jmb chapter 6 lecture 3 egr 252 spring 2011 slide 1 continuous probability distributions many...
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JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 1
Continuous Probability Distributions
• Many continuous probability distributions, including: UniformNormalGammaExponentialChi-SquaredLognormalWeibull
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 2
Standard Normal Distribution
Standard Normal Distribution
-5 -4 -3 -2 -1 0 1 2 3 4 5
Z
• Table A.3: “Areas Under the Normal Curve”
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 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
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?
-5 0 5
2
4044
Z
XZ
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 4
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 kk = -0.125 from table in back of book
Z = (x- μ) / σ
Z = -0.125 = (x-40) / 6
X = 39.25
-5 0 5
-5 0 5
39.25 40
X values
Z values
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 5
Normal Approximation to the Binomial• If n is large and p is not close to 0 or 1,
orif n is smaller but p is close to 0.5, then
the binomial distribution can be approximated by the normal distribution using the transformation:
• 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:
npq
npXZ
npq
npxZP
05.
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 6
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 then 30 survive?
n = 100 μ = np =____ σ = sqrt(npq)= ____
if x = 30, then z = _____________ (Don’t forget the continuity correction.)
and, P(X < 30) = P (Z < _________) = _________(Less than 30 for a discrete distribution is 29 or less. Adding .5 to 29 insures we “cover” the maximum value.)
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 7
The Normal Approximation: Your Turn
Using μ and σ from the previous example,
1. 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
DRAW THE PICTURE!!
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 8
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?
Hint: Find the area under the curve between two values.
2. What is the probability that exactly 45 survive if we use the binomial distribution? b(45;100,0.4)
-5 0 5
DRAW THE PICTURE!!
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 9
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 Lecture 3 EGR 252 Spring 2011 Slide 10
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 > 0.
μ = αβ
σ2 = αβ2
)!1()(
)(
1)( 1
nn
exxfx
Gamma Distribution
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8
x
f(x)
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 11
Exponential Distribution
0 5 10 15 20 25 30
• Special case of the gamma distribution with α = 1.
x > 0.
Describes the time until or time between Poisson events.
μ = β
σ2 = β2
x
exf
1
)(
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 12
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 Lecture 3 EGR 252 Spring 2011 Slide 13
Poisson Example Problem 1An 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
What is the value of P(X ≤ 1)?
Can we use a table? Must we integrate?
Can we use Excel?
)!1()()(
1)( 1
nnwhereexxf
x
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 14
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
= 0.7513
Excel = GAMMADIST (1, 2, 0.3704, TRUE)
1
0
7.27.22
7.21
0
2
7.2
7.2)1(
xx
x
eex
dxexXP
JMB Chapter 6 Lecture 3 EGR 252 Spring 2011 Slide 15
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.
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