#### Transcript Chap5-8x

Random Variables and Probability Distributions Chapters 5-8 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Learning Objectives Chapter 5 1. Define probability. 2. Define Random Variable. 3. Understand discrete random variable and continuous random variable. Chapter 6 Identify the characteristics of a probability distribution. Chapter 7 1. List the characteristics of the normal probability distribution. 2. Convert a normal distribution to the standard normal distribution. 3. Find the probability for a normally distributed random variable. 4. Find the Z-value. 5-2 Learning Objectives Chapter 8 1 Explain why a sample is often the only feasible way to learn something about a population. 4 Describe the sampling distribution of the sample mean. 5 Understand the central limit theorem. 6 Define the standard error of the mean. 7 Apply the central limit theorem to find probabilities. 8-3 A Survey of Probability Concepts Chapter 5 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Random Variables (r.v) RANDOM VARIABLE: A quantity resulting from an experiment that, by chance, can assume different values. Random variable is usually denoted by X. Discrete r.v.: A random variable that can assume only certain clearly separated values. 1. The outcome when throwing a die of six sides 2. The number of students in a class. 3. The number of cars entering a carwash in a hour. Continuous r.v.: can assume an infinite number of values within a given range. 1. The weight of each student in this class. 2. The temperature outside as you are reading this book. 3. The amount of money earned by each of the more than 750 players currently on Major League Baseball team rosters. Probability PROBABILITY: A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. A probability is the relative possibility that a random variable assumes certain values. Notations: For discrete r.v.: P(X = x) or P(x) For continuous r.v.: P(X > x) Discrete Probability Distributions Chapter 6 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Probability Distribution PROBABILITY DISTRIBUTION A listing of all the outcomes of an experiment and the probability associated with each outcome. Experiment: Outcome P(X=x) Toss a die of six sides. The outcome of the experiment is a r.v, denoted by X. The possible results are: 1, 2, 3, 4, 5, and 6. 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 What is the probability distribution for each of the outcome? 6 1/6 Total 1 Characteristics of a Probability Distribution 1.The probability of a particular outcome is between 0 and 1 inclusive. 2. The outcomes are mutually exclusive events. 3. The list is exhaustive. So the sum of the probabilities of the various events is equal to 1. 4. For continuous distributions, P(X = x) = 0 Continuous Probability Distributions Chapter 7 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Normal Probability Distribution 1. 2. 3. 4. It is bell-shaped and symmetrical about the mean. The area under the curve denotes probability. The total area under the curve is 1.00. The location of a normal distribution curve is determined by the mean , the shape of the curve is determined by the standard deviation,σ. For a r.v. that follows normal distribution with mean and s.d. σ, we denote X~(, σ). mean The Family of Normal Distribution The Standard Normal Probability Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is also called the z distribution. A r.v. that follows standard normal distribution is denoted by Z. Converting a normally distributed r.v., X to a standard normally distributed r.v., Z: subtract X by its mean and divide by its s.d. Z score Probabilities: Areas Under Curve P(Z > 1.80)= P(Z < -1.80) = .0359; P(Z < −1.52) = .0643; =NORMSDIST(-1.8) =NORMSDIST(-1.52) P(−1.30 < Z < 2.10) = P(Z<2.1) – P(Z< - 1.3) = .8853; =NORMSDIST(2.1) - NORMSDIST(-1.3) Excel function: =NORMSDIST(z) Normal distribution table Normal distribution table The Normal Distribution – Example The weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is between $840 and $1,200? the Z score is the Z score is P(840 < X <1,200) = P (-1.6 < Z < 2.0 ) =P (Z < 2.0) – P(Z < -1.6) = .9224 Excel function: =NORMDIST(mean, s.d, x, true) =NORMDIST(1000, 100, 1200, true) - NORMDIST(1000, 100, 840, true) Finding Values of Z For Given Areas Given an area (𝛼) under the curve, what is the corresponding value of z (z𝛼 ) on the horizontal axis such that the area to its right is 𝛼? P(Z > z𝜶) = 𝜶 Example: Z0.05=? 