Continuous Probability Distributions

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Continuous Probability Continuous Probability DistributionsDistributions

Chapter 8Chapter 8

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• A continuous random variable has an uncountably A continuous random variable has an uncountably infinite number of values in the interval (a,b).infinite number of values in the interval (a,b).

8.2 8.2 Continuous Probability DistributionsContinuous Probability Distributions

• The probability that a continuous variable The probability that a continuous variable XX will will assume any particular value is zero. Why?assume any particular value is zero. Why?

0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

The probability of each value

30 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1.

When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0.

The probability of each value

8.2 8.2 Continuous Probability DistributionsContinuous Probability Distributions

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• To calculate probabilities we define a probability To calculate probabilities we define a probability density function f(x).density function f(x).

• The density function satisfies the following conditionsThe density function satisfies the following conditions– f(x) is non-negative,f(x) is non-negative,– The total area under the curve representing f(x) equals 1.The total area under the curve representing f(x) equals 1.x1 x2

Area = 1P(x1<=X<=x2)

Probability Density FunctionProbability Density Function

• The probability that X falls between xThe probability that X falls between x11 and x and x22 is is found by calculating the area under the graph of f(x) found by calculating the area under the graph of f(x) between xbetween x11 and x and x22..

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– A random variable A random variable XX is said to be uniformly is said to be uniformly distributed if its density function isdistributed if its density function is

– The expected value and the variance areThe expected value and the variance are

12)ab(

)X(V2

baE(X)

.bxaab

1)x(f

2

Uniform DistributionUniform Distribution

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• Example 8.1Example 8.1– The daily sale of gasoline is uniformly distributed between The daily sale of gasoline is uniformly distributed between

2,000 and 5,000 gallons. Find the probability that sales are:2,000 and 5,000 gallons. Find the probability that sales are:– Between 2,500 and 3,500 gallonsBetween 2,500 and 3,500 gallons– More than 4,000 gallonsMore than 4,000 gallons– Exactly 2,500 gallonsExactly 2,500 gallons

2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x2500 3000

P(2500X3000) = (3000-2500)(1/3000) = .1667


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2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x4000

P(X4000) = (5000-4000)(1/3000) = .333


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2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x2500

P(X=2500) = (2500-2500)(1/3000) = 0


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8.3 Normal Distribution8.3 Normal Distribution

• This is the most important continuous distribution.This is the most important continuous distribution.

– Many distributions can be approximated by a normal Many distributions can be approximated by a normal distribution.distribution.

– The normal distribution is the cornerstone distribution The normal distribution is the cornerstone distribution of statistical inference.of statistical inference.

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• A random variable A random variable XX with mean with mean and and variancevariance is normally distributed if its is normally distributed if its probability density function is given byprobability density function is given by

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

Normal DistributionNormal Distribution

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The Shape of the Normal DistributionThe Shape of the Normal Distribution

The normal distribution is bell shaped, and symmetrical around

Why symmetrical? Let = 100. Suppose x = 110.

2210

)2/1(100110

)2/1(e

21

e2

1)110(f

Now suppose x = 9022

10)2/1(

10090)2/1(

e2

1e

21

)90(f

11090

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The effects of and The effects of The effects of and and

How does the standard deviation affect the shape of f(x)?

= 2

=3 =4

= 10 = 11 = 12How does the expected value affect the location of f(x)?

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• Two facts help calculate normal probabilities:Two facts help calculate normal probabilities:– The normal distribution is symmetrical.The normal distribution is symmetrical.– Any normal distribution can be transformed into a Any normal distribution can be transformed into a

specific normal distribution called…specific normal distribution called…• ““STANDARD NORMAL DISTRIBUTION”STANDARD NORMAL DISTRIBUTION”

ExampleExampleThe amount of time it takes to assemble a computer is normally The amount of time it takes to assemble a computer is normally

distributed, with a mean of 50 minutes and a standard distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?computer is assembled in a time between 45 and 60 minutes?

