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統計學 授課教師:統計系余清祥 日期:2014年10月30日 第六章:連續機率分配 Fall 2010 STATISTICS in PRACTICE • Procter & Gamble (P&G) produces and markets such products as detergents, disposable diapers, bar soaps, and paper towels. • The Industrial Chemicals Division of P&G is a supplier of fatty alcohols derived from natural substances such as coconut oil and from petroleum-based derivatives. • The division wanted to know the economic risks and opportunities of expanding its fatty-alcohol production facilities, so it called in P&G’s experts in probabilistic decision and risk analysis to help. 2 Chapter 6 Continuous Probability Distributions • Uniform Probability Distribution • Normal Probability Distribution • Normal Approximation of Binomial Probabilities • Exponential Probability Distribution f (x) f (x) Exponential Uniform f (x) Normal x x x 3 Continuous Probability Distributions -1 • 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. – Note: a single point is an interval of zero width, this implies that the probability of a continuous random variable assuming any particular value exactly is zero. 4 Continuous Probability Distributions -2 • The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2. f (x) f (x) Exponential Uniform f (x) x1 x2 Normal x1 xx12 x2 x x1 x2 x x 5 6.1 Uniform Probability Distribution -1 • 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 6 Uniform Probability Distribution -2 • Expected Value of x E(x) = (a + b)/2 • Variance of x Var(x) = (b - a)2/12 7 Uniform Probability Distribution -3 • Example: – Random variable x = the 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. – Assume every 1-minute interval being equally likely, x is said to have a uniform probability distribution and the probability density function is 1 /20 for 120 ≤ x ≤ 140 f (x) = elsewhere 0 8 Uniform Probability Distribution -4 • Example: Uniform Probability Density Function for Flight Time 9 Uniform Probability Distribution -5 • Example: – What is the probability that the flight time is between 120 and 130 minutes? That is, what is P(120 ≤ x ≤ 130) =? – Area provides probability of flight time between 120 and 130 minutes. 10 Uniform Probability Distribution -6 • 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. 11 Uniform Probability Distribution -7 • Example: Slater's Buffet – Uniform Probability Density Function f(x) = 1/10 for 5 < x < 15 =0 elsewhere where: x = salad plate filling weight 12 Uniform Probability Distribution -8 • Example: Slater's Buffet – Expected Value of x E(x) = (a + b)/2 = (5 + 15)/2 = 10 – Variance of x Var(x) = (b - a)2/12 = (15 – 5)2/12 = 8.33 13 Uniform Probability Distribution -9 • Example: Slater's Buffet – Uniform Probability Distribution for Salad Plate Filling Weight f(x) 1/10 0 5 10 Salad Weight (oz.) 15 x 14 Uniform Probability Distribution -10 • Example: Slater's Buffet – What is the probability that a customer will take between 12 and 15 ounces of salad? f(x) P(12 < x < 15) = 1/10(3) = 0.3 1/10 0 5 10 12 Salad Weight (oz.) 15 x 15 Area as a Measure of Probability • The area under the graph of f(x) and probability are identical. • This is valid for all continuous random variables. • The probability that x takes on a value between some lower value x1 and some higher value x2 can be found by computing the area under the graph of f(x) over the interval from x1 to x2. 16 6.2 Normal Probability Distribution -1 • Normal Curve • Standard Normal Probability Distribution • Computing Probabilities for Any Normal Probability Distribution • Grear Tire Company Problem 17 Normal Probability Distribution -2 • 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 measurements ‧Test scores ‧Scientific • Abraham de Moivre, a French mathematician, published The Doctrine of Chances in 1733. He derived the normal distribution. 