Statistics final exam, 40 multiple choice questions.

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Chapter ContentsChapter Contents

7 1 Describing a Continuous Distribution7 1 Describing a Continuous Distribution7.1 Describing a Continuous Distribution7.1 Describing a Continuous Distribution 7.2 Uniform Continuous Distribution 7.2 Uniform Continuous Distribution 7 3 N l Di t ib ti7 3 N l Di t ib ti7.3 Normal Distribution7.3 Normal Distribution 7.4 Standard Normal Distribution7.4 Standard Normal Distribution 7.5 Normal Approximations7.5 Normal Approximations 7.6 Exponential Distribution7.6 Exponential Distributionpp 7.7 Triangular Distribution (Optional)7.7 Triangular Distribution (Optional)

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Continuous Probability DistributionsContinuous Probability Distributions

Chapter Learning Objectives (LO’s)Chapter Learning Objectives (LO’s)

Chapter Learning Objectives (LO s)Chapter Learning Objectives (LO s)

LO7LO7 11LO7LO7--11: : Define a continuous random variable.Define a continuous random variable. LO7LO7--2: 2: Calculate uniform probabilities.Calculate uniform probabilities. LO7LO7--3: 3: Know the form and parameters of the normal distribution.Know the form and parameters of the normal distribution. LO7LO7--4:4: Find the normal probability for given z or x using tables or Excel.Find the normal probability for given z or x using tables or Excel.LO7LO7 4:4: Find the normal probability for given z or x using tables or Excel.Find the normal probability for given z or x using tables or Excel. LO7LO7--5:5: Solve for z or x for a given normal probability using tables or Excel.Solve for z or x for a given normal probability using tables or Excel.

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Continuous Probability DistributionsContinuous Probability Distributions

Chapter Learning Objectives (LO’s)Chapter Learning Objectives (LO’s)

Chapter Learning Objectives (LO s)Chapter Learning Objectives (LO s)

LO6LO6LO6:LO6: Use the normal approximation to a binomial or a PoissonUse the normal approximation to a binomial or a Poisson distribution.distribution.

LO7:LO7: Find the exponential probability for a given xFind the exponential probability for a given x.. LO8: LO8: Solve for x for given exponential probability.Solve for x for given exponential probability. LO9:LO9: Use the triangular distribution for “whatUse the triangular distribution for “what--if” analysis (optional).if” analysis (optional).

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7.1 Describing a Continuous Distribution7.1 Describing a Continuous DistributionLO7LO7--11 7

LO7LO7--1: 1: Define a continuous random variable.Define a continuous random variable.

Di t V i blDi t V i bl h l fh l f XX h it b bilith it b bilit PP((XX))

Events as IntervalsEvents as Intervals

•• Discrete VariableDiscrete Variable –– each value of each value of XX has its own probability has its own probability PP((XX).). •• Continuous VariableContinuous Variable –– events are events are intervalsintervals and probabilities are and probabilities are

areas under continuous curves A single point has no probabilityareas under continuous curves A single point has no probabilityareas under continuous curves. A single point has no probability.areas under continuous curves. A single point has no probability.

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7.1 Describing a Continuous Distribution7.1 Describing a Continuous DistributionLO7LO7--11

Continuous PDF’s:

PDF PDF –– Probability Density FunctionProbability Density Function

Continuous PDF’s: • Denoted f(x) • Must be nonnegative• Must be nonnegative • Total area under

curve = 1 • Mean, variance and

shape depend on the PDF parametersthe PDF parameters

• Reveals the shape of the distribution

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7.1 Describing a Continuous Distribution7.1 Describing a Continuous DistributionLO7LO7--11

CDF CDF –– Cumulative Distribution FunctionCumulative Distribution Function

Continuous CDF’s: • Denoted F(x) • Shows P(X ≤ x), the

cumulative proportion fof scores

• Useful for finding probabilitiesprobabilities

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7.1 Describing a Continuous Distribution7.1 Describing a Continuous DistributionLO7LO7--11

Probabilities as AreasProbabilities as Areas

Continuous probability functions:Continuous probability functions: •• Unlike discrete Unlike discrete

distributions, the distributions, the probability at any probability at any single point = 0single point = 0single point = 0.single point = 0.

•• The entire area under The entire area under any PDF, by definition, any PDF, by definition, y , y ,y , y , is set to 1.is set to 1.

•• Mean is the balanceMean is the balance point of the distribution.point of the distribution.