Finding Normal ProbabilitiesFinding Normal Probabilities

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• SolutionSolution– If If X X denotes the assembly time of a computer, we denotes the assembly time of a computer, we

seek the probability P(45<X<60).seek the probability P(45<X<60).– This probability can be calculated by creating a new This probability can be calculated by creating a new

normal variable the normal variable the standard normal variable.standard normal variable.

x

xXZ

x

xXZ

E(Z) = = 0 V(Z) = 2 = 1

Every normal variablewith some and, canbe transformed into this Z.

Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.


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• Example - continuedExample - continued

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-0.5 < Z < 1)

To complete the calculation we need to compute the probability under the standard normal distribution


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z 0 0.01 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.0636

. . . . .

. . . . .1.0 0.3413 0.3438 0.3531 0.3554

. . . . .

. . . . .1.2 0.3849 0.3869 ……. 0.3944 0.3962

. . . . . .

. . . . . .

Standard normal probabilities have been calculated and are provided in a table .

The tabulated probabilities correspondto the area between Z=0 and some Z = z0 >0 Z = 0 Z = z0

P(0<Z<z0)

Using the Standard Normal TableUsing the Standard Normal Table

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P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-.5 < Z < 1)

z0 = 1z0 = -.5

We need to find the shaded area


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P(-.5<Z<0)+ P(0<Z<1)

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .1.0 0.3413 0.3438 0.3531 0.3554

. . . . .

P(0<Z<1


= P(-.5<Z<1) =

z=0 z0 = 1z0 =-.5

.3413


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• The symmetry of the normal distribution makes it The symmetry of the normal distribution makes it possible to calculate probabilities for negative possible to calculate probabilities for negative values of Z using the table as follows:values of Z using the table as follows:

-z0 +z00

P(-z0<Z<0) = P(0<Z<z0)


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z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .0.5 0.1915 …. …. ….

. . . . .



.3413

.5-.5

.1915

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z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .0.5 0.1915 …. …. ….

. . . . .



.1915.1915.1915.1915

.3413

.5-.5

P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328

1.0

22

10%0%

20-2

(i) P(X< 0 ) = P(Z< ) = P(Z< - 2)0 - 10

5

=P(Z>2) =Z

X

• Example 8.2Example 8.2– The rate of return (X) on an investment is normally distributed The rate of return (X) on an investment is normally distributed

with mean of 10% and standard deviation of (i) 5%, (ii) 10%.with mean of 10% and standard deviation of (i) 5%, (ii) 10%.– What is the probability of losing money?What is the probability of losing money?

.4772

0.5 - P(0<Z<2) = 0.5 - .4772 = .0228


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10%0%

-1

(ii) P(X< 0 ) = P(Z< ) 0 - 10

10

= P(Z< - 1) = P(Z>1) =Z

X

• Example 8.2Example 8.2– The rate of return (X) on an investment is normally distributed The rate of return (X) on an investment is normally distributed

with mean of 10% and standard deviation of (i) 5%, (ii) 10%.with mean of 10% and standard deviation of (i) 5%, (ii) 10%.– What is the probability of losing money?What is the probability of losing money?

.3413

0.5 - P(0<Z<1) = 0.5 - .3413 = .1587


Find Normal Probabilities

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• Sometimes we need to find the value of Z for a Sometimes we need to find the value of Z for a given probabilitygiven probability

• We use the notation zWe use the notation zA A to express a Z value for to express a Z value for which P(Z > zwhich P(Z > zAA) = A) = A

Finding Values of ZFinding Values of Z

zA

A

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• Example 8.3 & 8.4Example 8.3 & 8.4– Determine z exceeded by 5% of the populationDetermine z exceeded by 5% of the population– Determine z such that 5% of the population is belowDetermine z such that 5% of the population is below

• SolutionSolutionzz.05.05 is defined as the z value for which the area on its right under the is defined as the z value for which the area on its right under the

standard normal curve is .05.standard normal curve is .05.

0.05

Z0.050

0.45

1.645

Finding Values of ZFinding Values of Z

0.05

-Z0.05

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Exponential DistributionExponential Distribution

• The exponential distribution can be used to modelThe exponential distribution can be used to model– the length of time between telephone callsthe length of time between telephone calls– the length of time between arrivals at a service stationthe length of time between arrivals at a service station– the life-time of electronic components.the life-time of electronic components.

