Chapter 5E-CRV%20-%20W0.pdf

1 Chapter E: Continuous Random Variables

Contents E.1 Continuous Random Variables E.2 The Density Function for a Continuous Random Variable E.3 The Uniform Probability Distribution E.4 Exponential Probability Distribution E.5 The Normal Probability Distribution E.6 The Gamma Probability Distribution E.7 The Beta Probability Distribution E.8 The Cauchy Probability Distribution E.9 The Weibull Probability Distribution Tutorial Essential Problems Mathematical Problems


E.1 Continuous Random Variables A variable is called a continuous variable if it takes uncountably infinite number of points on the real line. The points may be in one or more intervals. If the variable is defined on a sample space it is called a continuous random variable. It is impossible to assign a finite amount of probability to each of the points of a continuous random variable in such a way that the probabilities add to 1. Therefore the distinction between discrete and continuous random variables is usually based on the difference in their cumulative distributions. The cumulative distribution function (CDF) for a random variable X is given by

( ) ( )F x P X x= ≤ . In contrast to the cumulative distribution function (CDF) for a discrete random variable, ( )F x for a continuous random variable is a monotonically increasing continuous

function of x . This means that ( )F x is a continuous function such that if 1 2x x≤ , then

1 2( ) ( ).F x F x≤ That is as y increases, ( )F x never decreases. A continuous random variable has the following three properties. 1. It takes on an uncountably infinite number of values on the real line. 2. The CDF is continuous. 3. The probability that it equals a particular value is 0. E.2 The Density Function for a Continuous Random Variable In a relative frequency distribution we describe a data set neatly. If the data represent measurements on a continuous random variable and if the number of observations is very large, we can reduce the width of the class intervals until the distribution appears to be a smooth curve. A probability density function (PDF) is a theoretical model for this distribution. If )(uF is the CDF for a continuous random variable X at ,X u= then the probability density function for X at X u= is given by

( )( ) .XX

dF uf udu

=

It then follows that

( ) ( ) ( ) .u

X XF u P X u f x dx−∞

= ≤ = ∫ (E.1)

Thus the cumulative probability under the curve between ∞− and a point u is equal to )(uF . Properties of a Density Function 1. ( ) 0Xf x ≥ .

2. ( ) 1.Xf x dx∞

−∞

=∫


The probability of an event { : }X a X b≤ ≤ is given by

( ) ( ) .b

a

P a X b f x dx≤ ≤ = ∫ (E.2)

where a and b are constants. Remember that ( )P X a= is always zero in the continuous case. The integral in property (2) above can be represented by ( ).F ∞ The probability in (E.2) above can be represented by ( ) ( )F b F a− which is also ( )P a X b< < in most cases. Be warned that { : },X X a= or, { , , , }X a b c= are not probable events for a continuous variable, rather typical events are { : },X X a< { : },X X a> { : },X a X b< ≤ etc. whose probabilities are always defined by (D.1). Example E.2.1 Is the following function a probability density function?

2( ) 3(1 ) ,Xf x x= − 0 1;x< < ( ) 0,Xf x = elsewhere. Solution: The first condition ( ) 0Xf x ≥ is obviously true. The second condition is 1I =

where ( ) .XI f x dx∞

−∞

= ∫ By decomposing the domain, we have

0 1

0 1

( ) ( ) ( ) ,X X XI f x dx f x dx f x dx∞

−∞

= +∫ ∫ ∫

or, 0 1

2

0 1

0 3(1 ) 0 ,I dx x dx dx∞

−∞

= + − +∫ ∫ ∫

or, 1.I = Example E.2.2 The content of magnesium in an alloy is a random variable, given by the following pdf (probability density function)

( ) , 0 6, 18xf x x= ≤ ≤ and ( ) 0f x = elsewhere.

a. Find the probability that the content of magnesium in alloy is at most 3. b. Find the probability that the content of magnesium in alloy is at least 3. c. Find the probability that the content of magnesium in alloy at most k where k is a fixed

value of ,X and 0 6k< < . d. What content is exceeded 75% of the time? e. What content exceeds 25% of the time? f. Find 0.95 0.05x . Solution:

a. 33 2

0 0

1 1 1( 3) (9 0) 0.25.18 18 2 36 4x xP X dx

≤ = = = − = =

∫


b. 66 2

3 3

1 1 3( 3) (36 9) 0.75.18 18 2 36 4x xP X dx

≥ = = = − = =

∫

c. 2 2

2

0 0

1 1( ) ( 0) , 0 618 18 2 36 36

kk x x kP X k dx k k

≤ = = = − = < <

∫

Note that this function called CDF (Cumulative Distribution Function), usually denoted by ( )F x where x is a fixed value of the variable X such that ( ).P X x≤ The CDF ( )F x can be used to calculate both (a) and (b). d. Let k be a fixed value of X such that ( ) 0.75.P X k> = Then we have

66 22 21 1( ) (6 ) 0.75

18 18 2 36k k

y yP Y k dy k

> = = = − =

∫ . It then follows that 3k = .

e. This part is the same as part (d). f. Let 0.05 ,x k= a fixed value of X denoting 95th ( ) 0.05P X k> = percentile. Then or

equivalently ( ) 0.95P X k≤ = so that it follows from (c ) that

2

0.9536k

= , or, 5.85k ≈ .

E.3 The Uniform Distribution Consider an experiment that consists of observing events occurring in a certain time frame, such as buses arriving at a bus stop or telephone calls coming into a switchboard. Suppose we know that one such event has occurred in the time interval ( , )θ β . It may then be of interest to place a probability distribution on X , the actual time of occurrence of the event under consideration. A very simple model assumes that X is equally likely to lie in any small subinterval, say of length d , no matter where that subinterval lies within ( , ).θ β This assumption leads to the uniform probability distribution given by

1( ) , ,f x xθ ββ θ

= ≤ ≤−

and ( ) 0f x = elsewhere.

The density function will be denoted by ~ ( , ).X U θ β Let the subinterval be

) ,( dcc + contained entirely within ( , )θ β , then

1 ( )( ) ( ) c d c d

c c

c d c dP c X c d f x dy dxβ θ β θ β θ

+ + + −≤ ≤ + = = = =

− − −∫ ∫

Note that this probability does not depend on the location c but only on the length of the subinterval.


