1. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 1 Continuous Probability Distributions Many continuous probability distributions, including: Uniform Normal Gamma Exponential Chi-Squared Lognormal Weibull

2. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 2 Standard Normal Distribution Table A.3: “Areas Under the Normal Curve”

3. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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 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%

4. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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 k k = -0.125 from table in back of book Z = (x- μ ) / σ Z = -0.125 = (x-40) / 6 X = 39.25 39.25 40 X values Z values

5. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 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 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:

6. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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 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?

7. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 7 The Normal Approximation: Your Turn 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 ? Since we are looking for P (X>x) we add 0.5 to x. z = ((50 + 0.5) – 40)/4.899 = 2.14 Answer: _____________ DRAW THE PICTURE!!

8. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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. 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!!

9. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 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

10. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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 σ 2 = αβ 2

11. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 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 σ 2 = β 2

12. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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?

13. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 2011 Slide 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?

14. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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)

15. JMB Ch6 Lecture 3 revised 2 EGR 252 Fall 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.