Chapter 4

Continuous Random Variables and Probability Distributions

Learning Objectives• Determine probabilities from probability density

functions• Determine probabilities from cumulative

distribution functions• Calculate means and variances• Standardize normal random variables• Approximate probabilities for some binomial

and Poisson distributions• Calculate probabilities, determine means and

variances for the continuous probability distributions presented

Continuous Random Variables

• Discussed about discrete random variables

• Continuous random variable, X, has a distinctly different distribution from the discrete random variables

• Includes all values in an interval of real numbers

• Thought of as a continuum

Probability Distributions

• Describe the probability distribution of a continuous random variable X

• Probability that X is between a and b is determined as the integral of f(x) from a to b

Probability Density Function

• Continuous random variable X, a probability density function

1.

2.

3. area under f(x) from a to b

• Zero for x values that cannot occur

0)( xf

1)( dxxf

b

a

dxxfbXaP )()(

Important Point• f(x) is used to calculate an area that represents the

probability that X assumes a value in [a, b]• Probability at any point is zero, because every point

has zero width• P(X=x)=0• Not distinguish between inequalities such as < or

for continuous random variables• For any x1 and x2

• Not true for a discrete random variable! )( bXaP )( bXaP )( bXaP )( bXaP

Example• If a random variable

• Find the probability– between 1 and 3– greater than 0.5

• Solution

0for 00for x 2

)(2 xe

xf

135.02)31( 623

1

2 eedxeXP x

368.02)5(5.0

12

edxeXP x

Class Problem• Suppose that f(x)= e-(x-4) for x>4 • Determine the following probabilities

– P(1<X)– P(2X<5)– Determine such that P(X<x) =0.9

Solution

• Part a because f(x)=0 for x<4• Part b

• Part c

• Then, x = 4 – ln(0.10) = 6.303

1)1( 4)4(

4

)4(

xx edxeXP

6321.01)52( 15

4

54

)4()4( eedxeXP xx

90.01)( )4(4

)4(

4

)4( xxxx

x eedxexXP

Cumulative Distribution Function

• Stated in the same way as we did for the discrete random variable

• F(x) is defined for every number x by

x

duufxXPxF )()()( xfor

F(x) and f(x)

• Probability that the random variable will take on a value on the interval from a to b is F(b) – F(a)

• Fundamental theorem of integral calculus

)()( xfdxxdF

Example• Find the cumulative distribution function of the

following pdf

• Solution

• When x=1 yields

0for 00for x 2

)(2xe

xf

0for x 0

0for x e-1dt 2)(

2x-2

0

tx

exF

865.01)1( 2 eF

Class Problem• Suppose that f(x)=1.5 x2 for –1<x<1

– Determine the cumulative distribution function• Solution

Solution

• The cumulative distribution function for -1< x < 1• Then

5.05.05.05.1)( 31

3

1

2 xxdxxxF xx

x

x1 1,

1x1- 5.00.5x

-1x ,0

)( 3xF

Mean and Variance• Defined similarly to a discrete random variable• Mean or expected value of X

– E[X]=

• Variance

– V(X)=

dxxxf )(

dxxfx )()( 2

Example• If a random variable has

– Find the mean and variance of the given probability density function

• Solution

0for x 00for x 2

)(2 xe

xf

5.02.)(][0

2

dxexdxxxfxE x

4/1

2)2/1(

)()(][

0

22

2

dxex

dxxfxYV

x

Uniform Distribution• Simplest continuous distribution

• Probability Density

• Mean & Standard Deviation

• Proof

abxf

1)(

12 and

2abba

x

f(x)

12)(

)(3

)2

()2

(()(

2/)(5.0)(

232

2

abab

baxdx

ab

baxXV

baabXdx

abxXE

ba

b

a

ba

b

a

Example• Suppose X has a continuous uniform distribution

over the interval [1.5, 5.5]– Determine the mean, variance, and standard deviation

of X– What is P(X<2.5)?