The cdf of this distribution is given by

( ) ( ) , .kF k P X k kθ θ ββ θ−

= ≤ = ≤ ≤−

The 100(1 )-thα− percentile of X at x is given by 1 ,x θ αβ θ−

= −−

i.e., ( ).x β α β θ= − −

It is usually denoted by 1 xα α− or simply .xα Example E.3.1 Suppose that X is uniformly distributed between 0 and 1. Determine 95th percentile of .X Solution: Hence the 95th percentile X at x is given by ( ) 0.95.XF x = The cumulative

distribution function of X at x is given by ( ) .XF x x= Hence 0.95. Example E.3.2 Suppose that X is uniformly distributed between 40 and 60. Determine 95th percentile of .X Solution: Hence the 95th percentile X at x is given by ( ) 0.95.XF x = That is

100 0.05(100 40),x = − − or, 97.x = Example E.3.3 Subway trains on a certain line run every half hour between mid-night and 6 in the morning. What is the probability that a man entering the station at random during this period will have to wait at least twenty minutes. Solution: Let X denote the waiting time (in minutes) for the next train. Then

1( ) , 0 30,30

f x x= < < and ( ) 0f x = elsewhere.

The required probability is then

30 30

20 20

1( 20) ( ) ,30

P X f x dx dx≥ = =∫ ∫ which equals ]30

20

1 1 1 (30 20) .30 30 3

x = − =

Interested readers may go through Mendenhall, Wackerly and Scheaffer (1990, 156-159) for very interesting discussions. E.4 The Standard Normal Distribution The normal distribution was proposed by Carl F. Gauss (1777-1855) as a model for the relative frequency distribution of errors, such as errors of measurement. Amazingly, this bell-shaped curve provides an adequate model for the relative frequency distribution of data collected from many different areas of science, engineering and business. It has the following functional form:


∞<<∞= − zezf - , 21)( 2/z 2

π

The following graph of the distribution is from STATISTICA:

The random variable Z is called a standard normal variable. It has the expected value 0)( =ZE and variance 1)( =ZV . It is obvious that 50.0)0()0( =>=< ZPZP . It is also interesting to note that though theoretically 1)( =∞<<−∞ ZP but practically ( 4 4) 1P Z− < < ≈ , even

1)]1(30)1(30[ ≈+<<− ZP i.e. A Standard Normal Variable lies within 3 (times) standard deviation (1) of its mean (0). The cumulative distribution function of Z is given by

2 /21( ) .2

zt

ZF z e dtπ

−

−∞= ∫

The solution to z of the equation ( ) 1ZF z α= − is the 100(1 )-thα− percentile of Standard

Normal Distribution and is denoted by αz . For example, ( 1.959964) 0.975,P Z < = or,

equivalently 025.0)959964.1( =>ZP . That is 96.1959964.1025.0 ≈=z s the 97.5-th percentile of the Standard Normal Distribution. Since

25.0175.0)67449.0(50.0150.0)0()0(

75.0125.0)67449.0()67449.0(

−==<−==>=<

−==>=−<

ZPZPZP

ZPZP

Probability Density Functiony=normal(x,0,1)

0.00

0.15

0.30

0.45

0.60

-3.50 -1.75 0.00 1.75 3.50


the quartiles of the distribution are given by 67449.0 and 0 ,67449.0 25.050.075.0 ==−= zzz . See the Table of Normal Percentiles given below:

α αz 0.005 2.575829 0.010 2.326348 0.015 2.170090 0.020 2.053749 0.025 1.959964 0.030 1.880794 0.035 1.811911 0.040 1.750686 0.045 1.695398 0.050 1.644854 0.055 1.598193 0.060 1.554774 0.065 1.514102 0.070 1.475791 0.075 1.439531 0.080 1.405072 0.085 1.372204

0.090 1.340755 0.095 1.310579 0.100 1.281552 0.200 0.841621 0.250 0.674490 0.500 0.000000

Note that the pdf ( )Zf z is location symmetric around location 0, that is (0 ) (0 ).Z Zf z f z+ = −

Hence, 05.095.0 zz −= , and 01.099.0 zz −= . That is, in general, for normal distribution we have

αα −−= 1zz where 10 ≤≤α . The probability that Z is less than or equal to 2.0 is denoted by ( 2.0)P Z ≤ . It is 0.9772 tabulated in the intersection of the first column (at 2) and first row (at 0). We write

( 2) 0.9772.P Z ≤ = Since ( 2) 0P Z = = , we have ( 2) 0.9772.P Z < = The probability that Z is less than or equal to 2.05 is denoted by ( 2.05)P Z ≤ . It is 0.9798 tabulated in the intersection of the first column (at 2) and first row (at 0.05). The probability that Z is more than 2.05 is given by ( 2.05) 1 ( 2.05)P Z P Z> = − ≤ which equals

( 2.05) 1 0.9798 0.0202P Z > ≈ − = . See TUTORIAL for details. Moreover ( 2.0) 0.9772P Z ≤ = implies that 2z = is the 97.72nd percentile. To find the 97.72nd percentile of standard normal distribution, we write ( ) 0.9772P Z k≤ = . Then in the body of the table we look for 0.9772. By comparing with ( 2.0) 0.9772P Z ≤ = , we write 2.k = Hence

2z = is the 97.72nd percentile.