18 Normal Probability Distribution -3 • Normal Probability Density Function 1 − ( x − µ )2 /2σ 2 f (x) = e σ 2π where: µ = mean σ = standard deviation π = 3.14159 e = 2.71828 19 Normal Probability Distribution -4 • Characteristics The distribution is symmetric; its skewness measure is zero. x 20 Normal Probability Distribution -5 • Characteristics The entire family of normal probability distributions is defined by its mean µ and its standard deviation σ . Standard Deviation σ Mean µ x 21 Normal Probability Distribution -6 • Characteristics The highest point on the normal curve is at the mean, which is also the median and mode. Mean µ x 22 Normal Probability Distribution -7 • Characteristics The mean can be any numerical value: negative, zero, or positive. -10 0 x 20 23 Normal Probability Distribution -8 • Characteristics The standard deviation determines the width of the curve: larger values result in wider, flatter curves. σ = 15 σ = 20 Mean µ x 24 Normal Probability Distribution -9 • Characteristics Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (0.5 to the left of the mean and 0.5 to the right). 0.5 0.5 Mean µ x 25 Normal Probability Distribution -10 • Characteristics (basis for the empirical rule) 68.26% of values of a normal random variable are within +/- 1 standard deviation of its mean. 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. 26 Normal Probability Distribution -11 • Characteristics (basis for the empirical rule) 99.72% 95.44% 68.26% µ – 3σ µ – 1σ µ – 2σ µ µ + 3σ µ + 1σ µ + 2σ x 27 Standard Normal Probability Distribution -1 • Characteristics 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. 28 Standard Normal Probability Distribution -2 • Characteristics The letter z is used to designate the standard normal random variable. σ=1 0 z 29 Standard Normal Probability Distribution -3 • Standard Normal Density Function 1 − z2 /2 f ( z) = e 2π 30 Standard Normal Probability Distribution -4 • By looking in the body of the table, we find that the 1.0 row and the 0.00 column intersect at the value of 0.8413; thus, P(z ≤ 1.00) = 0.8413. The following excerpt from the probability table shows these steps. 31 Standard Normal Probability Distribution -5 • Cumulative Probabilities for The Standard Normal Distribution – Note: Entries in the table give the area under the curve to the left of the z value. For example, for z = −0.85, the cumulative probability is 0.1977. 32 Standard Normal Probability Distribution -6 33 Standard Normal Probability Distribution -7 – Entries in the table give the area under the curve to the left of the z value. For example, for z = 1.25, the cumulative probability is 0.8944. 34 Standard Normal Probability Distribution -8 35 Standard Normal Probability Distribution -9 • Three types of probabilities we need to compute 1. the probability that the standard normal random variable z will be less than or equal to a given value; P(z ≤ x) = ? 2. the probability that z will be between two given values; P(x1 ≤ z ≤ x2) = ? 3. the probability that z will be greater than or equal to a given value. P(z ≥ x) = ? 36 Standard Normal Probability Distribution -10 • Example: – P(z ≤ 1.00) = 0.8413 – P(− 0.50 ≤ z ≤ 1.25) = P(z ≤ 1.25) − P(z ≤ − 0.50) = 0.8944 − 0.3085 = 0.5859. 37 Standard Normal Probability Distribution -11 • Example: – P(− 1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) − P(z ≤ − 1.00) = 0.8413 − 0.1587 = 0.6826 38 Standard Normal Probability Distribution -12 • Example: – P(z ≥ 1.58) = 1 − 0.9429 = 0.0571 39 Standard Normal Probability Distribution -13 • Suppose we want to find a z value such that the probability of obtaining a larger z value is 0.10. The following figure shows this situation graphically. 40 Standard Normal Probability Distribution -14 • Converting to the Standard Normal Distribution z= x−µ σ – We can think of z as a measure of the number of standard deviations x is from µ. 41 Standard Normal Probability Distribution -15 • 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. – The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order. 42 Standard Normal Probability Distribution -16 • Example: Pep Zone – 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 exceed 20 gallons? P(x > 20) = ? 