continuous random variables. l. wang, department of statistics university of south carolina; slide 2...

Continuous Random Continuous Random VariablesVariables

L. Wang, Department of StatisticsL. Wang, Department of Statistics

University of South Carolina; Slide University of South Carolina; Slide 22

Continuous Random Continuous Random VariableVariable

A A continuous random variablecontinuous random variable is is one for which the outcome can be any one for which the outcome can be any value in an interval of the real number value in an interval of the real number line.line.

Usually a measurement.Usually a measurement. ExamplesExamples

– Let Y = length in mmLet Y = length in mm– Let Y = time in secondsLet Y = time in seconds– Let Y = temperature in Let Y = temperature in ºCºC



Continuous Random Continuous Random VariableVariable

We don’t calculate We don’t calculate P(Y = y)P(Y = y), we , we calculate calculate P(a P(a << Y Y << b) b), where , where aa and and bb are real numbers.are real numbers.

For a continuous random variableFor a continuous random variable

P(Y = y) = 0P(Y = y) = 0..



Continuous Random Continuous Random VariablesVariables

The The probability density function (pdf)probability density function (pdf) when plotted against the possible values when plotted against the possible values of Y forms a curve. The area under an of Y forms a curve. The area under an interval of the curve is equal to the interval of the curve is equal to the probability that Y is in that interval.probability that Y is in that interval.

f(y)

Y

0.40

a b



The entire area under a The entire area under a probability density curve for a probability density curve for a continuous random variablecontinuous random variable

A.A. Is always greater than 1.Is always greater than 1.

B.B. Is always less than 1.Is always less than 1.

C.C. Is always equal to 1.Is always equal to 1.

D.D. Is Is undeterminableundeterminable..



Properties of a Probability Density Function Properties of a Probability Density Function (pdf)(pdf)

1)1) f(y) f(y) >> 0 0 for all possible intervals of y. for all possible intervals of y.

2)2)

3)3) If If yy00 is a specific value of interest, then is a specific value of interest, then the cumulative distribution function (cdf) the cumulative distribution function (cdf) isis

4)4) If If yy11 and and yy22 are specific values of interest, are specific values of interest, thenthen

1)( dyyf

0

)()()( 00

y

dyyfyYPyF

2

1

)()()()( 1221

y

y

yFyFdyyfyYyP



Grams of lead per liter of Grams of lead per liter of gasoline has the probability gasoline has the probability density function: f(y) = 12.5y - density function: f(y) = 12.5y - 1.251.25for 0.1 for 0.1 << y y << 0.5 0.5What is the probability that the What is the probability that the next liter of gasoline has less next liter of gasoline has less than 0.3 grams of lead?than 0.3 grams of lead?



Suppose a random variable Y Suppose a random variable Y has the following probability has the following probability density function: f(y) = y if density function: f(y) = y if 0<y<10<y<1 2-y if 1 2-y if 1 << y<2 y<2 0 if 2 0 if 2 << y. y.Find the complete form of the Find the complete form of the cumulative distribution function cumulative distribution function F(y) for any real value y. F(y) for any real value y.



Expected Value for a Expected Value for a Continuous Random VariableContinuous Random Variable

Recall Expected Value for a discrete Recall Expected Value for a discrete random variable:random variable:

Expected value for a continuous random Expected value for a continuous random variable:variable:

)()( ypyYE

dyyyfYE )()(



Variance for Continuous Variance for Continuous Random VariableRandom Variable

)()()( 22 ypyYVar

dyyfyYVar )()()( 22

Recall: Variance for a discrete random variable:

Variance for a continuous random variable:



Difference between DiscreteDifference between Discreteand continuous random and continuous random

variablesvariables•Possible values that can be assumed•Probability distribution function•Cumulative distribution function•Expected value•Variance



Times Between Industrial Times Between Industrial AccidentsAccidents

The times between accidents for a The times between accidents for a 10-year period at a DuPont facility 10-year period at a DuPont facility can be modeled by the can be modeled by the exponential exponential distributiondistribution..

0 and 0 )( yeyf y

where λ is the accident rate (the expected number of accidents per day in this case)



Example of time between Example of time between accidentsaccidents

Let Y = the number of days between two Let Y = the number of days between two accidents.accidents.

TimeTime 12 days 35 days 5 days12 days 35 days 5 days

● ● ● ● ● ● ● ● ● ●

Accident Accident Accident Accident Accident Accident #1#1 #2 #2 #3 #3



Times Between Industrial Times Between Industrial AccidentsAccidents

Suppose in a 1000 day period there Suppose in a 1000 day period there were 50 accidents.were 50 accidents.

or

λλ = 50/1000 = 0.05 accidents per = 50/1000 = 0.05 accidents per dayday

1/λ = 1000/50 = 20 days between accidents



What is the probability that this What is the probability that this facility will go less than 10 days facility will go less than 10 days between the next two accidents?between the next two accidents?