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7.1 Describing a Continuous Distribution7.1 Describing a Continuous DistributionLO7LO7--11

Expected Value and VarianceExpected Value and Variance

The mean and variance of a continuous random variable are analogous to E(X) and Var(X ) for a discrete random variable, Here the integral sign replaces the summation sign. Calculus is required to compute the integrals. p g q p g

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7.2 Uniform Continuous Distribution7.2 Uniform Continuous DistributionLO7LO7--22 7

LO7LO7--2: 2: Calculate uniform probabilities.Calculate uniform probabilities.

Characteristics of the Uniform Characteristics of the Uniform DistributionDistribution

If If XX is a random variable that is is a random variable that is uniformly distributed between uniformly distributed between aa and and bb, its PDF has , its PDF has constant height.constant height.

• Denoted U(a, b) • Area =

base x height =base x height (b-a) x 1/(b-a) = 1

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7.2 Uniform Continuous Distribution7.2 Uniform Continuous DistributionLO7LO7--22

Characteristics of the Uniform DistributionCharacteristics of the Uniform Distribution

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7.2 Uniform Continuous Distribution7.2 Uniform Continuous DistributionLO7LO7--22

Example: Anesthesia EffectivenessExample: Anesthesia Effectiveness •• An oral surgeon injects a painkiller prior to extracting a tooth Given theAn oral surgeon injects a painkiller prior to extracting a tooth Given the

An oral surgeon injects a painkiller prior to extracting a tooth. Given the An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutesbetween 15 minutes and 30 minutesbetween 15 minutes and 30 minutes.between 15 minutes and 30 minutes.

•• XX is is UU(15, 30)(15, 30) •• aa = 15,= 15, bb = 30, find the mean and standard deviation.= 30, find the mean and standard deviation.aa 15, 15, bb 30, find the mean and standard deviation. 30, find the mean and standard deviation.

•• Find the probability that the effectiveness anesthetic takes between Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes.20 and 25 minutes.

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20 and 25 minutes.20 and 25 minutes.

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7.2 Uniform Continuous Distribution7.2 Uniform Continuous DistributionLO7LO7--22 7

Example: Anesthesia EffectivenessExample: Anesthesia Effectiveness

PP(20 < (20 < XX < 25) = (25 < 25) = (25 –– 20)/(30 20)/(30 –– 15) = 5/15 = 0.3333 = 33.33% 15) = 5/15 = 0.3333 = 33.33%

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7.3 Normal Distribution7.3 Normal DistributionLO7LO7--33 7

LO7LO7--3: 3: Know the form and parameters of the normal distribution.Know the form and parameters of the normal distribution.

Characteristics of the Normal DistributionCharacteristics of the Normal Distribution

• Normal or Gaussian (or bell shaped) distribution was named for German mathematician Karl Gauss (1777 – 1855).

• Defined by two parameters, µ and . • Denoted N(µ )• Denoted N(µ, ). • Domain is –  < X < +  (continuous scale). • Almost all (99.7%) of the area under the normal curve is included in the ( )

range µ – 3 < X < µ + 3. • Symmetric and unimodal about the mean.

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7.3 Normal Distribution7.3 Normal DistributionLO7LO7--33

Characteristics of the Normal DistributionCharacteristics of the Normal Distribution

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7.3 Normal Distribution7.3 Normal DistributionLO7LO7--33 7

Characteristics of the Normal DistributionCharacteristics of the Normal Distribution

•• Normal PDF Normal PDF ff((xx) reaches a maximum at ) reaches a maximum at µµ and has points of inflection at and has points of inflection at µµ ±± 

Bell-shaped curve

NOTE:NOTE: All normal All normal distributionsdistributionsdistributions distributions have the same have the same shape but differshape but differshape but differ shape but differ in the axis scales.in the axis scales.

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7.3 Normal Distribution7.3 Normal DistributionLO7LO7--33 7

Characteristics of the Normal DistributionCharacteristics of the Normal Distribution

•• Normal CDF Normal CDF

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--33

Characteristics of the Standard Normal DistributionCharacteristics of the Standard Normal Distribution

• Since for every value of µ and , there is a different normal distribution, we transform a normal y µ , , random variable to a standard normal distribution with µ = 0 and  = 1 using the formula.

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LO7LO7--33 7.4 Standard Normal Distribution7.4 Standard Normal Distribution

Characteristics of the Standard NormalCharacteristics of the Standard Normal St d d l PDF f( ) h i t 0 d h

• Standard normal PDF f(x) reaches a maximum at z = 0 and has points of inflection at +1.

•• Shape is unaffected by Shape is unaffected by the transformationthe transformationthe transformation. the transformation. It is still a bellIt is still a bell--shaped shaped curve.curve.

