Download EGR252S11 Chapter6 Lecture1 JMB

Continuous Probability Distributions
• Many continuous probability distributions,
including:
 Uniform
 Normal
 Gamma
 Exponential
 Chi-Squared
 Lognormal
 Weibull
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 1
Uniform Distribution
• Simplest – characterized by the interval
endpoints, A and B.
1
f ( x; A, B ) 
BA
=0
A≤x≤B
elsewhere
• Mean and variance:
2
AB
(
B

A
)
2



and
2
12
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 2
Example: Uniform Distribution
A circuit board failure causes a shutdown of a
computing system until a new board is delivered. The
delivery time X is uniformly distributed between 1 and 5
days.
What is the probability that it will take 2 or more days for
the circuit board to be delivered?
1
1
f ( x;1,5) 

5 1
4
P ( x  2)  
JMB Chapter 6 Part 1 v2
5
2
1
dx
4
P ( x  2)  5 / 4  2 / 4  0.75
EGR 252 Spring 2011
Slide 3
Normal Distribution
• The “bell-shaped curve”
• Also called the Gaussian distribution
• The most widely used distribution in statistical
analysis
 forms the basis for most of the parametric tests we’ll
perform later in this course.
 describes or approximates most phenomena in
nature, industry, or research
• Random variables (X) following this distribution
are called normal random variables.
 the parameters of the normal distribution are μ and σ
(sometimes μ and σ2.)
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 4
Normal Distribution
• The density function of the normal random
variable X, with mean μ and variance σ2, is
1
n( x; , ) 
e
2 
 ( x   )2
2 2
all x.
Normal Distribution
P(x)
(μ = 5, σ = 1.5)
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
12
x
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 5
Standard Normal RV …
• Note: 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 1 v2
EGR 252 Spring 2011
Slide 6
Standard Normal Distribution
• Table A.3: “Areas Under the Normal Curve”
Standard Normal Distribution
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 7
Examples
• P(Z ≤ 1) =
-5
0
5
-5
0
5
-5
0
5
• P(Z ≥ -1) =
• P(-0.45 ≤ Z ≤ 0.36) =
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 8
Name:________________________
• Use Table A.3 to determine (draw the picture!)
1. P(Z ≤ 0.8) =
2. P(Z ≥ 1.96) =
3. P(-0.25 ≤ Z ≤ 0.15) =
4. P(Z ≤ -2.0 or Z ≥ 2.0) =
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 9
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
-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 1 v2
EGR 252 Spring 2011
Slide 10
The Normal Distribution “In Reverse”
•
Example:
Given a normal distribution with μ = 40 and σ = 6, find
the value of X for which 45% of the area under the
normal curve is to the left of X.
1) If P(Z < k) = 0.45,
k = ___________
2) Z = _______
-5
0
5
X = _________
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 11
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 the transformation:
X  np
Z
npq
• NOTE: add or subtract 0.5 from X to be sure the value
of interest is included (draw a picture to know which)
• Look at example 6.15, pg. 191 (8th edition)
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 12
Look at example 6.15, pg. 191-192
p = 0.4
n = 100
μ = ____________
σ = ______________
if x = 30, then z = _____________________
and, P(X < 30) = P (Z < _________) = _________
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 13
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 1 v2
EGR 252 Spring 2011
Slide 14
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.
• Typical problem statement: The length of time in days
between student complaints about too much homework
follows a gamma distribution with α = 3 and β = 7. What
is the probability that up to two weeks will elapse before
6 complaints are heard?
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 15
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 1 v2
EGR 252 Spring 2011
Slide 16
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 1 v2
5
10
EGR 252 Spring 2011
15
20
25
30
Slide 17
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 3.5 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 1 v2
EGR 252 Spring 2011
Slide 18
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 1 v2
EGR 252 Spring 2011
Slide 19
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?
β = 1/ 2.7 = 0.3704 α = 2
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
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 20
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
β = 1/2.7
μ = β = 0.3407 min.
We expect the next call to arrive in 0.3407 minutes.
When α = 1 the gamma distribution is known
as the exponential distribution.
JMB Chapter 6 Part 1 v2
EGR 252 Spring 2011
Slide 21