Download EGR252F11_Chapter6_Lecture3_JMB publish

Continuous Probability Distributions
• Many continuous probability distributions,
including:
 Uniform
 Normal
 Gamma
 Exponential
 Chi-Squared
 Lognormal
 Weibull
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 1
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 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 2
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
We conclude that 97.72% will have a resistance less than 44 ohms.
What percentage will have a resistance greater than 44 ohms? 2.28%
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 3
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 Lecture 3 revised 2
EGR 252 Fall 2011
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 the transformation:
X  np
Z
npq
• NOTE: When we apply the theorem, we apply a continuity
correction: add or subtract 0.5 from x to be sure the value of
interest is included (drawing a picture helps you determine whether
to add or subtract)
• To find the area under the normal curve to the left of x+ 0.5, use:


x

0
.
5

np

P Z 


npq


JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 5
Look at example 6.15, pg. 191-192
The probability that a patient recovers is p = 0.4
If 100 people have the disease, what is the
probability that less than 30 survive?
n = 100 μ = np =____
σ = sqrt(npq)= ____
if x = 30, then z =((30-0.5) – 40)/4.899 = -2.14
(Don’t forget the continuity correction.)
and, P(X < 30) = P (Z < _________) = _________
(Subtract 0.5 from 30 because we are looking for P (X< x).
Less than 30 for a discrete distribution is 29 or less. How would the equation
change if we wanted the probability that 30 or less survived?
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 6
The Normal Approximation: Your Turn
DRAW THE
PICTURE!!
Using μ and σ from the previous example,
What is the probability that more than
50 survive?
More than 50 for a discrete distribution is 51 or greater.
How do we include 51 and not 50 ?
-5
0
5
Since we are looking for P (X>x) we add 0.5 to x.
z = ((50 + 0.5) – 40)/4.899 = 2.14
Answer: _____________
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 7
The Normal Approximation: Your Turn 2
Using μ and σ from the previous example,
1. What is the probability that exactly 45 survive if we use the normal
approximation to the binomial?
DRAW THE
PICTURE!!
Hint: Find the area under the curve between two values.
Calculate P (X>x) by adding 0.5 to x.
Calculate P (X<x) by subtracting 0.5 from x.
Determine P(X=x) by calculating the difference.
-5
0
5
2. What is the probability that exactly 45 survive if we use the
binomial distribution? b(45;100,0.4)
Note that n=100 is too large for the tables. Determine the value using the binomial
equation.
0 ….
44 45 46 ….. 100
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 8
Gamma & Exponential Distributions
• Sometimes we’re interested in the number of events that
occur in a certain time period.
 Related to the Poisson Process
 Number of occurrences in a given interval or region
 “Memoryless” process
 Discrete
• If we are interested in the time until a certain number of
events occur, we will use continuous distributions
(exponential and gamma).
 The time until a number of Poisson events uses gamma
distribution with alpha = number of events and beta = mean time.
 See Examples 6.17 and 6.18
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 9
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 Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 10
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 Ch6 Lecture 3 revised 2
5
10
EGR 252 Fall 2011
15
20
25
30
Slide 11
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 Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 12
Poisson Example Problem 1
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
 x=1
x
1
f ( x)  
x 1e 
 ( )
where (n)  (n  1)!
What is the value of P(X ≤ 1)?
Can we use a table? Must we integrate?
Can we use Excel?
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 13
Poisson Example Solution Based on
the Gamma Distribution
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
P ( X  1)  2.7
 2.7
Excel
2
xe
 2.7 x
2

1
0
 e
xe
 2.7 x

 2.7 x
dx 
1
0
= 0.7513
= GAMMADIST (1, 2, 0.3704, TRUE)
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 14
Poisson Example Problem 2
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?
α = next call =1 λ = 1/ β = 2.7
Expected value of time (Gamma) = μ = α β
Since α =1
μ = β = 1 / 2.7 = 0.3407 min.
Note that when α = 1 the gamma distribution is known as
the exponential distribution.
JMB Ch6 Lecture 3 revised 2
EGR 252 Fall 2011
Slide 15