• When the number of occurrences of an event When the number of occurrences of an event follows the Poisson distribution, the time between follows the Poisson distribution, the time between occurrences follows the exponential distribution.occurrences follows the exponential distribution.

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A random variable is exponentially distributed A random variable is exponentially distributed if its probability density function is given byif its probability density function is given by

f(x) = f(x) = ee--xx, x>=0., x>=0.is the distribution parameter (is the distribution parameter (>0).>0).

E(X) = 1/E(X) = 1/ V(X) = (1/ V(X) = (1/


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0

0.5

1

1.5

2

2.5f(x) = 2e-2x

f(x) = 1e-1x

f(x) = .5e-.5x

0 1 2 3 4 5

Exponential distribution for = .5, 1, 2

0

0.5

1

1.5

2

2.5

a bP(a<x<b) = e-a - e-b

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• Finding exponential probabilities is relatively Finding exponential probabilities is relatively easy:easy:– P(X > a) = eP(X > a) = e––aa..– P(X < a) = 1 – e P(X < a) = 1 – e ––aa

– P(aP(a11 < X < a < X < a22) = e ) = e – – a1)a1) – e – e – – a2)a2)


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• Example 8.5Example 8.5– The lifetime of an alkaline battery is exponentially The lifetime of an alkaline battery is exponentially

distributed with distributed with = .05 per hour.= .05 per hour.– What is the mean and standard deviation of the What is the mean and standard deviation of the

battery’s lifetime?battery’s lifetime?– Find the following probabilities:Find the following probabilities:

• The battery will last between 10 and 15 hours.The battery will last between 10 and 15 hours.• The battery will last for more than 20 hours?The battery will last for more than 20 hours?


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• SolutionSolution– The mean = standard deviation = The mean = standard deviation =

1/1/1/.05 = 20 hours.1/.05 = 20 hours.– Let X denote the lifetime.Let X denote the lifetime.

• P(10<X<15) = eP(10<X<15) = e-.05(10)-.05(10) – e – e-.05(15)-.05(15) = .1341 = .1341• P(X > 20) = eP(X > 20) = e-.05(20)-.05(20) = .3679 = .3679


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• Example 8.6Example 8.6– The service rate at a supermarket checkout is 6 The service rate at a supermarket checkout is 6

customers per hour. customers per hour. – If the service time is exponential, find the following If the service time is exponential, find the following

probabilities:probabilities:• A service is completed in 5 minutes,A service is completed in 5 minutes,• A customer leaves the counter more than 10 minutes after A customer leaves the counter more than 10 minutes after

arrivingarriving• A service is completed between 5 and 8 minutes.A service is completed between 5 and 8 minutes.


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• SolutionSolution– A service rate of 6 per hour = A service rate of 6 per hour =

A service rate of .1 per minute (A service rate of .1 per minute ( = .1/minute). = .1/minute).– P(X < 5) = 1-eP(X < 5) = 1-e--xx = 1 – e = 1 – e-.1(5)-.1(5) = .3935 = .3935– P(X >10) = eP(X >10) = e--xx = e = e-.1(10)-.1(10) = .3679 = .3679– P(5 < X < 8) = eP(5 < X < 8) = e-.1(5)-.1(5) – e – e-.1(8)-.1(8) = .1572 = .1572


Compute Exponential probabilities

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8.5 Other Continuous Distribution8.5 Other Continuous Distribution

• Three new continuous distributions:Three new continuous distributions:– Student t distributionStudent t distribution– Chi-squared distributionChi-squared distribution– F distributionF distribution

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The Student t DistributionThe Student t Distribution

• The Student t density functionThe Student t density function

is the parameter of the student t distributionis the parameter of the student t distribution

E(t) = 0 V(t) =E(t) = 0 V(t) =((– 2)– 2)

2/)1(2t1

)]!2[()]!1[(

)t(f

2/)1(2t1

)]!2[()]!1[(

)t(f

(for n > 2)(for n > 2)

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The Student t DistributionThe Student t Distribution

0

0.05

0.1

0.15

0.2

-6 -4 -2 0 2 4 6

0

0.05

0.1

0.15

0.2

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

= 3

= 10

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Determining Student t Values Determining Student t Values

• The student t distribution is used extensively in The student t distribution is used extensively in statistical inference.statistical inference.

