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1 1 Slide © 2009 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University

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2 2 Slide © 2009 Thomson South-Western. All Rights Reserved Chapter 6 Continuous Probability Distributions n Uniform Probability Distribution n Normal Probability Distribution n Normal Approximation of Binomial Probabilities n Exponential Probability Distribution f ( x ) x x Uniform x Normal x x Exponential

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3 3 Slide © 2009 Thomson South-Western. All Rights Reserved Continuous Probability Distributions n A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. n It is not possible to talk about the probability of the random variable assuming a particular value. n Instead, we talk about the probability of the random variable assuming a value within a given interval.

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4 4 Slide © 2009 Thomson South-Western. All Rights Reserved Continuous Probability Distributions n The probability of the random variable assuming a value within some given interval from x 1 to x 2 is defined to be the area under the graph of the probability density function between x 1 and x 2. f ( x ) x x Uniform x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x Normal x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 Exponential x x x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2

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5 5 Slide © 2009 Thomson South-Western. All Rights Reserved Uniform Probability Distribution where: a = smallest value the variable can assume b = largest value the variable can assume b = largest value the variable can assume f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere n A random variable is uniformly distributed whenever the probability is proportional to the interval’s length. n The uniform probability density function is:

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6 6 Slide © 2009 Thomson South-Western. All Rights Reserved Var( x ) = ( b - a ) 2 /12 E( x ) = ( a + b )/2 Uniform Probability Distribution n Expected Value of x n Variance of x

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7 7 Slide © 2009 Thomson South-Western. All Rights Reserved Uniform Probability Distribution n 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.

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8 8 Slide © 2009 Thomson South-Western. All Rights Reserved n Uniform Probability Density Function f ( x ) = 1/10 for 5 < x < 15 f ( x ) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere f ( x ) = 1/10 for 5 < x < 15 f ( x ) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere where: x = salad plate filling weight x = salad plate filling weight Uniform Probability Distribution

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9 9 Slide © 2009 Thomson South-Western. All Rights Reserved n Expected Value of x n Variance of x E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33 Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33 Uniform Probability Distribution

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10 Slide © 2009 Thomson South-Western. All Rights Reserved n Uniform Probability Distribution for Salad Plate Filling Weight f(x)f(x) f(x)f(x) x x 5 5 10 15 1/10 Salad Weight (oz.) Uniform Probability Distribution

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11 Slide © 2009 Thomson South-Western. All Rights Reserved f(x)f(x) f(x)f(x) x x 5 5 10 15 1/10 Salad Weight (oz.) P(12 < x < 15) = 1/10(3) =.3 What is the probability that a customer What is the probability that a customer will take between 12 and 15 ounces of salad? will take between 12 and 15 ounces of salad? 12 Uniform Probability Distribution

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12 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n The normal probability distribution is the most important distribution for describing a continuous random variable. n It is widely used in statistical inference. n It has been used in a wide variety of applications: Heights of people Heights of people Scientific measurements Scientific measurements Test scores Test scores Amounts of rainfall Amounts of rainfall

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13 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n Normal Probability Density Function = mean = standard deviation = 3.14159 e = 2.71828 where:

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14 Slide © 2009 Thomson South-Western. All Rights Reserved The distribution is symmetric; its skewness The distribution is symmetric; its skewness measure is zero. measure is zero. The distribution is symmetric; its skewness The distribution is symmetric; its skewness measure is zero. measure is zero. Normal Probability Distribution n Characteristics x

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15 Slide © 2009 Thomson South-Western. All Rights Reserved The entire family of normal probability The entire family of normal probability distributions is defined by its mean and its distributions is defined by its mean and its standard deviation . standard deviation . The entire family of normal probability The entire family of normal probability distributions is defined by its mean and its distributions is defined by its mean and its standard deviation . standard deviation . Normal Probability Distribution n Characteristics Standard Deviation Mean x

