Chapter 4Continuous Random Variables and their Probability Distributions

The Theoretical Continuous Distributions starring

The Rectangular The Normal The Exponential and The Weibull

Chapter 4B

Continuous Uniform DistributionA continuous RV X with probability density function

has a continuous uniform distribution or rectangular distribution

1( ) , f x a x b

b a

2

22 22 2

( )2( ) 2

( )( ) ( )

2 12

b b

aa

b b

a a

x x a bE X dx

b a b a

x a b b aV X x f x dx dx

b a

1 '( ) '

xx

aa

x x aF x dx

b a b a b a

a b

Rect( , )X a b

4-5 Continuous Uniform Random Variable

Mean and Variance

Using Continuous PDF’s

Given a pdf, f(x), a <= x <= b and and a <= m < n <= bP(m <= x <= n) =

( ) 1b

af x dx

( ) ( ) ( )n n

mmf x dx F x F n F m

10 10

55

20 20

1010

( ) 0.05, 0 20

(5 10) 0.05 0.05 0.05(10 5) 0.25

(10 30) 0.05 0.05 0.05(20 10) 0.50

If f x x

P x dx x

P x dx x

Problem 4-33

2 2

2

1 10

2 2

1 1 1 1, 0.577

12 12 3 3

b a

b a

1( ) 0.90 ( )

1 1 1 = ( )

1 1 2 20.90

x x

x x

x x

xx

P x X x f t dt dtb a

dt t x x x

x

Rect( 1,1)X

( 1) 1( )

1 ( 1) 2

x xF x

Let’s get Normal

Most widely used distribution; bell shaped curve

Histograms often resemble this shape Often seen in experimental results if a process is

reasonably stable & deviations result from a very large number of small effects – central limit theorem.

Variables that are defined as sums of other random variables also tend to be normally distributed – again, central limit theorem.

If the experimental process is not stable, some systematic trend is likely present (e.g., machine tool has worn excessively) a normal distribution will not result.

4-6 Normal Distribution

Definition 2( , )X n

4-6 Normal Distribution

The Normal PDF

http://www.stat.ucla.edu/~dinov/courses_students.dir/Applets.dir/NormalCurveInteractive.html

Normal IQs

4-6 Normal Distribution

Some useful results concerning the normal distribution

Normal Distributions

Standard Normal Distribution

A normal RV with is called a standard normal RV and is denoted as Z.Appendix A Table III provides probabilities of the form P(Z < z) where

You cannot integrate the normal density function in closed form.

Fig 4-13. Standard Normal Probability Function

20 and 1

( ) ( )z P Z z

Examples – standard normal

P(Z > 1.26) = 1 – P(Z 1.26) = 1 - .89616 = .10384

P(Z < -0.86) = .19490

P(Z > -1.37) = P(z < 1.37) = .91465

P(-1.25< Z<0.37) = P(Z<.0.37) – P(Z<-1.25) = .64431 - .10565 = .53866

P(Z < -4.6) = not found in table; prob calculator = .0000021

P(Z > z) = 0.05; P(Z < z) =.95; from tables z 1.65; from prob calc = 1.6449

P(-z < Z < z) = 0.99; P(Z<z) =.995; z = 2.58

Converting Normal RV’s to Standard Normal Variates (so you can use the tables!)

Any arbitrary normal RV can be converted to a standard normal RV using the following formula:

After this transformation, Z ~ N(0, 1)

2

2 2

[ ][ ] 0

1[ ] [ ] 1

XZ

X E XE Z E

XV Z V V X V

the number of standard deviations from the mean

4-6 Normal DistributionTo Calculate Probability

Converting Normal RV’s to Standard Normal Variates (an example)

For example, if X ~ N(10, 4)To determine P(X > 13):

XZ

13 1013 1.5

2

1 1.5 1 0.93319 .06681

XP X P P z

P z

from Table III

Converting Normal RV’s

A scaling and a shift are involved.

More Normal vs. Std Normal RV

X ~ N(10,4)

9 10 11 109 11 .5 .5

2 2

.5 .5 0.69146 0.30854 0.38292

XP X P P z

P z P z

Example 4-14 Continued(sometimes you need to work backward

Determine the value of x such that P(X x) = 0.98

10 10 10( ) 0.98

2 2 2

X x xP X x P P Z

II: P(Z ) 0.98

P(Z 2.05) 0.97982

10 ==> = 2.05

2 x = 14.1

That is, there is a 98% probability that a

current measurement is less than 14.1

Table z

x

X ~ N(10,4)

Check out this website

http://www.ms.uky.edu/~mai/java/stat/GaltonMachine.html

An Illustration of Basic Probability: The Normal Distribution

See the normal curve generated right in front of your very own eyes

4-8 Exponential Distribution

Definition ( )X Exp

The Shape of Things

Exponential Probability Distribution

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

X

f(X

)

lambda = .1 lambda = .5 lambda = 1.0

The Mean, Variance, and CDF

20 0

2 22 2

3 20

0 0

1 1

1 2 1 1

( ) 1 1

x x

x

xx uu x x

xe dx xe dx

x e dx

eF x e du e e

table ofdefiniteintegrals

What about the median?

