Download Chapter 4 Continuous Random Variables and their Probability

Chapter 4
Continuous Random Variables and
their Probability Distributions
The Theoretical Continuous Distributions starring




The Rectangular
The Normal
The Exponential
and The Weibull
Chapter 4B
Continuous Uniform Distribution
A continuous RV X with probability density function
X  Rect(a, b)
1
f ( x) 
, a xb
ba
has a continuous uniform distribution or rectangular
distribution
1
x'
xa
F ( x)  
dx ' 

ba
ba a ba
a
x
x
E( X )  
b
a
2
x
x
dx 
ba
2(b  a )
b
a
b
b
a
a
V ( X )   x 2 f ( x)dx   2  
ab

2
a
b
2
x
a

b
(
b

a
)


dx  
 
ba
12
 2 
2
2
4-5 Continuous Uniform Random
Variable
Mean and Variance
Using Continuous PDF’s
b
Given a pdf, f(x), a <= x <= b and  f ( x)dx  1
a
and a <= m < n <= b
n
n
P(m <= x <= n) =
f ( x)dx  F  x   F (n)  F (m)

m
m
If f ( x)  0.05, 0  x  20
10
P(5  x  10)   0.05dx  0.05 x 5  0.05(10  5)  0.25
10
5
20
P(10  x  30)   0.05dx  0.05 x 10  0.05(20  10)  0.50
10
20
Problem 4-33
X  Rect(1,1)
b  a 1 1


0
2
2
b  a

 
2
1  1


2
1
1
 , 
 0.577
12
12
3
3
x
x
1
P( x  X  x)  0.90   f (t )dt  
dt 
x
x b  a
x
1
1 x
1
=
dt  t  x  ( x  x)  x
x 11
2
2
 x  0.90
2
x  (1) x  1
F ( x) 

1  (1)
2
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
X
n( ,  2 )
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
  0 and  is
 1called a standard normal
A normal RV with
RV and is denoted as Z.
Appendix A Table III provides probabilities of the form P(Z < z)
where
( z )  P( Z  z )
2
You cannot integrate the normal density function in closed form.
Fig 4-13. Standard Normal Probability Function
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)
Z
X 

the number of standard
deviations from the mean
 X    E[ X ]   
E[ Z ]  E 

   0


  
  
2
X  1
  
V [Z ]  V 
 2 V [ X ] V    2  1

   
  
4-6 Normal Distribution
To Calculate Probability
Converting Normal RV’s to Standard Normal
Variates (an example)
Z
X 

For example, if X ~ N(10, 4)
To determine P(X > 13):
 X   13  10 
P  X  13  P 

  P  z  1.5
2 
 
 1  P  z  1.5  1  0.93319  .06681
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 X   11  10 
P 9  X  11  P 


  P .5  z  .5

2 
 2
 P  z  .5  P  z  .5  0.69146  0.30854  0.38292
Example 4-14 Continued
X ~ N(10,4)
(sometimes you need to work backward
Determine the value of x such that P(X  x) = 0.98
x  10 
 X  10 x  10 

P( X  x)  P 

  P Z 
  0.98
2 
2 
 2

Table II: P(Z  z )  0.98
P(Z  2.05)  0.97982
x  10
==>
= 2.05
2
x = 14.1
That is, there is a 98% probability that a
current measurement is less than 14.1
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
1.2
1
f(X)
0.8
0.6
0.4
0.2
0
0
1
2
3
4
X
lambda = .1
lambda = .5
lambda = 1.0
5
The Mean, Variance, and CDF

    xe
 x
0

dx    xe
2  x
0
x
F ( x)    e
0
 x
0
   x e
2

 u
 1  1
dx    2  
  
2
2
1
1
 2  1
dx       3      2

   
 u
x
e 
 x
 x
du   


e

1

1

e




0
table of
definite
integrals
What about the median?
F ( x)  1  e   x  .5
 x
e  .5
 x  ln .5
1
x   ln .5    ln .5  .6931472 