To find the 75th percentile of the distribution, we write ( ) 0.75P Z k≤ = . Then in the body of the table we look for 0.75 or any value closest to it. We find 0.7486. By comparing with

( 0.67) 0.7486P Z ≤ = , we write 0.67.k ≈ Hence 0.67z ≈ is the 74.86th or 75th percentile. The Normal Distribution It is characterized by two natural parameters mean and variance usually denoted by 2 and σµ . distributions. The pdf is given by

21 1( ) exp , 22X

xf x µσσ π

− = −

, , .x µ σ−∞ < < ∞ −∞ < < ∞ −∞ < < ∞

The parameter µ is real and 2σ is nonnegative. They are the mean and variance of the distribution. Though these symbols are also used to denote the mean and variance of other distributions, they should not be confused. The distribution is usually denoted by ) ,( 2σµN . The function ( )Xf x is location symmetric around µ , i.e., ( ) ( ).X Xf x f xµ µ+ = − If X is a random variable with mean µ and variance 2σ i.e. 2~ ( , )X N µ σ , then

2 ~ (0, )X Nµ σ− and ~ (0, 1).XZ Nµσ−

=

If 2~ ( , )X N µ σ , then µ is the center or mean of the distribution. Fifty percent of the area under the normal distribution is below the mean which implies that mean and median coincide. In fact mean, median and mode coincide for the normal distribution. Since the quartiles of the standard normal distribution (with 0=µ and )1=σ are given by

67449.0 and 0 ,67449.0 25.050.075.0 ==−= zzz , it follows that

0.75 0.75 0.750.67449, 0 and 0.67449x x xµ µ µσ σ σ− − −

= − = = so that

0.75 0.50 0.250.67449 , , 0.67449 ,x x xµ σ µ µ σ= − = = + are the quartiles of normal distribution with mean )(µ and standard deviation )(σ . Example E.4.1 Mid term grades of the students of Engineering Statistics have a mean of 20. The variance of the grades is 4. What is the probability that a randomly selected student got more than 16. Explain this probability.

( 16) ( 2) ( 2) 0.9772P X P Z P Z> = > − = < = That is 97.72% students got more than 16 in the mid term.


Example E.4.2 A manufacturing process has a machine that fills coke to 300 ml bottles. Over a long period of time, the average amount dispensed into the bottles was 300 mls, but there is a standard deviation of 5 mls in these measurement. If the amounts of fill per bottle can be assumed to be normally distributed, find the probability that the machine will dispense more than 310 mls of liquid in any one bottle. (cf. Scheaffer and McClave, 1995, 216-217) Solution: Let X be the amount of fill in a bottle. Then 2~ (300, 5 ).X N

300 310 300( 310) ( 2) 0.02285 5

XP X P P Z− − > = > = > =

Example E.4.2 Experience indicates that the development time )(Y for a photographic printing paper is distributed as normal with mean 30=µ seconds and standard deviation 1 second. Find the probability that (a) the development time ( )X differs from its expected value )(µ by more than 2 seconds. (b) the development time ( )X will be within 1.645 seconds of its expected value )(µ . (c ) the development time will be within 1.645 seconds of true development time Solution: (a) [ 2, , X 2] [ 28, , X 32]P X or P X orµ µ− < − − > = < > Note that if somebody uses the left hand side of the above identity he can do the problem without knowing the value of µ . In either case the required probability is

2 2( 2) ( 2) ,1 1 1 1

X Xp P X P X P Pµ µµ µ − − − = − < − + − > = < + >

which equals

( 2) ( 2) 2(0.02275).p P Z P Z= < − + > = b. ( 1.645 1.645) ( 1.645 1.645),p P X P Xµ µ µ= − < < + = − < − < which equals

( 1.645 1.645) 0.90.p P Z= − < < = c. Same as above. (cf Hines and Montgomery, 1990) Example E.4.3 The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square meter. (a) What is the probability that the strength of a sample is less than 6164.5 kg/cm 2 ? (b) What compressive strength is exceeded by 95% of the time? (c) What compressive strength exceeds 5% of the time? Solution:


a. 6000 6164.5 6000( 6164.5) ( 1.645) 0.95

100 100XP X P P Z− − < = < = < ≈

b. Let k be a fixed value of the variable X . Then we have ( ) 0.95,P X k> = or equivalently ( ) 0.05,P X k< = which can be standardized as

0.05,X kP µ µσ σ− − < =

or, 0.05.kP Z µ

σ− < =

From the Standard Normal Probability Table, we have ( 1.645) 0.05P Z < − = so that by comparison we have

1.645,k µσ−

= − or, 1.645 5835.5.k µ σ= − =

c. The same as (b) (Montgomery, Runger and Hubele, 1998, 74) Example E.4.4 A machine for filling cereal boxes has a standard deviation of 1 ounce on of fill per box. a. What setting of the mean ounces of fill per box allow 16-ounce boxes to overflow only 1% of the time? Assume that the ounces of fill per box are normally distributed. (Scheaffer and McClave, 1995, 228) b. Suppose that the standard deviation is not known but can be fixed at certain levels by carefully adjusting the machine. What is the largest value of the standard deviation that will allow the actual value dispensed to be within 1 ounce of the mean with probability at least 0.95? Solution: a. ( 16) 0.01P X > = which can be transformed as ( 16 )P X µ µ− > − which can be standardized as ( 16 ) 0.01P Z µ> − = so that 16 2.33µ− = and consequently 13.67.µ = b. ( 1 1) 0.95P Xµ µ− < < + ≥ which can be standardized as

1 1 0.95XP µσ σ σ− − < < ≥

which equals

1 1 0.95P Zσ σ− < < ≥

where ~ (0,1).Z N Then

from the standard normal table, 1 1.96σ≥ so that

1 .1.96

σ ≤

Therefore the largest value of the standard deviation would be 1/1.96. E.5 The Exponential Distribution Life lengths of batteries or light bulbs usually follow exponentially decreasing distribution known as Negative Exponential Distribution or simply Exponential Distribution. Relative frequency


histogram of samples of lifelengths or light bulbs can be approximated by exponential distribution. The probability density function of an exponential random variable is given by

/1( ) , (0 ,0 ),xf x e xβ ββ

−= ≤ < ∞ < < ∞ and ( ) 0f x = elsewhere.

