0.010.01SumSum 5,863,609 5,863,609 1.001.00 μμ=1.28=1.28 32.1632.16 σσ22=0.45=0.45 2.102.10

Page 9: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 9

Chebyshev’sChebyshev’s TheoremTheorem

•• Helpful in understanding or interpretingHelpful in understanding or interpreting•• Helpful in understanding or interpreting Helpful in understanding or interpreting a value of a standard deviationa value of a standard deviation

•• Empirical rule applies only to data sets Empirical rule applies only to data sets with a bellwith a bell--shaped distributionshaped distribution

•• Chebyshev’sChebyshev’s theorem applies to ANY theorem applies to ANY

1717

yy ppppdata set, but its result are very data set, but its result are very approximateapproximate

Chebyshev’sChebyshev’s TheoremTheorem

•• The proportion (or fraction) of any set of The proportion (or fraction) of any set of data lying within data lying within KK standard deviations of standard deviations of the mean is always at least 1 the mean is always at least 1 –– 1/1/KK22

(where (where KK > 1)> 1)•• For For KK = 2 and = 2 and KK = 3, the results are as = 3, the results are as

follows:follows:

1818

follows:follows:At least ¾ (or 75%) off all values lie within 2 At least ¾ (or 75%) off all values lie within 2 standard deviations of the meanstandard deviations of the meanAt least 8/9 (or 89%) off all values lie within 3 At least 8/9 (or 89%) off all values lie within 3 standard deviations of the meanstandard deviations of the mean

Page 10: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 10

Cherbyshev’s Rule and Empirical Rule Cherbyshev’s Rule and Empirical Rule for a Discrete Random Variablefor a Discrete Random Variable

Let x be a discrete random variable with a probabilityLet x be a discrete random variable with a probabilityLet x be a discrete random variable with a probability Let x be a discrete random variable with a probability distribution distribution pp((xx), mean ), mean μμ, and standard deviation , and standard deviation σσ. Then, . Then, depending on the shape of depending on the shape of pp((xx), the following probability ), the following probability statements can be made:statements can be made:

Chebyshev’s RuleChebyshev’s RuleApplies to any probability Applies to any probability

distribution (eg: telephones distribution (eg: telephones i W hi t St t )i W hi t St t )

Empirical RuleEmpirical RuleApplies to probability distributions Applies to probability distributions

that are moundthat are mound--shaped and shaped and t i ( i l b f 4t i ( i l b f 4

1919

in Washington State)in Washington State) symmetric (eg: girls born of 4 symmetric (eg: girls born of 4 children)children)

P(P(μμ -- σσ < x < < x < μμ + + σσ)) ≥≥00 ≈≈.68.68P(P(μμ -- 22σσ < x < < x < μμ + 2+ 2σσ)) ≥≥3/43/4 ≈≈.95.95P(P(μμ -- 33σσ < x < < x < μμ + 3+ 3σσ)) ≥≥8/98/9 ≈≈1.001.00

Data TypesData Types

Data

Numerical Qualitative

2020

Discrete Continuous

Page 11: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 11

Continuous Random Continuous Random VariableVariable

•• A variable with many possible values atA variable with many possible values at•• A variable with many possible values at A variable with many possible values at all intervals all intervals

2121

Continuous Random Continuous Random Variable ExamplesVariable Examples

ExperimentExperiment RandomRandom PossiblePossibleExperimentExperiment RandomRandomVariableVariable

PossiblePossibleValuesValues

Weigh 100 PeopleWeigh 100 People WeightWeight 45.1, 78, ...45.1, 78, ...

Measure Part LifeMeasure Part Life HoursHours 900, 875.9, ...900, 875.9, ...

Ask Food SpendingAsk Food Spending SpendingSpending 54 12 4254 12 42

2222

Ask Food SpendingAsk Food Spending SpendingSpending 54.12, 42, ...54.12, 42, ...

Measure TimeMeasure TimeBetween ArrivalsBetween Arrivals

InterInter--ArrivalArrivalTimeTime

0, 1.3, 2.78, ...0, 1.3, 2.78, ...