𝜶 Excel function =NORMSINV(1-𝛼) =NORMSINV(1-0.05) 𝜶 8.18 Z0.05=1.645 Finding Values of Z For Given Areas 0.025 -z0.025 Example: - Z0.025=? Excel function =NORMSINV(𝛼) =NORMSINV(0.025) - Z0.025= -1.96 Other commonly used z-values z0.10 = 1.28 -z0.10 = -1.28 z0.01 = 2.326 -z0.01= -2.326 Finding z values Using the Normal Table Example: z0.025 To find the z value in the normal table, we look for .025 (the subscript of Z) inside the table. Here we can find an exact match for .025. Thus the corresponding z value is 1.96. Example: z0.05 When we look for .05 (the subscript of Z) inside the table, we cannot find an exact match for .05. Instead we find two values that are close to .05: .495 and .505. In this class only: if we cannot find an exact match for the subscript of Z, we will take the average of the two corresponding Z values, 1.64 and 1.65. z0.05=(1.64+1.65)/2=1.645 z values found this way can be slightly different from the z values found using Excel. Using Z in Finding X Given Area - Example Layton Tire and Rubber Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 with a standard deviation of 2,050 miles and that the distribution of miles follows the normal probability distribution. Layton wants to set the minimum guaranteed mileage so that no more than 4 percent of the tires will have to be replaced. What minimum guaranteed mileage should Layton announce? 21 Using Z in Finding X Given Area - Example Solve X using the formula : z x- x 67,900 2,050 The value of z is found using the 4% informatio n - z 0 .04 - 1.75. - 1.75 Excel function =NORMINV(𝛼, mean, s. 𝑑.) x - 67,900 , then solving for x 2,050 - 1.75(2,050 ) x - 67,900 =NORMINV(0.04, 67900, x 67,900 - 1.75(2,050 ) 2050) x 64,312 The Empirical Rule About 68 percent of the area under the normal curve is within one standard deviation of the mean. About 95 percent is within two standard deviations of the mean. Practically all is within three standard deviations of the mean. 7-23 Sampling Methods and the Central Limit Theorem Chapter 8 McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Sampling Method—Simple Random Sample Simple Random Sample: A sample selected so that each item or person in the population has the same chance of being included. EXAMPLE: A population consists of 845 employees of Nitra Industries. A sample of 52 employees is to be selected from that population. The name of each employee is written on a small slip of paper and deposited all of the slips in a box. After they have been thoroughly mixed, the first selection is made by drawing a slip out of the box without looking at it. This process is repeated until the sample of 52 employees is chosen. 8-25 Sampling Distribution of the Sample Mean The sampling distribution of the sample mean is a probability distribution consisting of all possible sample means of a given sample size selected from a population. If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with larger samples. CENTRAL LIMIT THEOREM 8-26 Central Limit Theorem 8-27 The Quality Assurance Department for Cola, Inc., maintains records regarding the amount of cola in its Jumbo bottle. The actual amount of cola in each bottle is critical, but varies a small amount from one bottle to the next. Cola, Inc., does not wish to underfill the bottles. On the other hand, it cannot overfill each bottle. Its records indicate that the amount of cola follows the normal probability distribution. The mean amount per bottle is 31.2 ounces and the population standard deviation is 0.4 ounces. At 8 A.M. today the quality technician randomly selected 16 bottles from the filling line. The mean amount of cola contained in the bottles is 31.38 ounces. Is this an unlikely result? Is it likely the process is putting too much soda in the bottles? To put it another way, does the population represented by the sample of 16 bottles really has a population mean of only 31.2 ounces? 8-28 Using the Sampling Distribution of the Sample Mean (Sigma Known) - Example X 31.38 31.20 z 1.80 n $0.4 16 8-29 Using the Sampling Distribution of the Sample Mean (Sigma Known) - Example 8-30 Using the Sampling Distribution of the Sample Mean (Sigma Known) - Example What do we conclude? With less than a 4 percent chance, It is unlikely for us to observe such a sample of 16 observations from a normal population (which has a mean of 31.2 ounces and a population standard deviation of 0.4 ounces) that the sample mean is as high as 31.38 ounces. We conclude the process is putting too much cola in the bottles. 8-31