Chapter 6 - Continuous Probability s Chapter 6 Continuous Probability s Uniform Probability Normal Probability f () Uniform f () Normal Continuous Probability s A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval. Slide 1 Slide 2 Continuous Probability s The probability of the random variable assuming a value within some given interval from 1 to 2 is defined to be the area under the graph of the probability density function between 1 and 2. Uniform Probability A random variable is uniformly distributed whenever the probability is proportional to the interval s length. The uniform probability density function is: f () Uniform f () Normal f () = 1/(b a) for a < < b = 0 elsewhere 1 2 1 2 where: a = smallest value the variable can assume b = largest value the variable can assume Slide 3 Slide 4 Uniform Probability Epected Value of Variance of E() = (a + b)/2 Var() = (b - a) 2 /12 where: a = smallest value the variable can assume b = largest value the variable can assume Uniform Probability Eample Slater buffet 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. Uniform Probability Density Function f() = 1/10 for 5 < < 15 = 0 elsewhere where: = salad plate filling weight Slide 5 Slide 6 1
Chapter 6 - Continuous Probability s Uniform Probability Epected Value of E() = (a + b)/2 = (5 + 15)/2 = 10 Uniform Probability Uniform Probability for Salad Plate Filling Weight f() Variance of Var() = (b - a) 2 /12 = (15 5) 2 /12 = 8.33 1/10 0 5 10 15 Salad Weight (oz.) Slide 7 Slide 8 Uniform Probability What is the probability that a customer will take between 12 and 15 ounces of salad? 1/10 f() P(12 < < 15) = 1/10(3) =.3 Uniform Probability (Another Eample) Eample: Flight time of an airplane traveling from Chicago to New York Suppose the flight time can be any value in the interval from 120 minutes to 140 minutes. Uniform Probability Density Function 0 5 10 12 15 Salad Weight (oz.) f() = 1/20 for 120 < < 140 = 0 elsewhere where: = Flight time of an airplane traveling from Chicago to New York Slide 9 Slide 10 Uniform Probability Another Eample Uniform Probability Epected Value of E() = (a + b)/2 = (120 + 140)/2 = 130 Variance of Var() = (b - a) 2 /12 = (140 120) 2 /12 = 33.33 Slide 11 Eample: Flight time of an airplane traveling from Chicago to New York Slide 12 2
Chapter 6 - Continuous Probability s Uniform Probability Eample: Flight time of an airplane traveling from Chicago to New York Probability of a flight time between 120 and 130 minutes P(120 < < 130) = 1/20(10) =.5 Area as a Measure of Probability The area under the graph of f() and probability are identical. This is valid for all continuous random variables. The probability that takes on a value between some lower value 1 and some higher value 2 can be found by computing the area under the graph of f() over the interval from 1 to 2. Slide 13 Slide 14 Normal Probability 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 including: Heights of people Rainfall amounts Test scores Scientific measurements Abraham de Moivre, a French mathematician, published The Doctrine of Chances in 1733. He derived the normal distribution. Normal Probability Normal Probability Density Function f ( ) where: 1 e 2 = mean = standard deviation = 3.14159 e = 2.71828 2 2 ( ) / 2 Slide 15 Slide 16 Normal Probability The distribution is symmetric; its skewness measure is zero. Normal Probability The entire family of normal probability distributions is defined by its mean and its standard deviation. Standard Deviation Mean Slide 17 Slide 18 3
Chapter 6 - Continuous Probability s Normal Probability Normal Probability The highest point on the normal curve is at the mean, which is also the median and mode. The mean can be any numerical value: negative, zero, or positive. -10 0 25 Slide 19 Slide 20 Normal Probability The standard deviation determines the width of the curve: larger values result in wider, flatter curves. = 15 Normal Probability Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and.5 to the right). = 25.5.5 Slide 21 Slide 22 Normal Probability Normal Probability (basis for the empirical rule) 68.26% of values of a normal random variable are within +/- 1 standard deviation of its mean. (basis for the empirical rule) 99.72% 95.44% 68.26% 95.44% of values of a normal random variable are within +/- 2 standard deviations of its mean. 99.72% of values of a normal random variable are within +/- 3 standard deviations of its mean. 3 1 2 + 1 + 3 + 2 Slide 23 Slide 24 4