Probability Density Function

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40 50 60 70 80 90 100

Y = Time between accidents

f(y)

?

f(y) = 0.05e-0.05y



10

0

05.005.0)10()10( dyeFYP y

uu edue

39.0|)10( 100

05.0 yeF

Probability Density Function

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40 50 60 70 80 90 100

Y = Time between accidentsf(

y)

?

Recall:



In General…In General…

y

tdteyYP0

)(

yyt eeyFyYP 1|)()( 0

yeyFyYP )(1)(



Exponential DistributionExponential DistributionProbability Density Function

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40 50 60 70 80 90 100

Y = Time between accidents

f(y)

ye 1 ye



If the time to failure for an electrical If the time to failure for an electrical component follows an exponential component follows an exponential distribution with a mean time to distribution with a mean time to failure of 1000 hours, what is the failure of 1000 hours, what is the probability that a randomly chosen probability that a randomly chosen component will fail before 750 hours?component will fail before 750 hours?

Hint: λ is the failure rate (expected number of failures per hour).



Mean and Variance for an Mean and Variance for an Exponential Random VariableExponential Random Variable

1

)(0

dyeyYE y

20

222 11

)(

dyeyYVar y

Note: Mean = Standard Deviation



The time between accidents at a The time between accidents at a factory follows an exponential factory follows an exponential distribution with a historical distribution with a historical average of 1 accident every 900 average of 1 accident every 900 days. What is the probability that days. What is the probability that that there will be more than 1200 that there will be more than 1200 days between the next two days between the next two

accidents?accidents?



If the time between accidents If the time between accidents follows an exponential distribution follows an exponential distribution with a mean of 900 days, what is with a mean of 900 days, what is the probability that there will be the probability that there will be less than 900 days between the less than 900 days between the next two accidents?next two accidents?



Relationship between Relationship between Exponential & Poisson Exponential & Poisson

DistributionsDistributions Recall that the Poisson distribution is Recall that the Poisson distribution is

used to compute the probability of a used to compute the probability of a specific number of events occurring specific number of events occurring in a particular interval of time or in a particular interval of time or space.space.

Instead of the number of events Instead of the number of events being the random variable, consider being the random variable, consider the time or space between events as the time or space between events as the random variable.the random variable.



Relationship between Relationship between Exponential & PoissonExponential & Poisson

Exponential distribution models time (or space) between Poisson events.

TIME



Exponential or Poisson Exponential or Poisson Distribution?Distribution?

We model the number of industrial We model the number of industrial accidents occurring in one year.accidents occurring in one year.

We model the length of time between two We model the length of time between two industrial accidents (assuming an accident industrial accidents (assuming an accident occurring is a Poisson event).occurring is a Poisson event).

We model the time between radioactive We model the time between radioactive particles passing by a counter (assuming a particles passing by a counter (assuming a particle passing by is a Poisson event). particle passing by is a Poisson event).

We model the number of radioactive We model the number of radioactive particles passing by a counter in one hour particles passing by a counter in one hour



Recall: For a Poisson Recall: For a Poisson DistributionDistribution

!

)()()(

y

etypyYP

ty

y = 0,1,2,…

where λ is the mean number of events per base unit of time or space and t is the number of base units inspected.

The probability that no events occur in a span of time (or space) is:

ttty

eet

y

etp

!0

)(

!

)()0(

0



Now let T = the time (or space) Now let T = the time (or space) until the next Poisson event.until the next Poisson event.

tetTP )(

In other words, the probability that the probability that the length of time (or space) until the length of time (or space) until the next event is greater than the next event is greater than some given time (or space), t, is some given time (or space), t, is the same as the probability that no the same as the probability that no events will occur in time (or space) events will occur in time (or space) t.t.



Radioactive Particles Radioactive Particles The arrival of radioactive particles at The arrival of radioactive particles at

a counter are Poisson events. So the a counter are Poisson events. So the number of particles in an interval of number of particles in an interval of time follows a Poisson distribution. time follows a Poisson distribution. Suppose we average 2 particles per Suppose we average 2 particles per millisecond.millisecond.

What is the probability that no What is the probability that no particles will pass the counter in the particles will pass the counter in the next 3 milliseconds?next 3 milliseconds?

What is the probability that more than What is the probability that more than 3 millisecond will elapse before the 3 millisecond will elapse before the next particle passes?next particle passes?



