Continuous Probability Distributions

1. Continuous Probability Distributions • Many continuous probability distributions, including: • Uniform • Normal • Gamma • Exponential • Chi-Squared • Lognormal • Weibull EGR 252 Spring 2011

2. Standard Normal Distribution • Table A.3: “Areas Under the Normal Curve” EGR 252 Spring 2011

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: 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 2011

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 k k = -0.125 from table in back of book Z = (x- μ) / σ Z = -0.125 = (x-40) / 6 X = 39.25 Z values 39.25 40 X values EGR 252 Spring 2011

5. 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: When we apply the theorem, we apply a continuity correction: add or subtract 0.5 from xto 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: EGR 252 Spring 2011

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 than 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 < _________) = _________ (Subtract 0.5 from 30 because we are looking for P (X< x). Less than 30 for a discrete distribution is 29 or less. EGR 252 Spring 2011

7. The Normal Approximation: Your Turn DRAW THE PICTURE!! 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 ? Since we are looking for P (X>x) we add 0.5 to x. EGR 252 Spring 2011

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. 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. 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 DRAW THE PICTURE!! EGR 252 Spring 2011

9. 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 EGR 252 Spring 2011

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 EGR 252 Spring 2011

11. 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 2011

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? EGR 252 Spring 2011

13. 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 What is the value of P(X ≤ 1)? Can we use a table? Must we integrate? Can we use Excel? EGR 252 Spring 2011

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) EGR 252 Spring 2011

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= αβ 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. EGR 252 Spring 2011