• SolutionE(X) = (5.5+1.5)/2 = 3.5,

866.043

4/312

2)5.15.5()(

x

XV

25.05.25.125.0

5.2

5.125.0)5.2(

x

dxXP

Class Problem• The thickness of a flange on a aircraft component

is uniformly distributed between 0.95 and 1.05 millimeters– Determine the cumulative distribution function of flange

thickness– Determine the proportion of flanges that exceeds 1.02

millimeters– What thickness is exceeded by 90% of the flanges?

Solution

a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05Now

b) The probability

c) If P(X > x)=0.90, then 1 F(X) = 0.90 and F(X) = 0.10. Therefore, 10x - 9.5 = 0.10 and x = 0.96.

F x

x

x x

x

( )

, .

. , . .

, .

0 0 95

10 9 5 0 95 105

1 105

P X P X F( . ) ( . ) ( . ) . 102 1 102 1 102 0 3

Normal Distribution• Most widely used model for the distribution of a

random variable is a normal distribution• Describes many random processes or continuous

phenomena• Used to approximate discrete probability

distributions• Basis for classical statistical inference

Mean and Variance• Random variables with different means and

variances can be modeled by normal probability• E[X]= determines the center of the probability

density function• V[X]=2 determines the width• Illustrates several normal probability density

functions

Probability Density Function• X with probability density function

• Normal random variable with parameters - < < and >1

• Mean and variance– E[X]= , V[X]= 2

• N(, 2 ) used to denote the distribution

2

2

2)(

21)(

x

exf x

Useful Information• Total area under the curve is 1.0• Two tails of the curve extend indefinitely • Useful results

– P(-<X<+)=0.6827– P(-2<X<+2)=0.9545– P(-3<X<+3)=0.9973

Calculating the Probabilities• Normal distributions differ by mean &

standard deviation

• Each distribution would require its own table

• Infinite number of tables!

X

f(X)

Definition• Normal random variable with =0 and 2=1

• Called a standard normal random variable• Denoted as Z• Cumulative distribution function of a standard

normal random variable is denoted as

• Appendix Table II provides cumulative probability values

)()( zZPz

Standardize the Normal Distribution• Use the following random variable z to

standardize a normal distribution into standard normal distribution

• Calculate the probabilities

X

= 0

= 1

Z

Normal distribution Standardized Normal Distribution

)()()( zZPxXPxXP

Xz

Working with Table• Table II provides values of (z) for values of Z• Suppose Z=1.5

• Note that P(a<X<b)=F(a) – F(b)

Example

• Find probabilities that a random variable having the standard normal distribution will take on a value– between 0.87 and 1.28– between –0.34 and 0.62– greater than 0.85– greater than –0.65– less than –0.85– less than –4.6

Solution• F(1.28) – F(0.85) = 0.8997-0.8078 = 0.0919• F(0.62) - F(-0.34)= 0.7324 – (1-0.6331) =

0.3655• P(z>0.85)=1-P(z0.85)= 1-F(0.85) = 1-

0.85023 = 0.1977• P(z>-0.65) =1-F(-0.65)=1-[1-

F(0.65)]=F(0.65) = 0.7422• P(z<-0.85)= (1-0.8551) = 0.1949• P(z<-4.6)=

– P (z<-3.99) = 1-0.99967 = 0.000033– P(z<-4.6)< P(z<-3.99) = ~ 0

Class Problem• Assume X is normally distributed with a mean

of 5 and a standard deviation of 4• Determine the following

– P(X<11)– P(X>0)– P(3<X<7)– P(2 < X < 9)