This distribution is characterized by the only one scale parameter β and may be represented, for short, by )(βExp . The mean and variance of exponential distribution is given by

22 and βσβµ == respectively. Thus it is a nonnegative distribution whose mean and standard deviation are the same. Many authors prefer to use 1/λ β= as the parameter of the distribution. Many interesting problems for sums of exponential variables are variable in Hines and Montgomery (1990). Note that if ( )P X k α> = , it can be proved that ln(1/ )k β α= . The proof is outlined below:

( ) ,k

f x dx α∞

=∫ or, /1 .x

ke dxβ α

β∞ − =∫ By integration, we have / ,x

ke β α

∞− − = or,

/ 0 .ke β α− − = Then by taking natural logarithm to both sides, we have

ln ln .k e αβ−

= Since ln 1,e = we have ln .k β α= −

In particular, if 1( )2

P X k> = , then ln 2k β= which is the median of the distribution. Since

ln 2 0.69 ,µ β β β= > = it follows that the density function is skewed to the right. Example E.3 Lifetime of a particular type of battery follow exponential distribution with mean 2 hundred hours. Find the probability that a. the lifetime of a particular battery of this type is less than 2 hundred hours. b. the lifetime of a particular battery of this type is more than 4 hundred hours. c. the lifetime of a particular battery of this type is less than 2 hundred hours or more than 4

hundred hours. d. it will survive three hundred hours more if it has already survived more than 5 hundred hours. e. it will survive between 5 hundred and 8 hundred hours? f. What is the probability that a battery will survive more than 5 hundred hours or more than 8

hundred hours? Solution Let X = lifetime of a battery. Then ~ ( ) (2)X Exp Expβ = .

a. The event of interest is { : 2}.X X < The required probability is 2

0( 2) ( ) ,P X f x dx< = ∫ or,

2 /

0

1( 2) .xP X e dxβ

β−< = ∫

Then by integration, we have


22/ 2/

/

00

1( 21

) 1/

.x

xeP X e eββ

β

ββ

−− −

< = × = − = − −

Since 2,β = finally, the probability is 11 0.6321.e−− ≈

b. 4

( ) ( )4 ,P X f x dx∞

> = ∫ or,

/

4

1( ) .4 xP X e dxβ

β∞ −> = ∫

By integration, we have

4/( .4)P X e β−> = Since 2,β = finally, the probability is 2 0.1353.e− ≈ c. 1 2[{ 2}, or , { 4}] ( 2) ( 4) 1 , P X X P X P X e e− −< > = < + > = − + by (a) and (b). d. Let { : 8}A X X= > and { : 5}.B X X= > Then the required probability is

( )( | ) .( )

P ABP A BP B

=

Since ,A B⊂ it follows that ,AB A= and then

( )( | ) .( )

P AP A BP B

=

Now / 8/

8

1( ) { 8} ,xP A P X e dx eβ β

β∞ − −= > = =∫ and similarly 5/( ) { 5} .P B P X e β−= > = Then,

we have

8/3/

5/( | ) ,eP A B ee

ββ

β

−−

−= =

which is the same as ( 8 5).P X > − This property is called the memoryless property of an

exponential distribution. Finally, the probability is 3/2 0.2231.e− ≈ e. 5/ 8/ 5/2 8/2(5 8) ( 5) ( 8) 0.0737.P X P X P X e e e eβ β− − − −< < = > − > = − = − ≈


It is also the probability that a battery survives more than 5 hundred hours but fails by 8 hundred hours. f. The required probability is ( ).P A B∪ Since ,A B⊂ it follows that .A B B∪ = Then

( ) ( ),P A B P B∪ = or, 8/2( ) ,P A B e−∪ = or, ( ) 0.0183.P A B∪ ≈

The above can also be done by ( ) ( ) ( ) ( ).P A B P A P B P AB∪ = + − Example E.4 An electronic component is known to have a useful life represented by an exponential density with failure rate 510− per hour. a. Determine the mean time to failure. b. Determine the fraction of such components that would fail before the mean life. c. What is the median life time? Solution: a. Since the failure rate is 510− , the mean is given by 5 51/10 10 .β −= = b. Let X be the life time. Then the density function is given by 1 /( ) xf x e ββ − −= for 0x > and

( ) 0f x = for 0.x < The fraction of such components that would fail before the mean life is

given by 1 / 1

0 0( ) ( ) 1 0.6321.xP X f x dx e dx e

β β ββ β − − −< = = = − ≈∫ ∫

c. The median (0 1)k k< < is a fixed value such that ( ) 0.5P X k< = so that

1 / /

0 0( ) 1 0.5.

k k x kf x dx e dx eβ ββ − − −= = − =∫ ∫ Then / 0.5ke β− = which can be inversed as

/ 2ke β = so that 55 10ln 2 10 ln 2 ln 2 .k β= = =

The Waiting Time Distribution of a Poisson Process Time between incidents is a continuous variable and it may be proved that it follows an exponential distribution. If )(tY , the number of incidents per t unit of time, follows Poisson Process with t λ as the expected number of incidents in t unit of time, then T , the interarrival time follows an exponential distribution with ( ) 1/E T λ= as the expected interarrival time. Derivation of the density function Consider a Poisson process tX with a mean of ( ) , 0.tE X t tλ= > Then the probability that

there will be no count in t unit of time is given by 0( )( 0)

0!t

ttP X e λλ −= = which

is ( 0) .ttP X e λ−= = Now let W be the waiting time one must wait to see the next count. Them

the event { }W t> is equivalent to { 0}.tX = Hence ( ) ( 0)tP W t P X> = = so that

( ) .rP W t e λ−> = Then the cumulative distribution function is given by


( ) ( ) 1 , 0.tF t P W t e tλ−= ≤ = − > The derivative of this function is the density function of the waiting time given below:

( ) , 0 , 0 ,tWf t e tλλ λ−= ≤ < ∞ ≤ < ∞ and ( ) 0Wf t = elsewhere.

Obviously, we have ( ) 1/E W λ= . If L denotes the length of life or time to failure, then )( tLP > is called the reliability function, and its logarithmic derivative is called the hazard. If the time to failure has the exponential distribution with mean λ/1 , then the reliability and the hazard functions are given by

tetR λ−=)( and )()()(

TRtRtH

′−=

respectively (see Lindgren, McElrath and Berry, 1978, 109). The following example relates to Example C.20 of Chapter C. There is a nice discussion on waiting time in telephone switching in Lapin (1997, 239). Example E.5 Accidents occur at an intersection at an average rate of 0.3 per hour. a. What is the probability that no accident happens in an hour? b. Find the probability that two consecutive accidents occur more than 4 hours apart. c. What is the probability that no accident happens in 2 hours? Solution: a. Let X , the number of accidents per hour, follow a Poisson distribution with meanλ so that

( )!

x

P X x ex

λλ −= = .

