Download EGR252S13_Chapter6 Part 1 v5 JMB publish

Continuous Probability Distributions
• Many continuous probability distributions,
including:
 Uniform
 Normal
 Gamma
 Exponential
 Chi-Squared
 Lognormal
 Weibull
Weibull PDF
JMB Chapter 6 Part 1 v5
Source: www.itl.nist.gov
EGR 252 2013
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 v5
EGR 252 2013
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 v5
5
2
1
dx
4
P ( x  2)  5 / 4  2 / 4  0.75
EGR 252 2013
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 v5
EGR 252 2013
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 v5
EGR 252 2013
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 v5
EGR 252 2013
Slide 6
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 1 v5
EGR 252 2013
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 v5
EGR 252 2013
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 v5
EGR 252 2013
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
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 v5
EGR 252 2013
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.
Step 1
If P(Z < z) = 0.45,
z = _______ (from Table A.3)
Why is z negative?
Step 2
Z
X 

45%
therefore X  Z  
-5
0
5
X = _________
JMB Chapter 6 Part 1 v5
EGR 252 2013
Slide 11