Chapter 4 Continuous Random Variables and their Probability Distributions The Theoretical Continuous...

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.