Page 12: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 12

Continuous Probability Continuous Probability Density FunctionDensity Function

11 M th ti l F lM th ti l F l FrequencyFrequency1.1. Mathematical FormulaMathematical Formula

2.2. Shows All Values, Shows All Values, xx, & , & Frequencies, f(Frequencies, f(xx))

f(f(XX) Is ) Is NotNot ProbabilityProbability

3.3. PropertiesProperties

(Value, Frequency)(Value, Frequency)

FrequencyFrequency

f(x)f(x)

2323

Area under curve sums to 1Area under curve sums to 1

Can add up areas of function Can add up areas of function to get probability less than a to get probability less than a specific valuespecific value ValueValue

aa bbxx

Continuous Random Continuous Random Variable Probability Variable Probability

P b bilit I AP b bilit I A PP cc dd(( ))≤≤ ≤≤Probability Is Area Probability Is Area Under Curve!Under Curve!

PP cc xx dd(( ))≤≤ ≤≤

f(x)f(x)

2424 © 1984-1994 T/Maker Co.

Xc d

Page 13: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 13

Continuous Probability Continuous Probability Distribution ModelsDistribution Models

ContinuousProbabilityDistribution

2525

Uniform Normal Exponential

Importance of Importance of Normal DistributionNormal Distribution

11 Describes Many Random Processes orDescribes Many Random Processes or1.1. Describes Many Random Processes or Describes Many Random Processes or Continuous PhenomenaContinuous Phenomena

2.2. Can Be Used to Approximate Discrete Can Be Used to Approximate Discrete Probability DistributionsProbability Distributions

Example: BinomialExample: Binomial

2626

Example: BinomialExample: Binomial

3.3. Basis for Classical Statistical InferenceBasis for Classical Statistical Inference

Page 14: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 14

Normal DistributionNormal Distribution

1.1. ‘Bell‘Bell--Shaped’ & Shaped’ & f(X)SymmetricalSymmetrical

2.2. Mean, Median, Mean, Median, Mode Are EqualMode Are Equal

3.3. ‘Middle Spread’ ‘Middle Spread’ Is 1.33 Is 1.33 σσ

X

( )

2727

4.4. Random Variable Random Variable Has Infinite RangeHas Infinite Range

Mean Mean Median Median ModeMode

Normal Distribution Normal Distribution Useful PropertiesUseful Properties

•• About half of “weight” below About half of “weight” below (b(b

f(X)mean (because mean (because symmetrical)symmetrical)

•• About 68% of probability About 68% of probability within 1 standard deviation within 1 standard deviation of mean (at change in of mean (at change in curve)curve)

•• About 95% of probability About 95% of probability within 2 standard deviationswithin 2 standard deviations

X

( )

σμ +

σμ − σμ +σμ 2− σμ 2+ σμ 3+σμ 3−

2828

within 2 standard deviationswithin 2 standard deviations•• More than 99% of More than 99% of

probability within 3 standard probability within 3 standard deviations deviations

Mean Mean Median Median ModeMode

Page 15: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 15

Probability Probability Density FunctionDensity Function

2

211)(

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛−

σμx

f 2e2

)( ⎠⎝⎠⎝= σ

πσxf

xx == Value of Random Variable (Value of Random Variable (--∞∞ < < xx < < ∞∞))σσ == Population Standard DeviationPopulation Standard Deviationππ == 3.141593.14159

2929

ππ 3.141593.14159e = 2.71828e = 2.71828μμ == Mean of Random Variable xMean of Random Variable x

Don’t memorize this!Don’t memorize this!

NotationNotation

X is N(X is N(μμ σσ))X is N(X is N(μμ,,σσ))The random variable X has a normal The random variable X has a normal distribution (N) with mean distribution (N) with mean μμ and standard and standard deviation deviation σσ..

X is N(40,1)X is N(40,1)

3030

( , )( , )X is N(10,5)X is N(10,5)X is N(50,3)X is N(50,3)

Page 16: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 16

Effect of Varying Effect of Varying Parameters (Parameters (μμ & & σσ))

f(X)

CA

B

3131

X

Normal Distribution Normal Distribution ProbabilityProbability

?dProbability is Probability is

?? ?)()( dxxfdxcPd

c∫=≤≤

f(x)

area under area under curve!curve!

??

3232

c dx

Page 17: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 17

Infinite Number Infinite Number of Tablesof Tables

Normal distributions differ by Normal distributions differ by Each distribution would Each distribution would

f(X)

yymean & standard deviation.mean & standard deviation. require its own table.require its own table.

