1. Continuous Probability Distributions Part 2
Many continuous probability distributions, including:
Uniform
Normal
Gamma
Exponential
Chi-Squared
Lognormal
Weibull
Uniform
Normal –
Gamma
Exponential
Chi-Squared
Lognormal
Weibull -
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2. Review: Standard Normal Random Variable
Normal Distribution Review: the probability of X taking on any value between x1 and x2 is given by:
To ease calculations, we define a normal random variable
where Z is normally distributed with μ = 0 and σ2 = 1
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3. Review: Standard Normal Distribution
Table A.3 Pages : “Areas under the Normal Curve”
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4. 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) =
Therefore, we conclude that 97.72% will have a resistance less than 44 ohms What percentage will have a resistance greater than 44 ohms?
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5. Gamma & Exponential Distributions
Related to the Poisson Process: Number of occurrences (discrete Ch.5) in a given interval or region
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 (continuous). Gamma distribution may apply.
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6. 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
Describes the time until a specified # of Poisson events occurs.
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7. Exponential Distribution
Special case of the gamma distribution with α = 1.
x > 0.
Describes the time until Poisson event occurs
Describes the time between Poisson events
μ = β
σ2 = β2
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8. 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 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. How long before the next call?
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9. Poisson/Gamma 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 =
α = 2 calls
x = 1 minute
What is the value of P(X ≤ 1)?
Can we use a table? No We must use integration.
β = 1/λ = 1/(2.7) =
α = 2
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10. Poisson/Gamma 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? The time until a number of Poisson events occurs follows the gamma distribution.
β = (1/ 2.7) = α = 2 (calls)
P(X < 1) = (1/ β2) x e-x/ β dx
= x e -2.7x dx
= [-2.7xe-2.7x – e-2.7x] 01
= 1 – e-2.7 ( ) =
P = 75.13%
Using Excel: =GAMMADIST( 1, 2, 1/2.7, TRUE )
β = 1/λ = 1/(2.7) =
α = 2
P(X < 1) = 0∫1 (1/ β2) x e-x/ β dx = ∫1 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x]01
= 1 – e-2.7 ( ) =
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11. 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(call) β = 1/2.7
μ = α β = (1) (0.3704) min.
We expect the next call to arrive in minutes.
When α = 1 the gamma distribution is known
as the exponential distribution.
Exponential distribution
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