Then 0.3(1)( 0) 0.7408.tP X e eλ− −= = = ≈ (This is an example of Poisson Variable with unit time) b. Let X be the number of accidents per hour follow a Poisson distribution with mean 3.0=λ per hour. Then W , the time between consecutive accidents, measured in hours, follows an exponential distribution with

( ) 1/E W λ= i.e. ( ) , 0wf w e wλλ −= > and consequently

4 4 1.20

4 4

( 4) ( ) [ ] 0.3012.w wP W f w dw e dw e e eλ λ λλ∞ ∞

− − − −> = = = = = ≈∫ ∫

( X is a Poisson Variable with unit time)


c. Let tX , the number of accidents in t unit of time, follow a Poisson distribution with mean tλ so that

( )( )!

xt

ttP X x ex

λλ −= = .

Then

2 2(0.3)2( 0) 0.5488.P X e eλ− −= = = ≈

(This is an example of Poisson Variable with non-unit time) Example E.6 Accidents occur at an intersection at an average rate of 0.3 per 30 minutes. a. What is the probability that no accident happens in an hour? b. Find the probability that two consecutive accidents occur more than 4 hours apart. Solution: a. Let X , the number of accidents per hour, follow a Poisson distribution with mean

(0.3 / 30)(60) 0.6λ = = per hour so that

( )!

x

P X x ex

λλ −= = .

Then

0.6( 0) 0.5488.P X e eλ− −= = = ≈ (By scaling the average, we solved it by Poisson Variable with unit time) Alternative Solution: a. Let tX , the number of accidents in t unit of time, which is 30 minutes, follow a Poisson

distribution with mean tλ so that ( )( )

!

xt

ttP X x ex

λλ −= = . In terms of the unit time (30 minutes),

(30minutean hour an hour 2 30minutes

s) (30minutes)t = = × = so that

0.3(2)

2( 0) 0.5488.P X e−= = ≈ b. Let X be the number of accidents in an hour so that it follows a Poisson distribution with mean 0.3(2) 0.6.λ = = Then W , the time between consecutive accidents, measured in the unit of an hour, follows an exponential distribution with ( ) 1/E W λ= i.e.,

( ) , 0 ,tWf t e tλλ −= ≤ < ∞ and ( ) 0Wf t = elsewhere. It follows that


4 4(0.6)

44 4

( 4) ( ) 0.0907.t tWP W f t dw e dt e e eλ λ λλ

∞ ∞∞− − − −> = = = = = ≈∫ ∫

Alternatively Let tX , the number of accidents in t unit of time, which is 30 minutes, follow a

Poisson distribution with mean tλ so that

( )( )!

xt

ttP X x ex

λλ −= =

Then W , the time between consecutive accidents, measured in the unit of half an hour, follows an exponential distribution with ( ) 1/E W λ= i.e., ( ) , 0 ,t

Wf t e tλλ −= ≤ < ∞ and

( ) 0Wf t = elsewhere. In terms of the unit time (30 minutes), 4 hours is

( )4 hours 8 ( )half half an hour half a

ann

h

hourour× =

so that the event of interest is { : 8}W W > and consequently

8 8(0.3)

88 8

( 8) ( ) 0.0907.t tWP W f t dw e dt e e eλ λ λλ

∞ ∞∞− − − −> = = = = = ≈∫ ∫

E.6 The Gamma Probability Distribution The probability density function for gamma distribution is given by

1 /

( ) , 0 ,0 ,0( )

xx ef x xα β

α α ββ α

− −

= ≤ < ∞ < < ∞ < < ∞Γ

where 1

0

( ) xx e dxαα∞

− −Γ = ∫ . It can be shown that ( ) ( 1) ( 1)α α αΓ = − Γ − and if α is a positive

integer then ( ) ( 1)!α αΓ = − The expected value and variance of the gamma distribution are,

respectively, ( )E Yµ αβ= = and 2 2 2( )E Yσ µ αβ= − = . Note that the expected value of gamma distribution can be derived, without resorting to integration by parts, by reparameterization technique (see Mendenhall and Sincich, 1995, pp238-239) A chi-square random variable with ν degrees of freedom is a gamma variable with / 2α ν= and

2β = . The related probability density function is given by

( /2) 1 /2

/2( ) , 2 ( / 2)

x

Xy ef x

ν

ν ν

− −

=Γ

for 0 ,0 ;x ν≤ < ∞ < < ∞

( ) 0,Xf x = elsewhere.


A variation of the above pdf is more meaningful in real world applications. Consider a sequence of arrivals over time. Let T be the time until α th arrival and for each t , let tN be the number of

arrivals in the time interval 0, ]x . Suppose that ( )tN Poisson tλ for each t . Then for 0t < it

is trivially true that ( ) 1P T t> = while for 0t ≥ ,

1

0

( ) ( -th claim arrives after )(fewer than arrivals in [0, ])( 0,1, , 1)

( ) .!

tn tr

n

P T t P r tP r tP N r

t en

λλ −−

=

≥ === = −

=∑

Since the above is a survival function by differentiating ( )P T t− ≥ with respect to t we have the density function

1

for 0( ) ( 1)!

0 for 0

r r tt e tf t r

t

λλ − −≥= −

<

(Bean, 2001, 232). We will denote this distribution by ( , )G r λ Example E.6.1 Instantaneous surges of electrical current occur randomly and independently on a particular line at an expected rate of 0.1 per hour. The electrical system will fail after four surges. Determine the probability that the system is functioning in 10 hours. (bean, 2001, 232) Solution: Let jT be the time in hours between the ( 1)j − th and the j th surge. It has

exponential distribution with mean rate 0.1λ = . Then 1 2 3 4T T T T T= + + + , the future lifetime of

the system measured in hours, has a gamma distribution (4, )G λ where expected surging rate is with mean and the desired probability is

103

0

(10 )( 10) ,!

n

n

ep P Tn

λλ −

=

= > =∑ which equals 10 0.13

0

(10 0.1) ,!