3333

XThat’s an That’s an infinite infinite number!number!

Standardize theStandardize theNormal DistributionNormal Distribution

Z X − μ Z is N(0,1)Z is N(0,1)

σ

Normal DistributionNormal Distribution

σ = 1

Z = μσ Standardized

Normal DistributionStandardized

Normal Distribution

( , )( , )

3434

XμOne table!One table!

μ = 0 Z

Page 18: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 18

Standardizing ExampleStandardizing Example

Z X − −μ 6 2 5 12.

σ = 10

Normal DistributionNormal Distribution

Z = = =σ 10

12.

σ = 1

Standardized Normal Distribution

Standardized Normal Distribution

3535Xμ= 5 6.2 Zμ= 0 .12

Obtaining Obtaining the Probabilitythe Probability

Standardized Normal Standardized Normal Standardized Normal Standardized Normal

σ = 1Z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438.0478.0478.0478.0478

.02.02

0.10.1 .0478

Probability Table (Portion)Probability Table (Portion)Probability Table (Portion)Probability Table (Portion)

3636

Zμ= 0 .120.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

0 8

ProbabilitiesProbabilitiesProbabilitiesProbabilitiesShaded area Shaded area exaggeratedexaggeratedShaded area Shaded area exaggeratedexaggerated

Page 19: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 19

ExampleExampleP(3.8 P(3.8 ≤≤ XX ≤≤ 5)5)

Z X= − = − = −μ 3 8 510

12. .

Z .00 .01 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

σ = 10

Normal Normal DistributionDistributionNormal Normal DistributionDistribution

σ 10

σ = 1

Standardized Standardized Normal DistributionNormal Distribution

Standardized Standardized Normal DistributionNormal Distribution

3737Xμ = 53.8 Zμ = 0-.12

.0478.0478

Shaded area exaggeratedShaded area exaggeratedShaded area exaggeratedShaded area exaggerated

ExampleExampleP(2.9 P(2.9 ≤≤ XX ≤≤ 7.1) 7.1)

Z X= − = − = −μ 2 9 5 21. .

Z .00 .01 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

σ = 10

Normal Normal DistributionDistributionNormal Normal DistributionDistribution

Z X= − = − =

σμ

σ

1071 5

1021. .

σ = 11664166416641664

Standardized Standardized Normal DistributionNormal Distribution

Standardized Standardized Normal DistributionNormal Distribution

383852.9 7.1 X 0-.21 Z.21

.1664.1664.1664.1664

.0832.0832.0832.0832

Shaded area exaggeratedShaded area exaggeratedShaded area exaggeratedShaded area exaggerated

Page 20: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 20

ExampleExampleP(P(XX ≥≥ 8)8)

Z X − −μ 8 5 30

Z .00 .01 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

σ = 10

Normal Normal DistributionDistributionNormal Normal DistributionDistribution

Standardized Standardized Normal DistributionNormal Distribution

Standardized Standardized Normal DistributionNormal Distribution

Z = = =μσ 10

30.

σ = 1

3939Xμ = 5 8 Zμ = 0 .30

.1179.1179.1179.1179

.5000.5000.3821.3821.3821.3821

Shaded area exaggeratedShaded area exaggeratedShaded area exaggeratedShaded area exaggerated

ExampleExampleP(7.1 P(7.1 ≤≤ XX ≤≤ 8)8)

Z X= − = − =μ 71 5 21. .

Z .00 .01 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

σ = 10

Normal Normal DistributionDistributionNormal Normal DistributionDistribution

Z X= − = − =

σμ

σ

108 510

30.

σ = 1

Standardized Standardized Normal DistributionNormal Distribution

Standardized Standardized Normal DistributionNormal Distribution

4040μ = 5 87.1 X μ = 0 .30 Z.21

.0832.0832

.1179.1179 .0347.0347.0347.0347

Shaded area exaggeratedShaded area exaggeratedShaded area exaggeratedShaded area exaggerated

Page 21: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 21

Travel Time and Travel Time and the Normal Distributionthe Normal Distribution

To help people plan their travel, WSDOT estimates To help people plan their travel, WSDOT estimates th t t i f S ttl t B ll t 5 40th t t i f S ttl t B ll t 5 40that average trip from Seattle to Bellevue at 5:40 pm that average trip from Seattle to Bellevue at 5:40 pm (at peak) takes 11 minutes and with a standard (at peak) takes 11 minutes and with a standard deviation of 10. They also believe this travel time deviation of 10. They also believe this travel time approximates a normal distribution.approximates a normal distribution.