Solution• P(X < 11) = = P(Z < 1.5) = 0.93319

• P(X > 0) = P(Z > (5/4)) = P(Z > 1.25) = 1 P(Z < 1.25) = 0.89435

• P(3 < X < 7) = = P(0.5 < Z < 0.5)=P(Z < 0.5) P(Z < 0.5)= 0.38292

• P(2 < X < 9) = =P(1.75 < Z < 1) = [P(Z < 1) P(Z < 1.75)] = 0.80128

4511ZP

4

574

53 ZP

4

594

52 ZP

Normal Approximation of Binomial Distribution

• Difficult to calculate probabilities when n is large

• Used to approximate binomial probabilities for cases in which n is large

• Gives approximate probability only

Approximation• If X is a binomial random variable

• is approximately a standard normal random variable

• Good for np>5 and n(1-p)>5• Accuracy of approximation

– Calculate the interval:

– If Interval lies in range 0 to n, normal approximation can be used

pnpnpXz 1

133 pnpnp

Normal Approximation of Poisson Distribution

• If X is a Poisson random variable with E(X)= and V(X)=

• Approximated a standard normal random variable

• Good for >5

Xz

Exponential Distribution• Poisson distribution defined a random variable to

be the number events during a given time interval or in a specified regions

• Time or distance between the events is another random variable that is often of interest

• Probability density function

• Mean and variance

xexf )(

121 ][ ,][ XVXE

Example• Suppose that the log-ons to a computer network

follow a Poisson process with an average of 3 counts per minute

a) What is the mean time between counts?b) What is the standard deviation of the time

between counts?c) Determine x such that the probability that at least

one count occurs before time x minutes is 0.95

Solutiona) E(X) = 1/ =1/3 = 0.333 minutesb) V(X) = 1/2 = 1/32 = 0.111, = 0.33c) Value of x

• Thus, x = 0.9986

95.01

3)(

3

0

3

0

3

x

xt

xt

ee

dtexXP

Class Problem• The time between the arrival of electronic

messages at your computer is exponentially distributed with a mean of two hours– (a) What is the probability that you do not

receive a message during a two-hour period?– (b) What is the expected time between your fifth

and sixth messages?

Solution• Let X denote the time until a message is

received. Then, X is an exponential random variable and

a) P(X > 2) =

b) E(X) = 2 hours.

12

2 2

2 2

1 0 3679e dx e ex x

/ / .

1 1 2/ ( ) /E X

Erlang Distribution• Describes the length until the first count is

obtained in a Poisson process• Generalization of the exponential distribution is

the length until r counts occur in a Poisson process.

• Random variable that equals the interval length until r counts occur in a Poisson process

• PDF

• Mean and Variance– E(X)=r/ and V(X)= r/2

... 1,2,3,r and 0for x ,)!1(

)(1

rexxf

xrr

Example• Errors caused by contamination on optical disks

occur at the rate of one error every bits. Assume the errors follow a Poisson distribution.

a) What is the mean number of bits until five errors occur?

b) What is the standard deviation of the number of bits until five errors occur?

c) The error-correcting code might be ineffective if there are three or more errors within 105 bits. What is the probability of this event?

Solutiona) X denote the number of bits until five errors occurb) Then, X has an Erlang distribution with r = 5 and =10-5

error per bit– E(X) = – V(X) =

c) Y denote the number of errors in 105 bits. Then, Y is a Poisson random variable with 10-5 error per bit which equals 1 error per 105 bits

r 5 105

r

2105 10 X 5 10 22360710

0803.0

1)2(1

)3(

!21

!11

!01 211101

eeeYP

YP

Gamma Function• Generalization of factorial function leads• Definition of gamma function for r > 0 • Integral (r) is finite• Using integration by parts it can be shown

– (r)=(r-1) (r-1)– if r is a positive integer (r)=(r-1)!

• Used in development of the gamma distribution

,)(0

1 dxexr xr

Gamma Distribution• X with the following PDF

• x > 0

• Gamma random variable with parameters >0 and r>0• If r is an integer, X has an Erlang distribution• Erlang is a special case of the gamma distribution• Mean and Variance

– E(X)=r/ and V(X)= r/2

forrexxf

xrr

)()(

1