Transcript EGR252S11_Chapter6Lecture2_JMB publish

Continuous Probability Distributions
• Many continuous probability distributions,
including:
 Uniform
 Normal
 Gamma
 Exponential
 Chi-Squared
 Lognormal
 Weibull
JMB Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 1
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 Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 2
Normal Distribution
The density function of the normal random variable X, with mean μ and
variance σ2, has the following properties:
 peak is both the mean and the mode and occurs at x = μ
 curve is symmetrical about a vertical axis through the mean
 total area under the curve and above the horizontal axis = 1.
 points of inflection are at x = μ + σ
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 Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 3
Standard Normal Random Variable
• 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
X 
Z

where Z is normally distributed with μ = 0 and σ2 = 1
JMB Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 4
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 Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 5
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 Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 6
The Normal Distribution “In Reverse”
•
Example:
Given a normal distribution with = 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.
•
Solution:
If P(Z < k) = 0.45, what is the value of k?
45% of area under curve less than k
k = -0.125 from table in back of book
-5
0
5
Z values
Z = (x- μ) / σ
Z = -0.125 = (x-40) / 6
X = 39.25
-5
0
5
39.25 40
X values
JMB Ch6 Lecture2 Review
EGR 252 Spring 2011
Slide 7
Your Turn
•
P(Z ≤ 1) =
Area to the left of the blue line
-5
•
0
5
P(Z ≥ -1) =
Area to the right of the blue line
-5
•
0
5
P(-2.5 ≤ Z ≤ 2.8) =
Area between the two lines
-5
•
0
5
For a normally distributed
variable for which the
mean is 30 and standard
deviation is 10, P(X < 40) =
Area to the left of the blue line
JMB Ch6 Lecture2 Review
-5
EGR 252 Spring 2011
0
30 40
x values for µ = 30
5
Slide 8