Figure 7 11

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Figure 7.11

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LO7LO7--33 7.4 Standard Normal Distribution7.4 Standard Normal Distribution

Characteristics of the Standard NormalCharacteristics of the Standard Normal •• Standard normal CDFStandard normal CDF

A common scale•• Standard normal CDFStandard normal CDF • A common scale from -3 to +3 is used.

• Entire area under the curve is unity.

• The probability of an event P(z < Z < z )event P(z1 < Z < z2) is a definite integral of f(z).

• However, standard normal tables or Excel functions can be used to find the desired probabilities.

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LO7LO7--33 7.4 Standard Normal Distribution7.4 Standard Normal Distribution

Normal Areas from Appendix CNormal Areas from Appendix C--11 • Appendix C-1 allows you to find the area under the curve

• Appendix C-1 allows you to find the area under the curve from 0 to z.

• For example, find P(0 < Z < 1.96):p , ( )

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LO7LO7--33 7.4 Standard Normal Distribution7.4 Standard Normal Distribution

Normal Areas from Appendix CNormal Areas from Appendix C--11 •• Now findNow find PP((--1 96 <1 96 < ZZ < 1 96)< 1 96)

•• Now find Now find PP((--1.96 < 1.96 < ZZ < 1.96).< 1.96). •• Due to symmetry, Due to symmetry, PP((--1.96 < 1.96 < ZZ) is the same as ) is the same as PP((ZZ < 1.96).< 1.96).

• So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500 or 95% of the area under the curve.

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LO7LO7--33 7.4 Standard Normal Distribution7.4 Standard Normal Distribution

Basis for the Empirical RuleBasis for the Empirical Rule

• Approximately 68% of the area under the curve is between + 1 • Approximately 95% of the area under the curve is between + 2 • Approximately 99.7% of the area under the curve is between + 3

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--44

N l A f A di CN l A f A di C 22

LO7LO7--4: 4: Find the normal probability for given z or x using tables or Excel.Find the normal probability for given z or x using tables or Excel.

Normal Areas from Appendix CNormal Areas from Appendix C--22 •• Appendix CAppendix C--2 allows you to find the area under the curve from the left of 2 allows you to find the area under the curve from the left of

zz (similar to Excel)(similar to Excel)z z (similar to Excel).(similar to Excel).

•• For example, For example,

PP((ZZ < < --1.96)1.96)PP((ZZ < 1.96< 1.96) PP((--1.96 < 1.96 < ZZ < 1.96)< 1.96)

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--44

Normal Areas from Appendices CNormal Areas from Appendices C--1 or C1 or C--22 •• Appendices CAppendices C--1 and C1 and C--2 yield identical results2 yield identical results

•• Appendices CAppendices C--1 and C1 and C--2 yield identical results.2 yield identical results. •• Use whichever table is easiest.Use whichever table is easiest.

Finding Finding zz for a Given Areafor a Given Area

• Appendices C-1 and C-2 can be used to find the z-value corresponding to a given probability. For e ample hat al e defines the top 1% of a normal• For example, what z-value defines the top 1% of a normal distribution?

• This implies that 49% of the area lies between 0 and z whichThis implies that 49% of the area lies between 0 and z which gives z = 2.33 by looking for an area of 0.4900 in Appendix C-1.

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--44

Finding Areas by using Standardized VariablesFinding Areas by using Standardized Variables

•• Suppose John took an economics exam and scored 86 points The classSuppose John took an economics exam and scored 86 points The classSuppose John took an economics exam and scored 86 points. The class Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is That is, what is PP((XX < 86) where X represents the exam scores?< 86) where X represents the exam scores?

•• So John’s score is 1.57 standard deviations about the mean. So John’s score is 1.57 standard deviations about the mean.

•• PP((XX < 86) = < 86) = PP((ZZ < 1.57) = .9418 (from Appendix C< 1.57) = .9418 (from Appendix C--2)2)

•• So John is approximately in the 94So John is approximately in the 94thth percentilepercentile•• So, John is approximately in the 94So, John is approximately in the 94thth percentilepercentile..

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--44

•• Finding Areas by using Standardized VariablesFinding Areas by using Standardized Variables

NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these probabilities directly.

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--55 7

LO7LO7--5: 5: Solve for z or x for a normal probability using tables or Excel.Solve for z or x for a normal probability using tables or Excel.