Chapter 6 - Continuous Probability s 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. The letter z is used to designate the standard normal random variable. 1 Converting to the Standard Normal z 0 z We can think of z as a measure of the number of standard deviations is from. Slide 25 Slide 26 Using ecel to compute standard normal probabilities Ecel has two functions for computing probabilities and z values for a standard normal probability distribution. Using Ecel to compute Standard Normal Probabilities Ecel Formula Worksheet NORM.S.DIST function computes the cumulative probability given a z value. NORM.S.INV function computes the z value given a cumulative probability. S in the function names reminds us that these functions relate to the standard normal probability distribution. Slide 27 Slide 28 Using Ecel to compute Standard Normal Probabilities Ecel Formula Worksheet Eample: Grear Tire Company Problem Grear Tire company has developed a new steel-belted radial tire to be sold through a chain of discount stores. But before finalizing the tire mileage guarantee policy, Grear s managers want probability information about the number of miles of tires will last. It was estimated that the mean tire mileage is 36,500 miles with a standard deviation of 5000. The manager now wants to know the probability that the tire mileage will eceed 40,000. Slide 29 P( > 40,000) =? Slide 30 5
Chapter 6 - Continuous Probability s Eample: Grear Tire Company Problem Solving for the Probability Step 1: Convert to standard normal distribution. z = ( - )/ = (40,000 36,500)/5,000 =.7 Step 2: Find the area under the standard normal curve to the left of z =.7. Eample: Grear Tire Company Problem Cumulative Probability Table for the Standard Normal z.00.01.02.03.04.05.06.07.08.09............5.6915.6950.6985.7019.7054.7088.7123.7157.7190.7224.6.7257.7291.7324.7357.7389.7422.7454.7486.7517.7549.7.7580.7611.7642.7673.7704.7734.7764.7794.7823.7852.8.7881.7910.7939.7967.7995.8023.8051.8078.8106.8133.9.8159.8186.8212.8238.8264.8289.8315.8340.8365.8389........... P(z <.7) =.7580 Slide 31 Slide 32 Eample: Grear Tire Company Problem Solving for the Probability Step 3: Compute the area under the standard normal curve to the right of z =.7 P(z >.7) = 1 P(z <.7) = 1-.7580 =.2420 Eample: Grear Tire Company Problem Slide 33 Slide 34 Eample: Grear Tire Company Problem Eample: Grear Tire Company Problem Area =.7580 Area = 1 -.7580 =.2420 What should be the guaranteed mileage if Grear wants no more than 10% of tires to be eligible for the discount guarantee? 0.7 z (Hint: Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.) Slide 35 Slide 36 6
Chapter 6 - Continuous Probability s Eample: Grear Tire Company Problem Solving for the guaranteed mileage Eample: Grear Tire Company Problem - Solving for the guaranteed mileage Step 1: Find the z value that cuts off an area of.1 in the left tail of the standard normal distribution. z.00.01.02.03.04.05.06.07.08.09........... -1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 Slide 37-1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681-1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823-1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170........... Slide 38 From the table we see that z = -1.28 cuts off an area of 0.1 in the lower tail. Step 2: Convert z.1 to the corresponding value of. = + z.1 = 36500-1.28 (5000) = 30,100 Thus a guarantee of 30,100 miles will meet the requirement that approimately 10% of the tires will be eligible for the guarantee. Slide 39 Using Ecel to Compute Normal Probabilities Ecel has two functions for computing cumulative probabilities and values for any normal distribution: NORM.DIST is used to compute the cumulative probability given an value. NORM.INV is used to compute the value given a cumulative probability. Slide 40 Eample: Pep Zone Pep Zone 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. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment 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 during replenishment lead-time. In other words, what is the probability that demand during lead-time will eceed 20 gallons? P( > 20) =? Solving for the Stockout Probability Step 1: Convert to the standard normal distribution. z = ( - )/ = (20-15)/6 =.83 Step 2: Find the area under the standard normal curve to the left of z =.83. Slide 41 Slide 42 7
Chapter 6 - Continuous Probability s Solving for the Stockout Probability 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) = 1-.7967 =.2033 Area =.7967 Area = 1 -.7967 =.2033 Probability of a stockout P( > 20) 0.83 z Slide 43 Slide 44 If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than.05, what should the reorder point be? Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of. z = ( - )/ z ( - ) = + z = + z.05 = 15 + 1.645(6) = 24.87 or 25 A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than).05. Slide 45 Slide 46 Normal Probability Solving for the Reorder Point Solving for the Reorder Point Probability of no stockout during replenishment lead-time =.95 Probability of a stockout during replenishment lead-time =.05 By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about.20 to.05. 15 24.87 This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer s desire to make a purchase. Slide 47 Slide 48 8