• Thus, it is important to determine values of tThus, it is important to determine values of tAA associated with a given number of degrees of freedom.associated with a given number of degrees of freedom.

• We can do this usingWe can do this using– t tables t tables – ExcelExcel– MinitabMinitab

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Degrees of Freedom1 3.078 6.314 12.706 31.821 63.6572 1.886 2.92 4.303 6.965 9.925. . . . . .. . . . . .

10 1.372 1.812 2.228 2.764 3.169. . . . . .. . . . . .

200 1.286 1.653 1.972 2.345 2.6011.282 1.645 1.96 2.326 2.576

tA

t.100 t.05 t.025 t.01 t.005

A = .05A = .05

-tA

The t distribution issymmetrical around 0

=1.812=-1.812

• The table provides the t values (tThe table provides the t values (tAA) for which P(t) for which P(t > t > tAA) = A) = AUsing the t TableUsing the t Table

tttttttt

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The Chi – Squared Distribution The Chi – Squared Distribution

• The Chi – Squared density function:The Chi – Squared density function:

• The parameter The parameter is the number of degrees of is the number of degrees of freedom.freedom.

0)(2]!1)2/[(

1)( 221)2/(2

2/2 2

ef 0)(

2]!1)2/[(

1)( 221)2/(2

2/2 2

ef

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The Chi – Squared Distribution The Chi – Squared Distribution

00.00020.00040.00060.0008

0.0010.00120.00140.00160.0018

0 5 10 15 20 25 30 35

= 5

= 10

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• Chi squared values can be found from the chi squared Chi squared values can be found from the chi squared table, from Excel, or from Minitab.table, from Excel, or from Minitab.

• The The 22-table entries are the -table entries are the values of the right hand values of the right hand tail probability (A), for which P(tail probability (A), for which P(= A.= A.

Determining Chi-Squared ValuesDetermining Chi-Squared Values

0 5 10 15 20 25 30 35

A

2A

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Degrees offreedom

1 0.0000393 0.0001571 . 3.84146 6.6349 7.87944..

10 2.15585 2.55821 . 18.307 23.2093 25.1882. . . . .. . . . . . . .

Degrees offreedom

1 0.0000393 0.0001571 . 3.84146 6.6349 7.87944..

10 2.15585 2.55821 . 18.307 23.2093 25.1882. . . . .. . . . . . . .

0 5 10 15 20 25 30 35

=.05A =.99

Using the Chi-Squared TableUsing the Chi-Squared Table

A

To find 2 for which P(2

<2)=.01, lookup the column labeled2

1-.01 or 2.99

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The F DistributionThe F Distribution

• The density function of the F distribution:The density function of the F distribution:

11 and and 2 2 are the numerator and denominator are the numerator and denominator degrees of freedom.degrees of freedom.

0FF

1

F

22

22

22

)F(f2

2

1

22

2

2

1

21

21

21

11

0FF

1

F

22

22

22

)F(f2

2

1

22

2

2

1

21

21

21

11

!!

!

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0

0.002

0.004

0.006

0.008

0.01

0 1 2 3 4 5

• This density function generates a rich family of This density function generates a rich family of distributions, depending on the values of distributions, depending on the values of 11 and and 22

The F DistributionThe F Distribution

1 = 5, 2 = 10

1 = 50, 2 = 10

00.0010.0020.0030.0040.0050.0060.0070.008

0 1 2 3 4 5

1 = 5, 2 = 10

1 = 5, 2 = 1

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Determining Values of FDetermining Values of F

• The values of the F variable can be found in the The values of the F variable can be found in the F table, Excel, or from Minitab.F table, Excel, or from Minitab.

• The entries in the table are the values of the F The entries in the table are the values of the F variable of the right hand tail probability (A), for variable of the right hand tail probability (A), for which P(Fwhich P(F1,1,22>F>FAA) = A.) = A.

Continuous Probability Distributions

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