Machine FailuresMachine Failures

If the number of machine failures in a If the number of machine failures in a given interval of time follows a given interval of time follows a Poisson distribution with an average Poisson distribution with an average of 1 failure per 1000 hours, what is of 1 failure per 1000 hours, what is the probability that there will be no the probability that there will be no failures during the next 2000 hours?failures during the next 2000 hours?

What is the probability that the time What is the probability that the time until the next failure is more than until the next failure is more than 2000 hours?2000 hours?



Number of failures in an interval of time Number of failures in an interval of time follows a Poisson distribution. If the mean follows a Poisson distribution. If the mean time to failure is 1000 hours, what is the time to failure is 1000 hours, what is the probability that more than 2500 hours probability that more than 2500 hours will pass before the next failure occurs?will pass before the next failure occurs?

A. e-4

B. 1 – e-4

C. e-2.5

D. 1 – e-2.5



If ten of these components are used in If ten of these components are used in different devices that run different devices that run independently, what is the probability independently, what is the probability that at least one will still be operating that at least one will still be operating at 2500 hours?at 2500 hours?

What about he probability that exact 3 What about he probability that exact 3 of them will be still operating after of them will be still operating after 2500 hours?2500 hours?

Challenging questions



Normal DistributionNormal Distribution

f(y) =f(y) =

E[Y] = μ and Var[Y] = σE[Y] = μ and Var[Y] = σ22

f(y)

y

ye y , 2

1 22 2/)(



Normal DistributionNormal Distribution CharacteristicsCharacteristics

– Bell-shaped curveBell-shaped curve– -- << y y << + +– μ determines distribution location μ determines distribution location

and is the highest point on curveand is the highest point on curve– Curve is symmetric about Curve is symmetric about μμ – σ determines distribution spreadσ determines distribution spread– Curve has its points of inflection at Curve has its points of inflection at μ μ

++ σ σ




-4 -3 -2 -1 0 1 2 3 4

σ

σσ

σ

μ

σ




-4 -3 -2 -1 0 1 2 3 4 5 6 7 8

N(μ = 0, σ = 1)

f(y)

y

N(μ = 5, σ = 1)




-4 -3 -2 -1 0 1 2 3 4

N(μ = 0,σ = 0.5)

f(y)

y

N(μ = 0,σ = 1)




-4 -3 -2 -1 0 1 2 3 4 5 6 7 8

f(y)

y

N(μ = 0, σ = 1)N(μ = 5, σ = 0.5)



68-95-99.7 Rule68-95-99.7 Rule

-4 -3 -2 -1 0 1 2 3 4

μ + 1σ covers approximately 68%

μ + 2σ covers approximately 95%

μ + 3σ covers approximately99.7%

0.680.95

0.997

µ µ+1σ µ+2σ µ+3σµ-1σµ-2σµ-3σ



Earthquakes in a California Earthquakes in a California TownTown

Since 1900, the magnitude of Since 1900, the magnitude of earthquakes that measure 0.1 or earthquakes that measure 0.1 or higher on the Richter Scale in a higher on the Richter Scale in a certain location in California is certain location in California is distributed approximately distributed approximately normally, with normally, with μ = 6.2μ = 6.2 and and σ = σ = 0.50.5, according to data obtained , according to data obtained from the United States Geological from the United States Geological Survey.Survey.



Earthquake Richter Scale Earthquake Richter Scale ReadingsReadings

-4 -3 -2 -1 0 1 2 3 46.25.7 6.7 7.2

15957

5.2

68%

34% 34%

95%

13.5% 13.5%2.5% 2.5%



Approximately what percent of the Approximately what percent of the earthquakes are above 5.7 on the Richter earthquakes are above 5.7 on the Richter Scale?Scale?

-4 -3 -2 -1 0 1 2 3 46.25.7 6.7 7.25.2

68%

34% 34%

95%

13.5% 13.5%2.5% 2.5%



The highest an earthquake can The highest an earthquake can read and still be in the lowest read and still be in the lowest 2.5% is _.2.5% is _.

-4 -3 -2 -1 0 1 2 3 46.25.7 6.7 7.25.2

68%

34% 34%

95%

13.5% 13.5%2.5% 2.5%



The approximate probability an The approximate probability an earthquake is above 6.7 is earthquake is above 6.7 is ______.______.

-4 -3 -2 -1 0 1 2 3 46.25.7 6.7 7.25.2

68%

34% 34%

95%

13.5% 13.5%2.5% 2.5%



Standard Normal DistributionStandard Normal Distribution Standard normal distribution is the Standard normal distribution is the

normal distribution that has a mean normal distribution that has a mean of 0 and standard deviation of 1.of 0 and standard deviation of 1.

