...C H A P T E R 6 The Normal Distribution Objectives After completing this chapter, you should be able to Outline Introduction 6–1 Normal Distributions 1 2 3 4 5 6 7 Identify distributions as symmetric or skewed. Identify the properties of a normal distribution. Find the area under the standard normal distribution, given various z values. Find probabilities for a normally distributed variable by transforming it into a standard normal variable. Find specific data values for given percentages, using the standard normal distribution. Use the central limit theorem to solve problems involving sample means for large samples. Use the normal approximation to compute probabilities for a binomial variable. 6–2 Applications of the Normal Distribution 6–3 The Central Limit Theorem 6–4 The Normal Approximation to the Binomial Distribution Summary 6–1 300 Chapter 6 The Normal Distribution Statistics Today What Is Normal? Medical researchers have determined so-called normal intervals for a person’s blood pressure, cholesterol, triglycerides, and the like. For example, the normal range of systolic blood pressure is 110 to 140. The normal interval for a person’s triglycerides is from 30 to 200 milligrams per deciliter (mg/dl). By measuring these variables, a physician can determine if a patient’s vital statistics are within the normal interval or if some type of treatment is needed to correct a condition and avoid future illnesses. The question then is,...
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...The probability distribution of the population data is called the population distribution. Tables 7.1 and 7.2 on page 309 of the text provide an example of such a distribution. The probability distribution of a sample statistic is called its sampling distribution. Tables 7.3 to 7.5 on page 311 of the text provide an example of the sampling distribution of the sample mean. 1. Sampling error is the difference between the value of the sample statistic and the value of the corresponding population parameter, assuming that the sample is random and no nonsampling error has been made. Example 7–1 on page 312 of the text exhibits sampling error. Sampling errors occur only in sample surveys. 2. Nonsampling errors are errors that may occur during collection, recording, and tabulation of data. The second part of Example 7–1 on pages 312 and 313 of the text exhibits nonsampling error. Nonsampling errors occur both in sample surveys and censuses. 3. a. [pic] b. [pic][pic] Sampling error = [pic] c. Liza’s incorrect [pic][pic] [pic] Sampling error (from part b) = –1.83 Nonsampling error [pic] d. | Sample | [pic] |[pic] | |15, 13, 8, 17 |13.25 |.92 | |15, 13, 8, 9 |11.25 |–1.08 | |15, 13, 17, 9 |13.50 |1.17 | |15, 8, 17, 9 |12...
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...City Cents City Cents 1 Vancouver, BC 5.83 15 Edmonton, AB 10.12 2 Saskatoon, SK 6.77 16 Calgary, AB 15.32 3 Abbotsford, BC 6.32 17 Oshawa, ON 6.23 4 St. John's, NL 9.68 18 Halifax, NS 8.36 5 Winnipeg, MB 5.41 19 Moncton, NB 8.83 6 Sudbury, ON 5.83 20 Price George, AB 6.19 7 Ottawa, ON 8.08 21 Brantford, ON 6.65 8 Sherbrooke, QC 6.56 22 St-Jean, QC 5.25 9 Quebec, QC 5.74 23 Regina, SK 7.23 10 Whitehorse, YT 6.06 24 Montreal, QC 6.74 11 Toronto, ON 8.62 25 Yellowknife, NWT 6.43 12 Saguenay, QC 6.20 26 Kewlowna, BC 6.83 13 Hamilton, ON 6.91 27 Windsor, ON 6.69 14 Charlottetown, PE 10.60 28 Thunder Bay, ON 6.12 1. Determining the mean, variance and standard deviation of the provided sample: Mean: 7.3429 Variance: 4.4322 Standard Deviation: 2.1053 The mean, variance, and standard deviation for electricity prices of this sample is 7.3429, 4.4322 & 2.1053 cents respectively. 2. 95% confidence interval for the mean of the population utility rates: Confidence Interval: 0.77979 We're 95% confident that the true mean falls somewhere between 6.5631and 8.1227. 3. 90% confidence interval for the mean of the population utility rates, asssuming population standard deviation 2.09 Confidence Interval: 0.6497 ...