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16 Slide © 2009 Thomson South-Western. All Rights Reserved The highest point on the normal curve is at the The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. The highest point on the normal curve is at the The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. Normal Probability Distribution n Characteristics x

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17 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n Characteristics -10020 The mean can be any numerical value: negative, The mean can be any numerical value: negative, zero, or positive. zero, or positive. The mean can be any numerical value: negative, The mean can be any numerical value: negative, zero, or positive. zero, or positive. x

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18 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n Characteristics = 15 = 25 The standard deviation determines the width of the curve: larger values result in wider, flatter curves. The standard deviation determines the width of the curve: larger values result in wider, flatter curves. x

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19 Slide © 2009 Thomson South-Western. All Rights Reserved Probabilities for the normal random variable are Probabilities for the normal random variable are given by areas under the curve. The total area given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and under the curve is 1 (.5 to the left of the mean and.5 to the right)..5 to the right). Probabilities for the normal random variable are Probabilities for the normal random variable are given by areas under the curve. The total area given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and under the curve is 1 (.5 to the left of the mean and.5 to the right)..5 to the right). Normal Probability Distribution n Characteristics.5.5 x

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20 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n Characteristics of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean. of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean.68.26%68.26% +/- 1 standard deviation of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean. of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean. 95.44%95.44% +/- 2 standard deviations of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean. of values of a normal random variable of values of a normal random variable are within of its mean. are within of its mean.99.72%99.72% +/- 3 standard deviations

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21 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Probability Distribution n Characteristics x – 3 – 1 – 2 + 1 + 2 + 3 68.26% 95.44% 99.72%

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22 Slide © 2009 Thomson South-Western. All Rights Reserved Standard Normal Probability Distribution A random variable having a normal distribution A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability said to have a standard normal probability distribution. distribution. A random variable having a normal distribution A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability said to have a standard normal probability distribution. distribution.

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23 Slide © 2009 Thomson South-Western. All Rights Reserved 0 z The letter z is used to designate the standard The letter z is used to designate the standard normal random variable. normal random variable. The letter z is used to designate the standard The letter z is used to designate the standard normal random variable. normal random variable. Standard Normal Probability Distribution

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24 Slide © 2009 Thomson South-Western. All Rights Reserved n Converting to the Standard Normal Distribution Standard Normal Probability Distribution We can think of z as a measure of the number of standard deviations x is from .

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25 Slide © 2009 Thomson South-Western. All Rights Reserved Standard Normal Probability Distribution n Example: 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.

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26 Slide © 2009 Thomson South-Western. All Rights Reserved The store manager is concerned that sales are being The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It ha 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). Standard Normal Probability Distribution n Example: Pep Zone

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27 Slide © 2009 Thomson South-Western. All Rights Reserved z = ( x - )/ z = ( x - )/ = (20 - 15)/6 = (20 - 15)/6 =.83 =.83 z = ( x - )/ z = ( x - )/ = (20 - 15)/6 = (20 - 15)/6 =.83 =.83 n Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. see next slide see next slide Standard Normal Probability Distribution

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28 Slide © 2009 Thomson South-Western. All Rights Reserved n Cumulative Probability Table for the Standard Normal Distribution P ( z <.83) Standard Normal Probability Distribution

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29 Slide © 2009 Thomson South-Western. All Rights Reserved P ( z >.83) = 1 – P ( z.83) = 1 – P ( z <.83) = 1-.7967 = 1-.7967 =.2033 =.2033 P ( z >.83) = 1 – P ( z.83) = 1 – P ( z <.83) = 1-.7967 = 1-.7967 =.2033 =.2033 n Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. Probability of a stockout of a stockout P ( x > 20) Standard Normal Probability Distribution

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30 Slide © 2009 Thomson South-Western. All Rights Reserved n Solving for the Stockout Probability 0.83 Area =.7967 Area = 1 -.7967 =.2033 =.2033 z Standard Normal Probability Distribution