Next Example
Let X = a continuous random variable, the time to
failure in operating hours of an electronic circuit
f(x) = (1/25) e-x/25
X
Exp(  1/ 25 hr)
F(x) = 1 - e-x/25
 = 1/ = E[X] = 25 hours
median = .6931472 (25) = 17.3287 hours
2 = V[X] = 252
 = 25
Example
X
Exp(  1/ 25 hr)
What is the probability there are no failures for 6 hours?
P( X  6)  

6
6

1  25x
e dx  e 25  0.7866
25
What is the probability that the time until the next
failure is between 3 and 6 hours?

x
25
F ( x)  1  e
P(3  X  6)  F (6)  F (3)  .2134  .1131  .1003
Exponential & Lack of Memory
Property: If X ~ exponential
P( X  t1  t2 X  t1 )  P ( X  t2 )
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.
Chapter Two stuff!
P(A | B)  P(A  B) / P(B)
Proof of Memoryless Property
 Pr t1  X  t1  t2  
Pr  X  t1  t2 | X  t1  Pr 

Pr  X  t1


  t1
  ( t1  t2 )




1

e

1

e
F (t1  t2 )  F (t1 ) 




1  F (t1 )
e  t1

e
  t1
  t1   t2
e e
e  t1

e  t1 1  e  t2 
e
  t1
 F  t2   Pr  X  t2 
A – the event that X < t1 + t2 and B – the event that X > t1
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 0
Time between events has
exponential distribution
Time T
Exponential and Poisson
Let X(t) = the number of events that occur in time t;
assume X(t) ~ Pois(t) then E[X(t)] = t
Pr  X (t )  n 
 t  e  t
n
n!
, for n  0,1, 2,...;
Let T = the time until the next event;
assume T ~ Exp() then E[T] = 1/ 
Pr T  t  1  F (t )  et  Pr  X (t )  0
4-10 Weibull Distribution
Definition
X
W ( ,  )
The PDF in Graphical Splendor
1.0
f(t)
0.9
Beta
0.8
0.5
0.7
1.5
2.0
0.6
4.0
Delta = 2
0.5
0.4
0.3
0.2
0.1
0.0
0.0
-0.1
1.0
2.0
3.0
4.0
5.0
6.0
t
More Splendor
1.6
f(t)
1.4
Delta
1.2
0.5
1.0
1
2.0
0.8
0.6
Beta = 1.5
0.4
0.2
0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t
4-10 Weibull Distribution
The Gamma Function
 (x) = the gamma function =
zy

0
x-1 -y
 (x) = (x - 1) (x - 1)
fine print: easier method is to use the prob calculator
e dy
4-10 Weibull Distribution
Example 4-25
The Mode of a Distribution
a measure of central tendency
f(t)
0.06
0.05
0.04
0.03
0.02
0.01
0
0
10
20
30
40
50
60
The Mode of a Distribution
a measure of central tendency
f(t)
0.06
0.05
0.04
0.03
0.02
0.01
0
0
10
20
30
40
50
60
The Mode of a Distribution
MAX

f(x) =
x0

df(x)  (  - 1)  x 
=
 
2
dx
 

e
x
- 
 

 
x  -2

 -2

2

x
(  - 1)     = 0
 
e
 -1

 x  - x 
  e  
 
x
- 
 

 x
- 2 
  
2
2  -2
e
x
- 
 

0
 


x
(  -1)  
=0



1

  -1
Mode =  
 for   1
  
A Weibull Example

X
W (80, 2.4)
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.

Pr{ X  100}  1  F (100)  1  1  e

1 

  80 1 
  80 1.42   70.92 yr.
 2.4 
 100 


 80 
1
2.4
 2.4 - 1 
Mode = 80 
  63.91yr.
 2.4 
2.4

  .1812

Other Continuous Distributions Worth
Knowing

Gamma

Erlang is a special case of the gamma


Beta



Like the triangular – used in the absence of data
Used to model random proportions
Lognormal



Used in queuing analysis
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.