•• Inverse NormalInverse Normal • How can we find the various normal percentiles (5th 10th 25th 75th• How can we find the various normal percentiles (5th, 10th, 25th, 75th,

90th, 95th, etc.) known as the inverse normal? That is, how can we find X for a given area? We simply turn the standardizing transformation around:

Solving for x in z = (x − μ)/ gives x = μ + zσSolving for x in z = (x μ)/ gives x = μ + zσ

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--55 7

•• Inverse NormalInverse Normal

• For example, suppose that John’s economics professor has decided that any student who scores below the 10th percentile must retake the exam.

• The exam scores are normal with μ = 75 and σ = 7. • What is the score that would require a student to retake the exam? • We need to find the value of x that satisfies P(X < x) = 10We need to find the value of x that satisfies P(X < x) .10. • The z-score for with the 10th percentile is z = −1.28.

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--55

Inverse NormalInverse Normal

•• Inverse NormalInverse Normal

• The steps to solve the problem are:The steps to solve the problem are:

• Use Appendix C or Excel to find z = −1.28 to satisfy P(Z < −1.28) = .10. • Substitute the given information into z = (x μ)/σ to get• Substitute the given information into z = (x − μ)/σ to get −1.28 = (x − 75)/7

• Solve for x to get x = 75 − (1.28)(7) = 66.03 (or 66 after rounding) S• Students who score below 66 points on the economics exam will be required to retake the exam.

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7.4 Standard Normal Distribution7.4 Standard Normal DistributionLO7LO7--55

•• Inverse NormalInverse Normal

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66 7

LO7LO7--6: 6: Use the normal approximation to a binomial or a Poisson.Use the normal approximation to a binomial or a Poisson.

Normal Approximation to the BinomialNormal Approximation to the Binomial Bi i l b biliti diffi lt t l l t hBi i l b biliti diffi lt t l l t h i li l•• Binomial probabilities are difficult to calculate when Binomial probabilities are difficult to calculate when nn is large.is large.

•• Use a normal approximation to the binomial distribution.Use a normal approximation to the binomial distribution. •• AsAs nn becomes large, the binomial bars become smaller and continuity isbecomes large, the binomial bars become smaller and continuity isAs As nn becomes large, the binomial bars become smaller and continuity is becomes large, the binomial bars become smaller and continuity is

approached.approached.

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66

Normal Approximation to the BinomialNormal Approximation to the Binomial

• Rule of thumb: when n ≥ 10 and n(1- ) ≥ 10, then it is appropriate to use the normal approximation to the binomial distribution.

• In this case the mean and standard deviation for the binomial distribution• In this case, the mean and standard deviation for the binomial distribution will be equal to the normal µ and , respectively.

Example Coin FlipsExample Coin Flips

If t fli i 32 ti d 50 th• If we were to flip a coin n = 32 times and  = .50, are the requirements for a normal approximation to the binomial distribution met?

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66

Example Coin FlipsExample Coin Flips

• n = 32 x .50 = 16 n(1- ) = 32 x (1 - .50) = 16

• So a normal approximation can be usedSo, a normal approximation can be used. • When translating a discrete scale into a continuous scale,

care must be taken about individual points. • For example, find the probability of more than 17 heads in

32 flips of a fair coin. • This can be written as P(X  18). ( ) • However, “more than 17” actually falls between 17 and 18

on a discrete scale.

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66

Example Coin FlipsExample Coin Flips •• Since the cutoff point for “more than 17” is halfway between 17 and 18 weSince the cutoff point for “more than 17” is halfway between 17 and 18 we

Since the cutoff point for more than 17 is halfway between 17 and 18, we Since the cutoff point for more than 17 is halfway between 17 and 18, we add 0.5 to the lower limit and find add 0.5 to the lower limit and find PP((XX > 17.5).> 17.5).

•• This addition to This addition to XX is called the is called the Continuity CorrectionContinuity Correction.. •• At this point, the problem can be completed as any normal distribution At this point, the problem can be completed as any normal distribution

problem.problem.

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Example Coin FlipsExample Coin Flips

P(X > 17) P(X ≥ 18) P(X ≥ 17 5)P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) = 0.2981

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66

Normal Approximation to the PoissonNormal Approximation to the Poisson • The normal approximation to the Poisson distribution works best

• The normal approximation to the Poisson distribution works best when  is large (e.g., when  exceeds the values in Appendix B).

• Set the normal µ and  equal to the mean and standard deviation µ q for the Poisson distribution.

Example Utility BillsExample Utility Billsp yp y • On Wednesday between 10A.M. and noon customer billing

inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. What is the probability of receiving more than 50 calls in an hour?

