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Continuous Probability Distributions

Continuous Probability Distributions. Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull. Uniform Distribution. Simplest – characterized by the interval endpoints, A and B. A ≤ x ≤ B = 0 elsewhere

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Continuous Probability Distributions

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  1. Continuous Probability Distributions • Many continuous probability distributions, including: • Uniform • Normal • Gamma • Exponential • Chi-Squared • Lognormal • Weibull EGR 252 Spring 2009

  2. Uniform Distribution • Simplest – characterized by the interval endpoints, A and B. A ≤ x ≤ B = 0 elsewhere • Mean and variance: and EGR 252 Spring 2009

  3. 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? EGR 252 Spring 2009

  4. 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 μand σ(sometimes μand σ2.) EGR 252 Spring 2009

  5. (μ = 5, σ = 1.5) Normal Distribution • The density function of the normal random variable X, with mean μ and variance σ2, is all x. EGR 252 Spring 2009

  6. Standard Normal RV … • Note: 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 EGR 252 Spring 2009

  7. Standard Normal Distribution • Table A.3: “Areas Under the Normal Curve” EGR 252 Spring 2009

  8. Examples • P(Z ≤ 1) = • P(Z ≥ -1) = • P(-0.45 ≤ Z ≤ 0.36) = EGR 252 Spring 2009

  9. Your turn … • 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 orZ≥ 2.0) = EGR 252 Spring 2009

  10. 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: Xis 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? EGR 252 Spring 2009

  11. 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. • If P(Z < k) = 0.45, k = ___________ • Z = _______ X = _________ EGR 252 Spring 2009

  12. 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: • NOTE: add or subtract 0.5 from X to be sure the value of interest is included (draw a picture to know which) • Look at example 6.15, pg. 191 (8th edition) EGR 252 Spring 2009

  13. Look at example 6.15, pg. 191-192 p = 0.4 n = 100 μ= ____________ σ= ______________ if x = 30, then z = _____________________ and, P(X < 30) = P (Z < _________) = _________ EGR 252 Spring 2009

  14. Your Turn DRAW THE PICTURE!! • Refer to the previous example, • What is the probability that more than 50 survive? • What is the probability that exactly 45 survive? EGR 252 Spring 2009

  15. 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? EGR 252 Spring 2009

  16. 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 EGR 252 Spring 2009

  17. Exponential Distribution • Special case of the gamma distribution with α = 1. x > 0. • Describes the time until or time between Poisson events. μ = β σ2= β2 EGR 252 Spring 2009

  18. 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? EGR 252 Spring 2009

  19. Poisson 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 = 0.3704 • α = 2 What is the value of P(X ≤ 1) Can we use a table? No We must use integration. EGR 252 Spring 2009

  20. Poisson 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? β = 1/ 2.7 = 0.3704 α = 2 P(X < 1) = (1/ β2) x e-x/ β dx = 2.72 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x] 01 = 1 – e-2.7 (1 + 2.7) = 0.7513 EGR 252 Spring 2009

  21. 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= αβα = 1 β = 1/2.7 μ = β = 0.3407 min. We expect the next call to arrive in 0.3407 minutes. When α = 1 the gamma distribution is known as the exponential distribution. EGR 252 Spring 2009

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