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31 Slide © 2009 Thomson South-Western. All Rights Reserved n Standard Normal Probability Distribution If the manager of Pep Zone wants the probability of a stockout to be no more than.05, what should the reorder point be? Standard Normal Probability Distribution

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32 Slide © 2009 Thomson South-Western. All Rights Reserved n Solving for the Reorder Point 0 Area =.9500 Area =.0500 z z.05 Standard Normal Probability Distribution

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33 Slide © 2009 Thomson South-Western. All Rights Reserved n 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 in the right tail of the standard normal distribution. distribution. Step 1: Find the z -value that cuts off an area of.05 in the right tail of the standard normal in the right tail of the standard normal distribution. distribution. We look up the complement of the tail area (1 -.05 =.95) Standard Normal Probability Distribution

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34 Slide © 2009 Thomson South-Western. All Rights Reserved n Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of x. x = + z.05 x = + z.05 = 15 + 1.645(6) = 24.87 or 25 = 24.87 or 25 x = + z.05 x = + z.05 = 15 + 1.645(6) = 24.87 or 25 = 24.87 or 25 A reorder point of 25 gallons will place the probability A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than).05. of a stockout during leadtime at (slightly less than).05. Standard Normal Probability Distribution

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35 Slide © 2009 Thomson South-Western. All Rights Reserved n Solving for the Reorder Point By raising the reorder point from 20 gallons to 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 Pep 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. Standard Normal Probability Distribution

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36 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Approximation of Binomial Probabilities When the number of trials, n, becomes large, 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 The normal probability distribution provides an easy-to-use approximation of binomial probabilities where n > 20, np > 5, and n (1 - p ) > 5.

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37 Slide © 2009 Thomson South-Western. All Rights Reserved Normal Approximation of Binomial Probabilities Set = np Set = np n Add and subtract 0.5 (a continuity correction factor) because a continuous distribution is being used to because a continuous distribution is being used to approximate a discrete distribution. For example, approximate a discrete distribution. For example, P ( x = 10) is approximated by P (9.5 < x < 10.5). P ( x = 10) is approximated by P (9.5 < x < 10.5).

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38 Slide © 2009 Thomson South-Western. All Rights Reserved Exponential Probability Distribution n The exponential probability distribution is useful in describing the time it takes to complete a task. n The exponential random variables can be used to describe: Time between vehicle arrivals at a toll booth Time between vehicle arrivals at a toll booth Time required to complete a questionnaire Time required to complete a questionnaire Distance between major defects in a highway Distance between major defects in a highway

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39 Slide © 2009 Thomson South-Western. All Rights Reserved n Density Function Exponential Probability Distribution where: = mean e = 2.71828 e = 2.71828 for x > 0, > 0

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40 Slide © 2009 Thomson South-Western. All Rights Reserved n Cumulative Probabilities Exponential Probability Distribution where: x 0 = some specific value of x x 0 = some specific value of x

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41 Slide © 2009 Thomson South-Western. All Rights Reserved Exponential Probability Distribution n Example: Al’s Full-Service Pump The time between arrivals of cars at Al’s full-service 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.

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42 Slide © 2009 Thomson South-Western. All Rights Reserved x x f(x)f(x) f(x)f(x).1.3.4.2 1 2 3 4 5 6 7 8 9 10 Time Between Successive Arrivals (mins.) Exponential Probability Distribution P ( x < 2) = 1 - 2.71828 -2/3 = 1 -.5134 =.4866 P ( x < 2) = 1 - 2.71828 -2/3 = 1 -.5134 =.4866

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43 Slide © 2009 Thomson South-Western. All Rights Reserved Relationship between the Poisson and Exponential Distributions The Poisson distribution provides an appropriate description of the number of occurrences per interval 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 The exponential distribution provides an appropriate description of the length of the interval between occurrences

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44 Slide © 2009 Thomson South-Western. All Rights Reserved End of Chapter 6

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