n

n

epn

− ×

=

×=∑ which equals

13

0,

!n

epn

−

=

=∑ which equals 0.98101118.p ≈

Example E.6.2 You are waiting in line at the express checkout of the grocery store. Three customers are waiting ahead of you and one customer is being served. The checkout time for an individual customer has an exponential distribution with mean 20 seconds. a. Determine the probability that you will be waiting for service to begin 2 minutes from now. b. What is the probability that you will be completely served 2 minutes from now? (Bean, 2001, 233)


Solution: Let 1T be the remaining service time in minutes for the customer currently being

served; let 2 3 4, ,T T T be the service times in minutes for the three customers ahead of you; and

let 5T be your service time in minutes. Since each ( )iT Exp λ i.e. (1, )iT G λ and they are

independent , it follows that 1

( , )r

ii

T G r λ=∑ and hence

a. 24 3

1 0

(2 )2 0.1512!

n

ii n

eP Tn

λλ −

= =

> = ≈

∑ ∑

b. 25 5 4

1 1 0

(2 )2 1 2 1 0.7149!

n

i ii i n

eP T P Tn

λλ −

= = =

≤ = − ≥ = − ≈

∑ ∑ ∑

Note that 1T represents the remaining service time, not the total service time , for the customer currently being served. However since the total service time is exponentially distributed, the remaining service time must also be exponentially distributed by the memoryless property. The probability density function of a Gamma random variable X with two parameters, 0α > and

0λ > is sometimes written as

1( )( ) ,( )

x

Xx ef x

α λλ λα

− −

=Γ

0 x< < ∞

where 1

0( ) ,yy e dyαα

∞ − −Γ = ∫ 0.α >

a. Verify that the PDF integrates to 1.

Solution: 1

0 0

( )( ) ,( )

x

Xx eI f x dx dx

α λλ λα

− −∞ ∞= =

Γ∫ ∫

Substituting x yλ = we have

1 1

0 0(

( ) ( ),) x yI dyx e dx y eα λ αλ λλ

α α λ∞ ∞− − − −= =

Γ Γ∫ ∫

1

0

1 ,( )

yI y e dyα

α∞ − −=

Γ ∫

The integral 1

0,yy e dyα∞ − −∫ is a well known special function, usually denoted by ( ).αΓ

Hence 1.I = b. Prove that ( 1) ( )α α αΓ + = Γ for any 0.α >


c. Prove that ( ) ( 1) ( 1)α α αΓ = − Γ − for any 0.α > e. Use (c) and (d) to prove that ( 1) !α αΓ + = for any non-negative integer .α E.7 The Beta Probability Distribution The probability density function for beta distribution is given by

1 1( )( ) (1 ) , 0 1, 0 , 0( ) ( )

f x x x xα βα β α βα β

− −Γ += − ≤ ≤ < < ∞ < < ∞Γ Γ

with expected value ( )E Y αµα β

= =+

and variance

2

( )( 1)α βσ

α β α β α β=

+ + + +.

There are excellent examples in Bean (2001, 263-265). E.8 Cauchy Distribution The Cauchy random variable X assumes value over the entire real line and has the pdf:

2 2

1( ) ,Xf xx

απ α

=+

,x−∞ < < ∞ and 0α > is any constant.

Let us verify that the pdf integrates to 1.

2 2

1( ) .XI f x dx dxx

απ α

∞ ∞

−∞ −∞= =

+∫ ∫

Substituting ,x yα= we have

2

1 .1

I dyy

π∞

−∞=

+∫

That is 1.I = Let us derive the cdf of X and plot it. For ,x−∞ < < ∞ the CPF (Cumulative Probability

Function) ( ) ( )XF x P X x= ≤ of X is given by


0( ) ( ) ,

t x

X XtF x f t dt

=

== ∫ or,

2 20

1( ) .t x

X tF x dt

tπ α

α=

==

+∫

Substituting ,t yα= we have

/

2

1( ) .1

y x

X yF x dy

yα

π=

=−∞=

+∫

Finally the CDF is

11 1( ) tan ,2X

xF xπ α

− = +

0.α >

E.9 The Weibull Probability Distribution The probability density function for Weibull distribution is given by

1 /( ) x , 0 ,0 ,0yf x e xαα βα α β

β− −= ≤ < ∞ < < ∞ < < ∞

The expected value and variance of the gamma distribution are, respectively,

1/( ) (1 1/ )E Y αµ β α= = Γ + and ( )2 2/ 2(1 2 / ) (1 1/ )ασ β α α= Γ + −Γ + .

E.10 The Lognormal Probability Distribution The probability density function for lognormal distribution is given by

2 2(ln ) /(2 )1( ) , 0 ,0 ,0 .2

xf x e xx

µ σ α βσ π

− −= ≤ < ∞ < < ∞ < < ∞

Note that 2ln ( , ).Y X N µ σ=


Tutorial E.1 The total time, measured in units of 100 hours , that a student runs his computer over a period of one year is a continuous random variable X that has the density function

, 0 1( ) 2 , 1 2

0, elsewhere

x xf x x x

< <= − ≤ ≤

a. What is the probability that a student runs the computer between 75 hours to 150 hours per year. b. What is the first quartile of the total time a student runs his computer? c. How much time, on an average, a student runs his computer? d. What is the median amount of time a student runs his computer? e. What is the third quartile? Solution:

a. 1 1.5

0.75 1(0.75 1.5) ( ) ( ) ,P X f x dx f x dx< < = +∫ ∫ or,

1 1.5

0.75 1(0.75 1.5) (2 ) ,P X xdx x dx< < = + −∫ ∫ or, (0.75 1.5) 0.594.P X< < ≈

b. Since 1 1

0 0( ) 0.5 0.25f x dx xdx= = >∫ ∫ , the first quartile is a fixed value (0 1)k k< < such

that0

( ) 0.25.k

f x dx =∫ This yields 0

0.25,k

xdx =∫ or, 2

0

1 0.25,2

kx = or, 2 1/ 2,k = or,

1 .2

k = Hence the first quartile is given by 0.7071.k ≈

c. The mean is given by 2 1 2

0 0 1( ) ( ) ( ) ( ) ,E f x dx xf x dx xf x dxX x= = +∫ ∫ ∫ or,

1 2

0 1( ) { } {2 } 1.E X x x dx x x dx= + − =∫ ∫

d. Since 1 1

0 0( ) 0.5,f x dx xdx= =∫ ∫ the median is given by 1.

