Chapter 6 Continuous Probability Distributions. Learning objectives

Transcription

1 Chapter 6 Continuous s Slide 1 Learning objectives 1. Understand continuous probability distributions 2. Understand Uniform distribution 3. Understand Normal distribution 3.1. Understand Standard normal distribution 3.2. Understand Normal approximation of binomial distribution 4. Understand Exponential distribution 4.1. Understand relationship between Poisson and Exponetial distribution. Slide 2 1

2 Continuous s A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. Probability density function f(x) does not directly provide probability (i.e., f(c) = 0 for any c). Instead, the area under the graph of f(x) corresponding to a given interval does provide the probability. i.e., we alwayscompute the probability of interval, for example between 10 and 20. Slide 3 Approximation Uniform A random variable is uniformly distributed whenever the probability is proportional to the interval s length. The uniform probability density function is: f (x) = 1/(b a) for a < x < b = 0 elsewhere where: a = smallest value the variable can assume b = largest value the variable can assume Slide 4 2

3 Approximation Uniform Expected Value of x E(x) = (a + b)/2 Variance of x Var(x) = (b - a) 2 /12 Slide 5 Approximation Uniform Example: Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Slide 6 3

4 Approximation Uniform Uniform Probability Density Function f(x) = 1/10 for 5 < x < 15 = 0 elsewhere where: x = salad plate filling weight Expected value and variance of x E(x) = (a + b)/2 = (5 + 15)/2 = 10 Var(x) = (b - a) 2 /12 = (15 5) 2 /12 = 8.33 Slide 7 Approximation Uniform What is the probability that a customer will take between 12 and 15 ounces of salad? 1/10 f(x) P(12 < x < 15) = 1/10(3) = Salad Weight (oz.) x Slide 8 4

5 Approximation In-class exercise #1 (p. 228) #2 (p. 228) Slide 9 Approximation Normal The normal probability distribution is the most important distribution for describing a continuous random variable. It is widely used in statistical inference. It has been used in a wide variety of applications: Heights of people Test scores Amounts of rainfall Slide 10 5

6 Approximation Normal Normal Probability Density Function fx ( ) = σ 1 2π e 2 2 ( x µ ) /2σ where: µ = mean σ = standard deviation π = e = Slide 11 Approximation Normal Characteristics The range of x is from - to Symmetric; skewness = 0 The highest point = Mean =Median = Mode The total area under the curve is 1 (.5 to the left of the mean and.5 to the right) x Mean = median = mode Slide 12 6

7 Approximation Normal Characteristics (continued) Defined by its mean µ and its standard deviation σ. Mean determines the center (location) of the curve. Standard deviation σ determines how flat the curve is: larger values result in flatter curves. s = 15 s = 25 x Slide 13 Approximation Normal Characteristics (continued) 68.26% of values of a normal random variable are within +/- 1 standard deviation ofits mean % of values of a normal random variable are within +/- 2 standard deviations ofits mean % of values of a normal random variable are within +/- 3 standard deviations ofits mean. Slide 14 7

8 Approximation Normal Characteristics (continued) 99.72% 95.44% 68.26% µ 3σ µ 1σ µ 2σ µ µ + 1σ µ + 3σ µ + 2σ x Slide 15 Approximation A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. Converting Normal Distribution x to the Standard Normal Distribution z z = x σ µ x x Slide 16 8

9 Approximation Example: sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. Slide 17 Approximation Example: The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stockout, P(x > 20). Slide 18 9

10 Approximation Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. z = (x - µ)/σ = (20-15)/6 =.83 Step 2: Find the area under the standard normal curve to the left of z =.83. see next slide Slide 19 Approximation Cumulative Probability Table for the Distribution z P(z <.83) Slide 20 10

11 Approximation Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z =.83. P(z >.83) = 1 P(z <.83) = =.2033 Probability of a stockout P(x > 20) Slide 21 Approximation Solving for the Stockout Probability Area =.7967 Area = = z Slide 22 11

12 Approximation If the manager of wants the probability of a stockout to be no more than.05, what should the reorder point be? Slide 23 Approximation Solving for the Reorder Point Area =.9500 Area = z.05 z Slide 24 12

13 Approximation Solving for the Reorder Point Step 1: Find the z-value that cuts off an area of.05 in the right tail of the standard normal distribution. z We.9738 look up.9744 the.9750 complement of the. tail area. ( =..95).. Slide 25 Approximation Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of x. x = µ + z.05 σ = (6) = or 25 A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than).05. Slide 26 13

14 Approximation Solving for the Reorder Point By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about.20 to.05. This is a significant decrease in the chance that will be out of stock and unable to meet a customer s desire to make a purchase. Slide 27 Approximation Normal Approximation of Binomial Probabilities When the number of trials, n, becomes large, evaluating the binomial probability function by hand or with a calculator is difficult The normal probability distribution provides an easy-to-use approximation of binomial probabilities where n > 20, np > 5, and n(1 - p) > 5. Slide 28 14

15 Approximation Normal Approximation of Binomial Probabilities Set µ = np σ = np(1 p) Add and subtract 0.5 (a continuity correction factor) because a continuous distribution is being used to approximate a discrete distribution. For example, P(x = 10) is approximated by P(9.5 < x < 10.5). Slide 29 Approximation In-class exercise Standard normal distribution #11 (p. 240) #14 (p. 241) Normal approximation #26 (p. 245) #27 (p. 245) Slide 30 15

16 Approximation Exponential The exponential probability distribution is useful in describing the time it takes to complete a task. The exponential random variables can be used to describe: Time between vehicle arrivals at a toll booth Time required to complete a questionnaire Distance between major defects in a highway SLOW Slide 31 Approximation Exponential Density Function 1 f ( x ) = µ e x / µ for x > 0, µ > 0 where: µ = mean e = Cumulative Probabilities / 0 1 o µ Px ( x ) = e x where: x 0 = some specific value of x Slide 32 16

17 Approximation Exponential Example: Al s Full-Service Pump The time between arrivals of cars at Al s full-service gas pump follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less. Slide 33 Approximation Exponential f(x).4 P(x < 2) = /3 = = x Time Between Successive Arrivals (mins.) Slide 34 17

18 Approximation Exponential A property of the exponential distribution is that the mean, µ, and standard deviation, σ, are equal. Thus, the standard deviation, σ, and variance, σ 2, for the time between arrivals at Al s full-service pump are: σ = µ = 3 minutes σ 2 = (3) 2 = 9 Slide 35 Approximation Relationship between the Poisson and Exponential Distributions The Poisson distribution provides an appropriate description of the number of occurrences per interval The exponential distribution provides an appropriate description of the length of the interval between occurrences Slide 36 18

19 Approximation Relationship between the Poisson and Exponential Distributions Poisson distribution x e f ( x) = x! (The average number of cars that arrive at a car wash during 1 hour = 10) The average time between cars arriving is 1 hour = 0.1 hours/car 10 cars The corresponding exponential distribution for the time between the arrivals 1 f ( x) = e = 10e.1 x/.1 10 x Slide 37 Approximation In-class exercise #32 (p. 249) #36 (p. 249) Slide 38 19

20 End of Chapter 6 Slide 39 20