43 Standard Normal Probability Distribution -17 • Example: Pep Zone – Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. z = (x - µ)/σ = (20 - 15)/6 = 0.83 Step 2: Find the area under the standard normal curve to the left of z = 0.83. see next slide 44 Standard Normal Probability Distribution -18 • Example: Pep Zone – Cumulative Probability Table for the Standard Normal Distribution P(z ≤ 0.83) 45 Standard Normal Probability Distribution -19 • Example: Pep Zone – Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z = 0.83. P(z > 0.83) = 1 – P(z ≤ 0.83) = 1 – 0.7967 = 0.2033 Probability of a stockout P(x > 20) 46 Standard Normal Probability Distribution -20 • Example: Pep Zone – Solving for the Stockout Probability Area = 1 – 0.7967 Area = 0.7967 = 0.2033 0 0.83 z 47 Standard Normal Probability Distribution -21 • Example: Pep Zone – If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than 0.05, what should the reorder point be? --------------------------------------------------------------– (Hint: Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.) 48 Standard Normal Probability Distribution -22 • Example: Pep Zone. Solving for the Reorder Point Area = 0.9500 Area = 0.0500 0 z0.05 z 49 Standard Normal Probability Distribution -23 • Example: Pep Zone. Solving for the Reorder Point Step 1: Find the z-value that cuts off an area of 0.05 in the right tail of the standard normal distribution. We look up the complement of the tail area (1 - 0.05 = 0.95) 50 Standard Normal Probability Distribution -24 • Example: Pep Zone. Solving for the Reorder Point Step 2: Convert z0.05 to the corresponding value of x. x = µ + z0.05σ = 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) 0.05. 51 Standard Normal Probability Distribution -25 • Example: Pep Zone. Solving for the Reorder Point Probability of no stockout during replenishment lead-time = 0.95 Probability of a stockout during replenishment lead-time = 0.05 15 24.87 x 52 Standard Normal Probability Distribution -26 • Example: Pep Zone. 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 0.20 to 0.05. – 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. 53 Standard Normal Probability Distribution -27 • Example: Grear Tire Company Problem – The Grear Tire Company developed a new steelbelted radial tire to be sold through a national chain stores. Grear’s managers believe that the mileage guarantee offered with the tire will be an important factor in the acceptance of the product. – Grear’s managers want probability information about x number of miles the tires will last. 54 Standard Normal Probability Distribution -28 • Example: Grear Tire Company Problem – Grear’s engineering group estimated that the mean tire mileage is μ = 36,500 miles and the standard deviation is σ = 5000. The data collected indicate that a normal distribution. – What percentage of the tires can be expected to last more than 40,000 miles? 55 Standard Normal Probability Distribution -29 • Example: Grear Tire Company Problem 56 Standard Normal Probability Distribution -30 • Example: Grear Tire Company Problem – At x = 40,000, we have x − µ 40, 000 − 36, 500 3500 = z = = = 0.70 σ 5000 5000 – P(z ≥ 0.70) = 0.2420. We can conclude that about 24.2%of the tires will exceed 40,000 in mileage. – Assume that Grear is considering a guarantee that will provide a discount on replacement tires if the original tires do not provide the guaranteed mileage. What should the guarantee mileage be if Grear wants no more than 10% of the tires to be eligible for the discount guarantee? 57 Standard Normal Probability Distribution -31 • Example: Grear Tire Company Problem – Using the standard normal probability table, we see that z = − 1.28 cuts off an area of 0.10 in the lower tail. 58 Standard Normal Probability Distribution -32 • Example: Grear Tire Company Problem – To find the value of x corresponding to z = − 1.28, z = − 1.28, then x = μ − 1.28σ , with μ = 36,500 , σ = 5000, then x = 36,500 − 1.28(5000) = 30,100. – A guarantee of 30,100 miles will meet the requirement that approximately 10%of the tires will be eligible for the guarantee. 59 6.3 Normal Approximation of Binomial Probabilities -1 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 np ≥ 5 and n(1 - p) ≥ 5. In the definition of the normal curve, set µ = np and σ = np (1 − p ) 60 Normal Approximation of Binomial Probabilities -2 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 = 12) for the discrete binomial probability distribution is approximated by P(11.5 ≤ x ≤ 12.5) for the continuous normal distribution. 