Download EGR252S12_Chapter6 Part 2 v4 JMB 2012

Continuous Probability Distributions
Part 2
• Many continuous probability distributions,
including:
 Uniform
 Normal
 Gamma
 Exponential
 Chi-Squared
 Lognormal
 Weibull
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 1
Review:
Standard Normal Random Variable
• Normal Distribution Review: the probability of X
taking on any value between x1 and x2 is given by:
x2
P ( x1  X  x2 )   n( x; , )dx 
x1
x2

x1
1
e
2 
 ( x   )2
2 2
dx
• To ease calculations, we define a normal
random variable
Z
X 

where Z is normally distributed with μ = 0 and σ2 = 1
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 2
Review:
Standard Normal Distribution
• Table A.3 Pages 735-736: “Areas under the
Normal Curve”
Standard Normal Distribution
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 3
Applications of the Normal Distribution
• A certain machine makes electrical resistors having a mean
resistance of 40 ohms and a standard deviation of 2 ohms. What
percentage of the resistors will have a resistance less than 44
ohms?
•
Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44
Z
Z
X 

44  40
2
2
-5
0
5
P(X<44) = P(Z< +2.0) = 0.9772
Therefore, we conclude that 97.72% will have a resistance less than 44
ohms.
What percentage will have a resistance greater than 44 ohms?
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 4
Normal Approximation to the Binomial
• If n is large and p is not close to 0 or 1,
or
if n is smaller but p is close to 0.5, then
the binomial distribution can be approximated by the
normal distribution using these equations:
P( X  x)


x

0
.
5

np

 P Z 


npq




x

0
.
5

np

P( X  x)  P Z 


npq


• Look at example 6.15, pg. 191 (9th edition)
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 5
Look at example 6.15, pg. 191-192
p = 0.4
n = 100
μ = np = ___________
σ = ____________
For X < 30 z = ((30-0.5) – 40)/4.899 )
P (Z < ((30-0.5) – 40)/4.899 )
= P (Z < -2.14)
= _________ (from table)
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 6
Your Turn
•
DRAW THE
PICTURE!!
Refer to the previous
example,
1. What is the probability that
more than 50 survive?
-5
0
5
-5
0
5
2. What is the probability that
exactly 45 survive?
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 7
Gamma & Exponential Distributions
• Related to the Poisson Process (discrete)
Number of occurrences in a given interval or
region
“Memoryless” process
• Sometimes we’re interested in the number of
events that occur in an area (eg flaws in a
square yard of cotton).
• Sometimes we’re interested in the time until a
certain number of events occur.
• Area and time are variables that are measured.
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 8
Gamma Distribution
• The density function of the random variable X with
gamma distribution having parameters α (number of
occurrences) and β (time or region).
x
1
f (x)  
x  1e 
 ( )
x > 0.
Gamma Distribution
(n )  (n  1)!
1
μ = αβ
σ2 = αβ2
f(x)
0.8
0.6
0.4
0.2
0
0
2
4
6
8
x
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 9
Exponential Distribution
• Special case of the gamma distribution with α = 1.
f (x) 
1

x
e
x > 0.
 Describes the time until or time between Poisson events.
μ=β
σ2 = β2
0
JMB Chapter 6 Part 2
5
10
EGR 252 Spring 2012
15
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25
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Slide 10
Is It a Poisson Process?
• For homework and exams in the introductory
statistics course, you will be told that the process
is Poisson.
 An average of 2.7 service calls per minute are
received at a particular maintenance center. The calls
correspond to a Poisson process. What is the
probability that up to a minute will elapse before 2
calls arrive?
 An average of 2.7 service calls per minute are
received at a particular maintenance center. The calls
correspond to a Poisson process. How long before the
next call?
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 11
Poisson Example Problem
An average of 2.7 service calls per minute are
received at a particular maintenance center. The
calls correspond to a Poisson process.
What is the probability that up to 1 minute will elapse
before 2 calls arrive?
 β = 1 / λ = 1 / 2.7 = 0.3704
 α=2
What is the value of P(X ≤ 1)
Can we use a table? No We must use integration.
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 12
Poisson Example Solution
An average of 2.7 service calls per minute are
received at a particular maintenance center. The calls
correspond to a Poisson process. What is the
probability that up to 1 minute will elapse before 2
calls arrive? The time until a number of Poisson
events occurs follows the gamma distribution.
β = 1/ 2.7 = 0.3704
α = 2 (calls)
1
P(X < 1) =  0 (1/ β2) x e-x/ β dx
1
2
= 2.7  0 x e -2.7x dx
= [-2.7xe-2.7x – e-2.7x] 01
= 1 – e-2.7 (1 + 2.7) = 0.7513
Using Excel: =GAMMADIST(1,2,1/2.7,TRUE)
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 13
Another Type of Question
An average of 2.7 service calls per minute are received at
a particular maintenance center. The calls correspond to a
Poisson process. What is the expected time before the next
call arrives?
Expected value = μ = α β
α = 1(call)
β = 1/2.7
μ = β = 0.3704 min.
We expect the next call to arrive in 0.3704 minutes.
When α = 1 the gamma distribution is known
as the exponential distribution.
JMB Chapter 6 Part 2
EGR 252 Spring 2012
Slide 14