Wh t ti f t i t k l th 27 i t ?Wh t ti f t i t k l th 27 i t ?

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What proportion of trips take less than 27 minutes?What proportion of trips take less than 27 minutes?

ProcessProcess

1.1. Draw a picture and write down the Draw a picture and write down the probability you need.probability you need.

2.2. Convert probability to standard scores.Convert probability to standard scores.3.3. Find cumulative probability in the table.Find cumulative probability in the table.

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Page 22: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 22

More Travel TimeMore Travel Time

Suppose we have only 10Suppose we have only 10--15 minutes to travel to Seattle15 minutes to travel to SeattleSuppose we have only 10Suppose we have only 10 15 minutes to travel to Seattle 15 minutes to travel to Seattle from Bellevue. What proportion of trips will make it in from Bellevue. What proportion of trips will make it in that time?that time?

( )4.1.0 <<−= ZP

( ) ⎟⎠⎞

⎜⎝⎛ −<<⎟

⎠⎞

⎜⎝⎛ −=<<

101115

1011101510 PZPXP

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( ) )4.(1.01 >−−<−= ZPZP( )

1952.)3446(.)4602(.1

)1554.5(.)0398.5(.1)4.(1.1

=−−=

−−−−=>−>−= ZPZP

Since normal curves Since normal curves are symmetrical:are symmetrical:

19.5% of trips will make it in between 10 and 15 minutes.19.5% of trips will make it in between 10 and 15 minutes.

Finding Z Values Finding Z Values for Known Probabilitiesfor Known Probabilities

Standardized Normal Standardized Normal Standardized Normal Standardized Normal What is Z given What is Z given What is Z given What is Z given

Z .00 0.2

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

σ = 1.1217.1217.1217.1217 .01.01

Probability Table (Portion)Probability Table (Portion)Probability Table (Portion)Probability Table (Portion)gg

P(Z) = .1217?P(Z) = .1217?gg

P(Z) = .1217?P(Z) = .1217?

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0.2 .0793 .0832 .0871

.1179 .1255

Zμ = 0 .310.30.3 .1217Shaded area Shaded area

exaggeratedexaggeratedShaded area Shaded area exaggeratedexaggerated

Page 23: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 23

Finding Finding ZZ Values Values for Known for Known

ProbabilitiesProbabilitiesStandardized Normal DistributionStandardized Normal DistributionStandardized Normal DistributionStandardized Normal Distribution

Z .00 .01 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

σ = 10Normal DistributionNormal DistributionNormal DistributionNormal Distribution

σ = 1

.1217.1217.1217.1217 .1217.1217.1217.1217

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Xμ = 5 ?

( )( ) 1.81031.5 =+=⋅+= σμ ZX

Zμ = 0 .31

Shaded areas exaggeratedShaded areas exaggeratedShaded areas exaggeratedShaded areas exaggerated

Travel Times Take 3Travel Times Take 3

How much time will the trip take 99% ofHow much time will the trip take 99% ofHow much time will the trip take 99% of How much time will the trip take 99% of the time?the time?

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Page 24: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 24

Finding Z Values Finding Z Values for Known Probabilitiesfor Known Probabilities

11 Write down probability statement and draw aWrite down probability statement and draw a1.1. Write down probability statement and draw a Write down probability statement and draw a picturepictureP(Z<____)=.99P(Z<____)=.99

2.2. Look up Z value in tableLook up Z value in tableP(Z<_____)=.99P(Z<_____)=.99

33 Convert Z value (SD units) to variable (X) byConvert Z value (SD units) to variable (X) by2.325

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3.3. Convert Z value (SD units) to variable (X) by Convert Z value (SD units) to variable (X) by using mean and SD.using mean and SD.X=X=μμ+Z+Zσσ so X=11+(_____)(10)=so X=11+(_____)(10)=2.325 34.25

So, the trip can be made 99% of the time in 34.25 minutes.

Assessing NormalityAssessing Normality

1.1. A histogram of the data is mound shaped and symmetrical A histogram of the data is mound shaped and symmetrical about the mean.about the mean.