-4 -3 -2 -1 0 1 2 3 4

N(µ = 0, σ = 1)



Z is Traditionally used as the Z is Traditionally used as the Symbol for a Standard Normal Symbol for a Standard Normal Random VariableRandom Variable

-4 -3 -2 -1 0 1 2 3 4

6.2 6.7 7.2 7.75.75.24.7

Z

Y



Normal Normal Standard Normal Standard Normal

y

Z

We can compare observations from two different normal distributions by converting the observations to standard normal and comparing the standardized observations.

Any normally distributed random variable can be converted to standard normal using the following formula:



What is the standard normal What is the standard normal value (or Z value) for a value (or Z value) for a Richter reading of 6.5?Richter reading of 6.5?Recall Y ~ N(Recall Y ~ N(µ=6.2, µ=6.2, σσ=0.5)=0.5)



ExampleExample Consider two towns in California. The Consider two towns in California. The

distributions of the Richter readings over 0.1 distributions of the Richter readings over 0.1 in the two towns are:in the two towns are:

Town 1:Town 1: X ~ N(X ~ N(µ = 6.2, µ = 6.2, σσ = 0.5) = 0.5)Town 2:Town 2: Y ~ N(µ = 6.2, Y ~ N(µ = 6.2, σσ = 1). = 1).

- What is the probability that Town 1 has an - What is the probability that Town 1 has an earthquake over 7 (on the Richter scale)?earthquake over 7 (on the Richter scale)?

- What is the probability that Town 2 has an - What is the probability that Town 2 has an earthquake over 7?earthquake over 7?



Town 1Town 1 Town 2 Town 2

Town 1:Town 1:

Town 2:Town 2:

X

Z

-4 -3 -2 -1 0 1 2 3 4

4.7 5.2 5.7 6.2 6.7 7.2 7.7

Z -4 -3 -2 -1 0 1 2 3 4

X

055.0)6.1(5.0

2.67)0.7(

ZPZPXP

212.0)8.0(0.1

2.67)0.7(

ZPZPYP

3.2 4.2 5.2 6.2 7.2 8.2 9.2Y

Z

0.055

0.212



Standard NormalStandard Normal

-4 -3 -2 -1 0 1 2 3 4

-2.576

0.0050.01

0.025

0.05

0.10

2.3261.96

1.645

1.282

0.010.005

2.576-2.326

-1.96

0.025

-1.645

0.05

-1.282

0.10



The thickness of a certain steel bolt The thickness of a certain steel bolt that continuously feeds a that continuously feeds a manufacturing process is normally manufacturing process is normally distributed with a mean of 10.0 mm distributed with a mean of 10.0 mm and standard deviation of 0.3 mm. and standard deviation of 0.3 mm. Manufacturing becomes concerned Manufacturing becomes concerned about the process if the bolts get about the process if the bolts get thicker than 10.5 mm or thinner than thicker than 10.5 mm or thinner than 9.5 mm.9.5 mm.

Find the probability that the thickness Find the probability that the thickness of a randomly selected bolt is of a randomly selected bolt is >> 10.5 10.5 or or << 9.5 mm. 9.5 mm.



Inverse Normal ProbabilitiesInverse Normal Probabilities Sometimes we want to answer a question Sometimes we want to answer a question

which is the reverse situation. Here we which is the reverse situation. Here we know the probability, and want to find the know the probability, and want to find the corresponding value of Y.corresponding value of Y.

-4 -3 -2 -1 0 1 2 3 4X=?

Area=0.025

y = ?



Inverse Normal ProbabilitiesInverse Normal Probabilities

Approximately 2.5% of the bolts produced Approximately 2.5% of the bolts produced will have thicknesses less than ______.will have thicknesses less than ______.

-4 -3 -2 -1 0 1 2 3 4Z

Y ?

0.025




Approximately 2.5% of the bolts produced Approximately 2.5% of the bolts produced will have thicknesses less than ______.will have thicknesses less than ______.

4.93.0

0.102

yY




Approximately 1% of the bolts produced Approximately 1% of the bolts produced will have thicknesses less than ______.will have thicknesses less than ______.

-4 -3 -2 -1 0 1 2 3 4Z

Y ?

0.01




Approximately 1% of the bolts produced Approximately 1% of the bolts produced will have thicknesses less than ______.will have thicknesses less than ______.

3.93.0

0.10326.2

yY

continuous random variables. l. wang, department of statistics university of south carolina; slide 2...

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