61 Normal Approximation of Binomial Probabilities -3 • Example: – To find the binomial probability of 12 successes in 100 trials and p = 0.1. – n = 100, p = 0.1, np = 10 > 5 and n(1 − p) = (100)(0.9) = 90 > 5 – Mean, np = (100)(0.1) = 10 > 5, Variance, np(1 − p) = (100)(0.1)(0.9) = 9 – Compute the area under the corresponding normal curve between 11.5 and 12.5 P(11.5 ≤ x ≤ 12.5) = ? 62 Normal Approximation of Binomial Probabilities -4 • Example: – Suppose that a company has a history of making errors in 10% of its invoices. A sample of 100 invoices has been taken, and we want to compute the probability that 12 invoices contain errors. – In this case, we want to find the binomial probability of 12 successes in 100 trials. So, we set: µ = np = 100(0.1) = 10 = σ np(1 − p) = [100(0.1)(0.9)]½ = 3 63 Normal Approximation of Binomial Probabilities -5 • Example: – Normal Approximation to a Binomial Probability Distribution with n = 100 and p = 0.1 σ=3 P(11.5 < x < 12.5) (Probability of 12 Errors) µ = 10 11.5 12.5 x 64 Normal Approximation of Binomial Probabilities -6 • Example: – Normal Approximation to a Binomial Probability Distribution with n = 100 and p = 0.1 P(x ≤ 12.5) = 0.7967 10 12.5 x 65 Normal Approximation of Binomial Probabilities -7 • Example: – Normal Approximation to a Binomial Probability Distribution with n = 100 and p = 0.1 P(x ≤ 11.5) = 0.6915 10 x 11.5 66 Normal Approximation of Binomial Probabilities -8 • Example: – The Normal Approximation to the Probability of 12 Successes in 100 Trials is 0.1052 P(x = 12) = 0.7967 - 0.6915 = 0.1052 10 11.5 12.5 x 67 Normal Approximation of Binomial Probabilities -9 • Example: Normal Approximation to a Binomial Probability Distribution with n = 100 and p = 0.1 Showing the Probability of 13 or Fewer Errors 68 6.4 Exponential Probability Distribution -1 • Computing Probabilities for the Exponential Distribution • Relationship Between the Poisson and Exponential Distributions 69 Exponential Probability Distribution -2 • 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 • In waiting line applications, the exponential distribution is often used for service times. 70 Exponential Probability Distribution -3 • A property of the exponential distribution is that the mean and standard deviation are equal. • The exponential distribution is skewed to the right. Its skewness measure is 2. 71 Exponential Probability Distribution -4 • Density Function f ( x) = 1 µ e− x /µ for x ≥ 0 where: µ = expected or mean e = 2.71828 72 Exponential Probability Distribution -5 • Cumulative Probabilities P ( x ≤ x0 ) = 1 − e − xo / µ where: x0 = some specific value of x 73 Exponential Probability Distribution -6 • Example: The Schips loading dock example. – x = loading time in minutes and μ = 15, which gives us P(x ≤ x0) = 1 − e −x0/15. What is the probability that loading a truck will take between 6 minutes and 18 minutes? – Since, P(x ≤ 6) = 1 − e −6/15 = 0.3297 and P(x ≤ 18) = 1 − e −18/15 = 0.6988 P(6 ≤ x ≤ 18) = 0.6988 − 0.3297 = 0.3691. 74 Exponential Probability Distribution -7 • Example: Al’s Full-Service Pump – The time between arrivals of cars at Al’s fullservice 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. 75 Exponential Probability Distribution -8 • Example: Al’s Full-Service Pump f(x) P(x < 2) = 1 - 2.71828-2/3 = 1 - 0.5134 = 0.4866 0.4 0.3 0.2 0.1 0 1 2 3 4 5 6 7 8 9 10 x Time Between Successive Arrivals (mins.) 76 Exponential Probability Distribution -9 • Example: – Suppose the number of cars that arrive at a car wash is a Poisson probability distribution with a 10 x e −10 mean of 10 cars per hour and f ( x ) = . x! – Because the average number of arrivals is 10 cars per hour, the average time between cars arriving is 1 hour/10 cars = 0.1 hour/car. – The corresponding exponential distribution that describes the time between the arrivals has a mean of µ = 0.1 hour per car; the exponential probability density function is f(x) = 10e–10x. 77 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 78 End of Chapter 6 79