2.2. Determine the percentage of measurements falling in each Determine the percentage of measurements falling in each of the intervals xof the intervals x±± s, xs, x±± 2s, and x2s, and x±± 3s. If the data are 3s. If the data are approximately normal, the percentages will be approximately normal, the percentages will be approximately equal to 68%, 95%, and 100% respectively.approximately equal to 68%, 95%, and 100% respectively.

3.3. Find the interquartile range, IQR, and standard deviation, s, Find the interquartile range, IQR, and standard deviation, s, for the sample, then calculate the ratio IQR/s. If the data for the sample, then calculate the ratio IQR/s. If the data

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ppare approximately normal, then IQR/S are approximately normal, then IQR/S ≈≈ 1.3.1.3.

4.4. Construct a normal probability plot for the data. If the data Construct a normal probability plot for the data. If the data are approximately normal, the points will fall are approximately normal, the points will fall (approximately) on a straight line.(approximately) on a straight line.

Page 25: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 25

Assessing Normality: Assessing Normality: Is Class Height Normally Distributed?Is Class Height Normally Distributed?

1.1. How does the histogram How does the histogram 7

gglook? look? SPSS can produce the line of SPSS can produce the line of the normal curve for you. In the normal curve for you. In SPSS select GRAPH, SPSS select GRAPH, HISTOGRAM. After you HISTOGRAM. After you choose the variable you want, choose the variable you want, click on the box “Display click on the box “Display

3

4

5

6

Frequency

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Normal Curve” and you’ll get Normal Curve” and you’ll get something that looks like this.something that looks like this.

60 62 64 66 68 70 72

Height 527 2005

0

1

2

Mean = 66.52Std. Dev. = 3.117N = 23

Assessing Normality: Assessing Normality: Is Class Height Normally Distributed?Is Class Height Normally Distributed?

Anticipated Anticipated ActualActual2 Compute the intervals:2 Compute the intervals:PercentPercent PercentPercent

xx±±ss [63.40,69.64][63.40,69.64] 68%68% 43%43%xx±±2s2s [60.29,72.75][60.29,72.75] 95%95% 96%96%xx±±3s3s [57.17,75.87][57.17,75.87] 100%100% 100%100%

2. Compute the intervals:2. Compute the intervals:

Height 527 2005

1 4.3 4.3 4.31 4.3 4.3 8.73 13.0 13.0 21.72 8.7 8.7 30.41 4.3 4.3 34.83 13 0 13 0 47 8

606263646566

ValidFrequency PercentValid Percent

CumulativePercent

5050

SPSS: ANALYZE,SPSS: ANALYZE,DESCRIPTIVE STATISTICS, DESCRIPTIVE STATISTICS, FREQUENCIESFREQUENCIES

3 13.0 13.0 47.82 8.7 8.7 56.52 8.7 8.7 65.25 21.7 21.7 87.01 4.3 4.3 91.31 4.3 4.3 95.71 4.3 4.3 100.0

23 100.0 100.0

66676869707172Total

Page 26: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 26

Assessing Normality: Assessing Normality: Is Class Height Normally Distributed?Is Class Height Normally Distributed?

3 D IQR/3 D IQR/ 1 3?1 3?Statistics

3. Does IQR/s3. Does IQR/s≈≈1.3?1.3?

IQR=69IQR=69--64=564=5IQR/s=5/3.117=1.6IQR/s=5/3.117=1.6

Height 527 200523

03.11764.0067.0069.00

ValidMissing

N

Std. Deviation255075

Percentiles

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SPSS: ANALYZE,SPSS: ANALYZE,DESCRIPTIVE STATISTICS, FREQUENCIES DESCRIPTIVE STATISTICS, FREQUENCIES then click on STATISTICS and choose the then click on STATISTICS and choose the ones you want.ones you want.

Assessing Normality: Assessing Normality: Is Class Height Normally Is Class Height Normally

Distributed?Distributed?

Normal Q-Q Plot of Height 527 2005

44 Wh t d th lWh t d th l

66

68

70

72

74

ected Normal Value

4. 4. What does the normal What does the normal probability plot look like?probability plot look like?