e. Since 1 1

0 0( ) 0.5 0.75f x dx xdx= = <∫ ∫ , the third quartile is a fixed value (1 2)k k< < such

that1

( ) 0.25.k

f x dx =∫ This yields 1

(2 ) 0.25,k

x dx− =∫ or, 21[2 0.5 ] 0.25kx x− = or,

2(2 0.5 ) (2 0.5) 0.25,k k− − − = 20.5 2 1.75 0,k k− + = 22 ( 2) 4(0.5)(1.75)

,2(0.5)

k− −

=


1.2929, 2.7071.k ≈ Hence the third quartile is 1.2929. Alternatively, since the density function is linear, the third quartile is approximately 2 0.7071 1.2929.− = Is it true in general? E.2 A box is to be constructed so that its height is 5 inches and its bases Y inches by Y inches , where Y is a random variable described by

( ) 6 (1 ), 0 1.f y y y y= − < < a. Find the expected value of 25 Y . b. Explain the quantity you found in part (a). (Larsen, Richard and Marx, Morris L. , 2001, 213). c. Derive 95th percentile of the above probability density function. Partial Solution:

a. 12

0(5 ) {6 (1 )} 1.5.E Y y y y dy= − =∫ (b) The expected volume.

E.3 Let Z be a standard normal distribution. Then determine the following probabilities: a. ( 1.96),P Z ≥ b. ( 1.96),P Z ≤ − c. ( 1.96),P Z ≥ − d. ( 1.96),P Z−∞ < ≤ − e. ( 1.96),P Z−∞ < ≤ f. ( 1.645 1.28),P Z− < ≤ − g. ( 1.645 1.96),P Z− < ≤ h. (| | 1.96),P Z ≤ i. (| | 1.96).P Z > Partial Solution by the CDF Method

)96.1(1)96.1( )( <−=≥ ZPZPa

))( (see )96.1()96.1( )( aZPZPb ≥=−≤

)96.1()96.1( )( ≤=−≥ ZPZPc


)96.1(1 )96.1()96.1()96.1( )( <−=≥=−≤=−≤<−∞ ZPZPZPZPd

)96.1()96.1( )( ≤=≤<−∞ ZPZPe

)28.1()645.1( )645.128.1()28.1645.1( )( ≤−≤=<≤=−≤<− ZPZPZPZPf

)]645.1(1[)96.1( )645.1()96.1()96.1645.1( )(

≤−−≤=−≤−≤=≤<−

ZPZPZPZPZPg

]50.0)96.1([21)96.1(2 )]96.1(1[)96.1()96.1()96.1(

)96.1()96.1()96.196.1( )96.1|(| )(

−≤=−≤=<−−≤=≥−≤=

−≤−≤=≤≤−=≤

ZPZPZPZPZPZP

ZPZPZPZPh

)]96.1(1[2 )]96.1()96.1([1

)96.196.1(1)96.1|(|1)96.1|(| )(

≤−=−≤−≤−=

≤≤−−=≤−=>

ZPZPZP

ZPZPZPi

Essential Problems E.8 The lifespan X , in hours, of a graphing card in a PC is thought to have a probability density function inversely proportional to the cube of x :

3( ) for 4000 5000f x x xβ −= ≤ ≤ where β is a constant. a. Determine the probability that the graphics card will last longer than 4500 hours after

installation. b. What guarantee period should the manufacturers offer if they wish to replace (under

guarantee) at most 1% of the graphics cards sold? (Smith, Peter, 1997, 186). Partial Solution: 89.88888888=β , ( ) ( 4500) 0.4170a P X > = ( ) ( ) 0.01b P X k< = where k is a fixed value. Then 4007.219k ≈ . E.17 A random variable Y , which represent the weight (in ounces) of an article, has probability density function given by

8 8 9

( ) 10 9 100

y for yf y y for y

otherwise

− ≤ ≤= − < ≤

a. Calculate the mean of the random variable Y . b. What is the probability that the weight of an article is more than 8.25 ounces?


c. The manufacture sells the article for a fixed price of $2.00. He guarantees to refund the purchase money to any customer who finds the weight of his article to be less than 8.25 ounces. His cost of production is related to the weight of the article by the relation ( /15) 0.35Y + . Find the expected profit per article. Partial Solution: a. 9, b. 0.969 approx. c. The sale price of the article is given by 2 for 8.25X ≥ and 0 for 8.25X < . The profit (sale price – cost) is given by

0 [( /15) 0.35] if 8.252 [( /15) 0.35] if 8.25

Y XZ

Y X− + <

= − + ≥

E.18 The pH of a reactant is measured by the random variable X whose density function is given by

25( 3.8) if 3.8 4( ) 25( 4.2) if 4 4.2

0 elsewhere

x xf x x x

− ≤ ≤= − − ≤ ≤

a. What is the median pH of the reactant b. What is the probability that the pH of the reactant exceeds 3.9?

Hints: (a) ( 4) 1/ 2P X ≤ = (b) ( 3.9) 1 ( 3.9) 0.875P X P X> = − ≤ =

E.21 An engineer designed a ball whose radius r changes from 10 inches to 20 inches with pdf given by

1 21500 (225 (3 45) ),10 20( )

0, otherwiser r

f r− − − ≤ ≤

=

a. Sketch the function ( )f r b. What proportion of balls have radius more than 15 inches? c. What is the expected radius of the balls? d. Find the expected surface area ( )S of the balls. e. Find the expected volume ( )V of the balls?

Partial Solution. 26 9 3( )5 50 500

f r r r= − + − , ( )3 45000 11600 63000 3600E R = − + − = ,

24S rπ= , 3 34 ( ) 4800 in3

V E Rπ π= =

E.45 Two instructors J and K have been teaching statistics course for 10 years. Final exam marks in the section taught by J usually varies around 75 with a standard deviation 9 while they are 70 and 10 respectively for the section taught by K. What is the probability that a randomly


selected student of J will have score more than the average score of a student of K? E.46 Let the diameter (in millimeters) of a hole drilled in a sheet metal can be modeled by the probability density function

20( 12.5)2 , 12.5( )

0 otherwise

xe xf x

− − ≤=

Note that the target diameter is 12.5 millimeters. If a part with a diameter larger than 12.6 millimeters is scrapped, what proportion of parts is scrapped? Find the median diameter of the sheet metal.