•  = 42 which is too big to use the Poisson table

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•  = 42 which is too big to use the Poisson table. • Use the normal approximation with  = 42 and  = 6.48074

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7.5 Normal Approximations7.5 Normal ApproximationsLO7LO7--66

Example Utility BillsExample Utility Bills

•• To find To find PP((XX > 50) calls, use the continuity> 50) calls, use the continuity--corrected cutoff point halfway corrected cutoff point halfway between 50 and 51 (i.e., between 50 and 51 (i.e., XX = 50.5).= 50.5).

•• At this point the problem can be completed as any normal distributionAt this point the problem can be completed as any normal distributionAt this point, the problem can be completed as any normal distribution At this point, the problem can be completed as any normal distribution problem.problem.

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--77 7

LO7LO7--7: 7: Find the exponential probability for a given xFind the exponential probability for a given x..

Characteristics of the Exponential DistributionCharacteristics of the Exponential Distribution

If t it f ti f ll P i di t ib ti th ti til thIf t it f ti f ll P i di t ib ti th ti til th•• If events per unit of time follow a Poisson distribution, the time until the If events per unit of time follow a Poisson distribution, the time until the next event follows the next event follows the Exponential distribution.Exponential distribution.

•• The time until the next event is a continuous variable.The time until the next event is a continuous variable.

NOTE HereNOTE HereNOTE: Here NOTE: Here we will findwe will find probabilitiesprobabilities > x or ≤ x.> x or ≤ x.

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--77

Characteristics of the Exponential DistributionCharacteristics of the Exponential Distribution

Probability of waiting more than xProbability of waiting less than or equal to x

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--77

Example Customer Waiting TimeExample Customer Waiting Time

• Between 2P.M. and 4P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls

iper minute. • What is the probability of waiting more than 30 seconds (i.e., 0.50

minutes) for the next call?minutes) for the next call? • Set  = 2.2 events/min and x = 0.50 min • P(X > 0 50) = e–x = e–(2.2)(0.5) = 3329P(X > 0.50) = e = e ( )( ) = .3329

or 33.29% chance of waiting more than 30 seconds for the next call.

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--77

Example Customer Waiting TimeExample Customer Waiting Time

P(X > 0.50) P(X ≤ 0.50)

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--88 7

LO7LO7--8: 8: Solve for Solve for x for given x for given exponential probability.exponential probability.

Inverse ExponentialInverse Exponential

If th i l t i 2 2 ll i t t th 90If th i l t i 2 2 ll i t t th 90thth•• If the mean arrival rate is 2.2 calls per minute, we want the 90If the mean arrival rate is 2.2 calls per minute, we want the 90thth percentile for waiting time (the top 10% of waiting time).percentile for waiting time (the top 10% of waiting time).

•• Find theFind the xx--valuevalueFind the Find the xx--value value that defines the that defines the upper 10%.upper 10%.

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--88

Inverse ExponentialInverse Exponential

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7.6 Exponential Distribution7.6 Exponential DistributionLO7LO7--88 7

Mean Time Between EventsMean Time Between Events

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7.7 Triangular Distribution7.7 Triangular DistributionLO7LO7--99

Ch t i ti f th T i l Di t ib tiCh t i ti f th T i l Di t ib ti

LO7LO7--9: 9: Use the triangular distribution for “whatUse the triangular distribution for “what--if” analysis (optional).if” analysis (optional).

Characteristics of the Triangular DistributionCharacteristics of the Triangular Distribution

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7.7 Triangular Distribution7.7 Triangular DistributionLO7LO7--99

Characteristics of the Triangular DistributionCharacteristics of the Triangular Distribution

The triang lar distrib tion is a a of thinking abo t ariation that• The triangular distribution is a way of thinking about variation that corresponds rather well to what-if analysis in business.

• It is not surprising that business analysts are attracted to the triangular model.

• Its finite range and simple form are more understandable than a normal distribution.

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7.7 Triangular Distribution7.7 Triangular DistributionLO7LO7--99

Characteristics of the Triangular DistributionCharacteristics of the Triangular Distribution

• It is more versatile than a normal, because it can be skewed in either direction. Y t it h f th i ti f l h di ti t d• Yet it has some of the nice properties of a normal, such as a distinct mode.

• The triangular model is especially handy for what-if analysis when the business case depends on predicting a stochastic variable (e.g., the price of a raw material, an interest rate, a sales volume).

• If the analyst can anticipate the range (a to c) and most likely value (b), it will be possible to calculate probabilities of various outcomes. p p

• Many times, such distributions will be skewed, so a normal wouldn’t be much help.

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