SPSS: Graphs>QSPSS: Graphs>Q--Q Test distribution is normal Q Test distribution is normal and click estimate distribution parameters from and click estimate distribution parameters from data.data.

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60 62 64 66 68 70 72 74

Observed Value

60

62

64Expe

Page 27: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 27

Exercise 1Exercise 1

•• Identify the given random variable as Identify the given random variable as being discrete or continuousbeing discrete or continuousa)a) The weight of the cola in a randomly The weight of the cola in a randomly

selected can.selected can.b)b) The cost of a randomly selected can of The cost of a randomly selected can of

Coke.Coke.

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c)c) The time it takes to fill a can of Pepsi.The time it takes to fill a can of Pepsi.

Exercise 2Exercise 2

•• Below is a case where a probability Below is a case where a probability di t ib ti i d ib d Fi d itdi t ib ti i d ib d Fi d it x P(x)distribution is described. Find its distribution is described. Find its mean and standard deviation.mean and standard deviation.

In a study of the In a study of the MicroSortMicroSort gendergender--selection method, couples in a control selection method, couples in a control group are not given a treatment, and they group are not given a treatment, and they

x P(x)0 0.1251 0.3752 0.3753 0.125

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g p g , yg p g , yeach have three children. The probability each have three children. The probability distribution for the number of girls is distribution for the number of girls is given.given.

Page 28: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 28

Exercise 3Exercise 3

•• Below is a case where a probability Below is a case where a probability di t ib ti i d ib d Fi d itdi t ib ti i d ib d Fi d it x P(x)distribution is described. Find its distribution is described. Find its mean and standard deviation.mean and standard deviation.

To settle a paternity suit, two different To settle a paternity suit, two different people are given blood tests. If people are given blood tests. If xx is the is the number having group A blood, then number having group A blood, then xx can can

x P(x)

5555

g g p ,g g p ,be 0, 1 or 2, and the corresponding be 0, 1 or 2, and the corresponding probabilities are 0.36, 0.48 and 0.16, probabilities are 0.36, 0.48 and 0.16, respectively.respectively.

Exercise 4Exercise 4

•• Let the random variable x Let the random variable x x P(x)represent the number of girls represent the number of girls

in a family of four children. in a family of four children. Construct a table describing Construct a table describing the probability distribution, the probability distribution, then find the mean and then find the mean and

x P(x)

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standard deviation.standard deviation.

Page 29: Chapter 2 Probability, Random Variables and Probability Distributions (1)

Student Lecture Notes 29

Exercise 5Exercise 5

•• Assume that the readings on the thermometers are Assume that the readings on the thermometers are normally distributed with a mean of 0ºnormally distributed with a mean of 0ºCC and a standardand a standardnormally distributed with a mean of 0normally distributed with a mean of 0 CC and a standard and a standard deviation of 1.00º deviation of 1.00º CC. A thermometer is randomly . A thermometer is randomly selected and tested. In each case, draw a sketch, and selected and tested. In each case, draw a sketch, and find the probability of each reading in degree.find the probability of each reading in degree.

a)a) Between 0 and 1.50Between 0 and 1.50b)b) Between Between --1.96 and 01.96 and 0

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c)c) Less than Less than --1.791.79d)d) Greater than 2.05Greater than 2.05e)e) Between 0.50 and 1.50Between 0.50 and 1.50f)f) PP((--1.96 < 1.96 < zz < 1.96)< 1.96)g)g) PP((zz > > --2.575)2.575)

Exercise 6Exercise 6

•• Assume that a test is designed to measure a person’s Assume that a test is designed to measure a person’s sense of humour and that scores on this test aresense of humour and that scores on this test aresense of humour and that scores on this test are sense of humour and that scores on this test are normally distributed with a mean of 10 and a standard normally distributed with a mean of 10 and a standard distribution of 2. draw a graph, find the relevant distribution of 2. draw a graph, find the relevant zz score, score, then find the indicated value.then find the indicated value.

a)a) Find the score separating the top 10% from the bottom Find the score separating the top 10% from the bottom 90%.90%.

b)b) Find the score separating the top 25% from the bottomFind the score separating the top 25% from the bottom

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b)b) Find the score separating the top 25% from the bottom Find the score separating the top 25% from the bottom 75%.75%.

c)c) Find the score separating the bottom 20% from the top Find the score separating the bottom 20% from the top 80%.80%.