E.48 At a certain location on high-way I-10, the number of cars exceeding the speed limit by more than 20 miles per hour is a random variable having a Poisson distribution with an average of 8.4 cars per an half hour. What is the probability of a waiting time of less than 5 minutes between cars exceeding the speed limit by more than 20 miles per hour? (Miller, Irwin and Miller, Meryless, 1999, 207) Partial Solution: ( 5 / 30)P T < where T , time between arrivals, is an exponential distribution with mean number of cars arriving 8.4λ = per 30 minutes. Then

1/ 6 / 6 1.4

0( 1/ 6) 1 1 0.7534.tP T e dt e eλ λλ − − −< = = − = − ≈∫

Or , ( 5)P T < where T , time between arrivals, is an exponential distribution with mean number of cars arriving 8.4λ = /30 per minute, or, 0.28 per minute. Then

5 5 1.4

0( 5) 1 1 0.7534.tP T e dt e eλ λλ − − −< = = − = − ≈∫

E.49 The number of planes per day at a small private airport is a random variable having a Poisson distribution with a mean of 28.8. What is the probability that the time between two consecutive arrivals is at least 1 hour? (Miller, Irwin and Miller, Meryless, 1999, 215) (Ans: 0.3012 approx.) Solution: The mean number of planes arriving per hour is 28.8 / 24λ = per hour, i.e. 1.2λ = per hour. Let T be the time between consecutive arrivals. Then

1.2

1( 1) 0.3012.tP T e dt e eλ λλ

∞ − − −≥ = = = ≈∫

Mathematical Problems E.54 Find the expected value of exponential random variable with mean 1. Solution: The density function is given by ∞<≤= − yeyf y 0 ,)( . The expected value is, therefore,


E.55 Consider the gamma probability function 1( )( ) .

( )

x

Xx ef x

α λλ λα

− −

=Γ

a. Prove that 2 2( ) ( 1) /E X α α λ= + and 2( ) / .Var X α λ= E.56 Let ~ ( ).X Expo λ a. Find the CDF ( )XF x of the distribution. Also derive the conditional CDF ( | ).XF x X t> Solution: The CDF of the exponential random variable is

00 0( ) ( ) ( ) 1 .

t x t x xt tX X

x

t tF P X x f t dt e dt e ex λ λ λλ

= = − − −

= = = ≤ = = = − = − ∫ ∫

The conditional CDF is given by ( )( | ) ,

1 ( )X

XX

f xF x X tF t

> =−

.x t≥ This is given by

( )( | ) ,

1 (1 )

xx t x t

X t

eF x X t e e ee

λλ λ λ

λ

λ λ λ−

− − −−> = = × =

− − .x t≥

Choose 3λ = say, and 0.10t = say. Then choose

0.10,0.20,0.30,0.40,0.40,0.50,x = and calculate values of ( )( | ) .x tXF x X t e λλ − −> =

Then plot ( )( | ) x tXF x X t e λλ − −> = in the vertical axis against x in the horizontal axis.

b. Prove that ( | ) ( ).P X t x X t P X x> + > = > Solution: For 0,t > consider the events { }A X t x= > + and { }.B X t= > Then the given problem reduces to

( ) ( )( | ) ,( ) ( )

P AB P AP A BP B P B

= =

Where we have used the fact AB A= . Note that if one set is a subset of the other, then the intersection is the smaller set and union is the bigger set. Notice that A B⊂ so that

.AB A=

( )( )( | ) ( ),( )

x tx

t

P A eP A B e P X xP B e

λλ

λ

− +−

−= = = = >

That is ( | ) ( ).P X t x X t P X x> + > = >


Since the left hand side does not depend on t , this property is called memory less property of exponential distribution. E.57 Prove that the cdf (Cumulative Distribution Function), denoted by ( ),F u of the exponential

distribution with mean β is given by /( ) 1 , 0 .uP Y u e uβ−≤ = − ≤ < ∞ E.58 The sdf (survival distribution function) of the exponential distribution with mean β is given

by is given by ∞<≤=≤−=>= − ueuYPuYPuS u 0 , )(1)()( / β . The probability of an event can be found by direct integration or by the use of cdf or sdf. E.58 Let Y have an Exponential Distribution with mean β . a. Solve the equation for )()( cYPcYP >=≤ for c . b. Also solve the equation 1)( 2 =≤ cYP for c c. Interpret your results in (a) and (b). E.59 Accidents occur at an intersection at an average rate of tλ per t unit of time. a. What is the probability that no accident happens in t unit of time? b. Find the probability that two consecutive accidents occur more than kt unit of time apart . Solution: a. ( 0) tP X e λ−= =

b. ( ) [ ( 0)]ktP W kt P X> = =

E.60 Let 2~ ( , ).X N µ σ Prove that x zα αµ σ= + where xα is the 100(1 )-thα− percentile of

the distribution of .X


Formulae for Chapter E E. Continuous Probability Distributions For a continuous random variable X with pdf ( )Xf x we have the following:

E.1 ( ) ( ) , ( ) ( ) .b k

Xa

P a X b f x dx F k f x dx−∞

< < = =∫ ∫

E.2 ( ) ( ) .E X x f x dxµ∞

−∞

≡ = ∫

E.3 2 2 2 2 2( ) ( ) , ( ) ( ) .E X x f x d x V X E Xσ µ∞

−∞

= ≡ = −∫

E.4 The Normal Distribution 2( , ) :N µ σ ( ) ,E X µ= 2( ) .V X σ=

E.5 The Exponential Distribution: /1( ) , 0,xf x e xβ

β−= ≥ ( ) ,E X β= 2( ) .V X β=

E.6 Waiting Time Distribution: ( ) , 0,tf t e tλλ −= ≥ ( ) 1/ ,E T λ= 2( ) 1/ ,V T λ=

E.7 The Gamma Distribution: 1 /1( ) x , 0 , 0 , 0( )

xf x e xα βα α β

β α− −= < < <

Γ( ) ,E X αβ=

2( ) .V X αβ=




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