( ) 1 .5

.5

ln .5

1ln .5 ln .5 .6931472

x

x

F x e

e

x

x

Next Example

Let X = a continuous random variable, the time to failure in operating hours of an electronic circuit ( 1/ 25hr)X Exp f(x) = (1/25) e-x/25

F(x) = 1 - e-x/25

= 1/ = E[X] = 25 hours

median = .6931472 (25) = 17.3287 hours

2 = V[X] = 252

= 25

Example

What is the probability there are no failures for 6 hours?

6

25 25

6

1( 6) 0.7866

25

x

P X e dx e

25( ) 1

(3 6) (6) (3) .2134 .1131 .1003

x

F x e

P X F F

( 1/ 25hr)X Exp

What is the probability that the time until the next failure is between 3 and 6 hours?

Exponential & Lack of Memory

Property: If X ~ exponential

This implies that knowledge of previous results (past history) does not affect future events.

An exponential RV is the continuous analog of a geometric RV & they both share this lack of memory property.

Example: The probability that no customer arrives in the next ten minutes at a checkout counter is not affected by the time since the last customer arrival. Essentially, it does not become more likely (as time goes by without a customer) that a customer is going to arrive.

1 2 1 2( ) ( )P X t t X t P X t

Proof of Memoryless Property

)B(P/)BA(P)B|A(P

A – the event that X < t1 + t2 and B – the event that X > t1

Chapter Two stuff!

1 1 2

1

1 21 1 2

1 1

1 1 21 2 1

1

( )

1 2 1

1

2 2

PrPr | Pr

Pr

1 1( ) ( )

1 ( )

1Pr

t t t

t

t tt t t

t t

t X t tX t t X t

X t

e eF t t F t

F t e

e ee e eF t X t

e e

Exponential as the Flip Side of the Poisson

If time between events is exponentially distributed, then the number of events in any interval has a Poisson distribution.

NT events till time T

Time between events has exponential

distribution

Time T

Time 0

Exponential and Poisson

Pr ( ) , for 0,1,2,...;

!

n tt eX t n n

n

Let X(t) = the number of events that occur in time t; assume X(t) ~ Pois(t) then E[X(t)] = t

Pr 1 ( ) Pr ( ) 0tT t F t e X t

Let T = the time until the next event; assume T ~ Exp() then E[T] = 1/

4-10 Weibull Distribution

Definition ( , )X W

The PDF in Graphical Splendor

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0t

f(t)

0.5

1.5

2.0

4.0

Beta

Delta = 2

More Splendor

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0.0 1.0 2.0 3.0 4.0 5.0 6.0t

f(t)

0.5

1.0

2.0

Delta

Beta = 1.5

4-10 Weibull Distribution

The Gamma Function

(x) = the gamma function = y e dy0x-1 -yz

(x) = (x -1) (x -1)

fine print: easier method is to use the prob calculator

4-10 Weibull DistributionExample 4-25

The Mode of a Distributiona measure of central tendency

f(t)

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40 50 60

The Mode of a Distributiona measure of central tendency

f(t)

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40 50 60

The Mode of a DistributionMAX -1

x-x

f(x) = ex 0

0-2 2 -22

x x- -

2 2

df(x) ( -1) x x = -e e

dx

2( 1) x

x -2- x- = 0e

( )x

-1 = 0

1

1

-1Mode = for

A Weibull Example The design life of the members used in constructing the

roof of the Weibull Building, a engineering marvel, has a Weibull distribution with = 80 years and = 2.4.

(80,2.4)X W

2.4100

80Pr{ 100} 1 (100) 1 1 .1812X F e

180 1 80 1.42 70.92 yr.

2.4

1

2.42.4 180 63.91yr.

2.4

-Mode =

Other Continuous Distributions Worth Knowing

Gamma Erlang is a special case of the gamma

Used in queuing analysis Beta

Like the triangular – used in the absence of data Used to model random proportions

Lognormal used to model repair times (maintainability) quantities that are a product of other quantities

(central limit theorem) Pearson Type V and Type VI

like lognormal – models task times

Picking a Distribution

We now have some distributions at our disposal.

Selecting one as an appropriate model is a combination of understanding the physical situation and data-fitting Some situations imply a distribution, e.g. arrivals

Poisson process is a good guess. Collected data can be tested statistically for a ‘fit’ to

distributions.

Next Week – Chapter 5

Double our pleasure by considering joint distributions.