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...4 problem set 1: Normal Probability Distributions Page.285 Ex 6,8,10,12 6. x = 80, z=80-10015 = -1.33 z= 0.0918 1-0.0918 = 0.9082 8. x = 110, z=110-10015 = 0.67 z= 0.7486 z= 75-10015 = -1.67 z= 0.0475 0.7486-0.0475= 0.7011 (shaded area) 10. z= 0.84 (shaded) z= -0.84 x= 100+(-0.84∙15) = 87 (rounded) 12. . z= 2.33 x= 100+(2.33∙15) = 135 (rounded) Page 288 Ex 34 34.Appendix B Data Set: Duration of Shuttle Flights a. Find the mean and standard deviation, and verify that the data have a distribution that is roughly normal. Mean= 25317115 = 220.15 Standard Deviation=115253172-(25317)2115(115-1) = 86 (rounded) The normal distribution is 115 b. Treat the statistics from part (a) as if they are population parameters and assume a normal distribution to find the values of the quartiles 1,2 and 3. Mean= 220.15 Standard Deviation= 86 Q1 = 220.5 + (-0.67 ∙ 86)= 162.53 Q2= 220.5 + (0.00 ∙ 86) = 220.5 Q3=220.5 + (0.67 ∙ 86) = 277.77 Page.300 Ex 20 Quality Control: Sampling Distribution of Proportion after constructing a new manufacturing machine. 5 prototype integrated circuit chips are produced and it is found that 2 are defective (D) and 3 are acceptable (A). Assume that two if the chips are randomly selected with replacement from this population a. After identifying the 25 different possible samples, find the proportion of defects in each of them, then use a table to describe the sampling distribution of the proportions...
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...standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution standard normal distribution...
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...Lecture 7. Sampling Distributions. Statistical Inference: Using statistics calculated from samples to estimate the values of population parameters. Select Random Sample Sample for (statistic) Calculate to estimate Becomes Population Parameter. BASIC Example: Soft Drink Bottler μ=600, σ=10. Normal Distribution. What is P(X>598)? p(x<598) . Sampling Dist.of the Mean – Distribution of all Possible Sample Means if you select a sample of a certain size. μX= μ. μ = i=1NXiN (formula for mean) . σ = i=1N(Xi-μ)2N Although you do not know how close the sample mean of any particular sample selected comes to the pop mean, you know that the mean of all possible sample means that could have been selected = the pop mean. Standard error is calculating the probability of a certain amount of error. EXAMPLE: Standard error is . As n increases decreases. CENTRAL LIMIT THEOREM: Regardless of shape of individual values in distribution; as long as sample size is large enough the sampling distribution of the mean will be approximately normally distributed with μX= μ and σX= σ . For most population distributions n ≥ 30 will be large enough. For symmetric population distributions, n ≥ 5 is sufficient. For normal population distributions, the sampling distribution of the mean is always normally distributed EXAMPLE: SAMPLING DISTRIBUTION OF THE PROPORTION. π is the proportion of items in the population with a characteristic of interest. p is the sample proportion...
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...easier and more efficient. If you had a good introductory quantitative methods course, and like the text that was used, you may want to refer to it whenever you feel in need of a refresher. If you feel uncomfortable with standard quantitative texts, this reference is for you. Our aim is to present the essential quantitative concepts and methods in a self-contained, nontechnical, and intuitive way. Our approach is structured in line with requirements for the CFA program. The material included is relevant to investment management by the ICFA, the Institute of Chartered Financial Analysts. We hope you find this appendix helpful. Use it to make your venture into investments more enjoyable. 1006 Appendix A 1007 A.1 PROBABILITY DISTRIBUTIONS Statisticians talk about “experiments,” or “trials,” and refer to possible outcomes as “events.” In a roll of a die, for example, the “elementary events” are the numbers 1 through 6. Turning up one side represents the most disaggregate mutually exclusive outcome. Other events are compound, that is, they consist of more than one elementary event, such as the result “odd number” or “less than 4.” In this case “odd” and “less than 4” are not mutually exclusive. Compound events can be mutually exclusive outcomes, however, such as “less than 4” and “equal to or greater than 4.”...
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.............................................................. 1 1 Introduction ............................................................................................................................. 2 2 Methodology ........................................................................................................................... 4 2.1 ARCH and GARCH ......................................................................................................... 4 2.2 EGARCH ......................................................................................................................... 4 2.3 GJR-GARCH ................................................................................................................... 5 2.4 Distributions ..................................................................................................................... 5 2.5 Information criterions ....................................................................................................... 6 3 Data ......................................................................................................................................... 7 3.1 Descriptive statistics ......................................................................................................... 7 4 Results ..................................................................................................................................... 9 4.1 Evaluation of the models ....................................
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...Some basic relationships of Probability Read ASW Chapter 4 or LR Chapter 4 | 2 | Theories of Probability - Classical theory, Relative Frequency theory, Axioms, Addition rule, Multiplication rule, Rule of at least one, Concept of Expected number of Success – Numerical Problems & Applications Case Problem: Hamilton County JudgesSIP: Morton International - Chicago, Illinois Read ASW Chapter 4 or LR Chapter 4 | 3 | Bayes Theorem – Theory, Problems & Applications, Probability revision using tabular approach Read ASW Chapter 4 or LR Chapter 4 | 4 | Probability Distribution - Meaning of Probability Distribution, Type of Probability Distribution, Need (Application) for Probability DistributionRead ASW Chapter 5 or LR Chapter 5 | 5 | Discrete Probability Distribution - Binomial Distribution – Applications, Numerical Problems, Excel & SPSS functions, Poisson Distribution...
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...Z-Scores A z-score, which is symbolized as z, is the statistic that relates the distance a score is relative to its mean when measured in standard deviation (SD) units (Heiman, 2012). The importance of a z-score is that it enables one to analyze data relative to scores. Z-Scores allow us to determine whether a particular score is equal to the mean, below the mean or above the mean of a bunch of scores, and how far a particular score is away from the mean. Evaluating the scores of Eric’s to determine different z-scores, we use the following computations that he computed where it takes a mean of 17 minutes with a standard deviation of 3 minutes to drive from home, park the car, and walk to his job. , Next we determine the z-score relative to the mean and the SD, to analyze the difference in time to accomplish these steps on different days. One day it took Eric 21 minutes to get to work, and computing the means from the minutes by the SD results in a z of +1.33, which tells you that Eric's time to get to work is 1.33 standard deviations from the mean. The z is positive because it is above the mean, and demonstrates that it took longer for Eric to leave home, park his car and walk to his job when the raw value of time was 21 minutes. On another day in which it took Eric 12 minutes to get to work, the z value resulted -1.66, demonstrating that the time it took Eric to get to work was below the mean, or it took less time. Observing the time in comparison to the mean, it is obvious...
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...78583 * 2 × Error (b)n for Factor M 2 × 9.78586 * 2 × Error (a)n for Factor PP 2 × 4.96756 * 2 critical-t values → t at 2 and t at 4 df i.e. 4.303 and 2.776 * Could ask: do ANOVA and t-test, or ANOVA and interpret results from F; Standard error for the difference (a or b); Conclusion: levels differ/do not differ at 1% etc. NS 13 – Non-parametric tests * Parametric tests for data with normal distribution (t, F or X2 distribution) * Non-parametric tests for * Categorical data, * Quantitative data divided into class intervals, * Small data sets, * Data sets without repetition of the TMTs. * Non- parametric tests * Medians, not Means * Usually rank your data * Single sample: * Sign test (No assumptions about distribution) * Rank test (assumes data comes from symmetrical distribution) * Wilcoxon’s symmetry test * For 2 independent samples * Mann-Whitney U test (Assumes distributions have same shape and equal...
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...Case Study#2 The XYZ Company Katharine Rally is the vice president of operations for the XYZ Company. She oversees operations at a plant that manufactures components for hydraulic systems. Katharine is concerned about the plant’s present production capability. She has reduced the decision situation to three alternatives. The first alternative, which is fully automation, would result in significant changes in present operations. The second alternative, which is semi-automation, involves fewer changes in present operations. The third alternative is to make no changes (do nothing). As a manager of the plant management team, you have been assigned the task of analyzing the alternatives and recommending a course of action. The capital investment and annual revenue for the first two alternatives are shown in the following table: |Alternative |Capital Investment |Future Sales |Annual Revenue | |A |$300,000 |Good |$250,000 | | | |Average |$100,000 | | | |Poor |$50,000 | |B |$85,000 |Good |$100,000 | | | ...
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...CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties 2. Which of the following about the normal distribution is NOT true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within 1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean, , and standard deviation, . ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between 1 standard deviation around the mean. b) 4 of every 5 observations would fall between 1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between 2 standard deviations around the mean. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 4. For some positive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770. The value of Z is a) 0.18. b) 0.81. c) 1.16. d) 1...
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...Week Three Quiz Comparing Variations: 1. For the following exercise, complete the following by using the data sets below (there are two separate sets of values, one shows the ages of Presidents from Washington through Jackson and the other shows the ages of the seven most recent Presidents): A. Find the mean, median, and range for each of the two data sets. First set of numbers: Mean: 58.28571 Median: 57 Range 4 Second set of numbers: Mean: 56.14286 Median: 54 Range 23 B. Find the standard deviation using the range rule of thumb for each of the data sets. Please show your work. (Please see Chapter 4, Section 4.3, page 173 of the text).Standard deviation = 61 – 57 = 4 4 / 4 = 1 1 is the standard deviation using the range rule of thumb. Standard deviation = 69 – 46 = 23 23 / 4 = 5.75 5.75 is the standard deviation using the range rule of thumb. C. Compare the two sets and describe what you discover. You could interpret that younger men are being elected president from past trends.The standard deviation shows an increase. The following data sets shows the ages of the first seven presidents (President Washington through President Jackson) and the seven most recent presidents including President Obama. Age is given at time of inauguration. First 7: 57 61 57 57 58 57 61 Second 7: 61 52 69 64 46 54 47 2. Any given data set consists of a set of numerical values. Please indicate by stating yes or no for each of the following statements whether or not it could be correct for...
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...© Orangetree Business Solutions Private Limited, 2012 No part of this book should be referenced or copied without the prior permission of the company. A FEW WORDS TO THE STUDENTS Analytics is becoming a popular tool for managerial decision making. It‘s still not so widespread in countries like India, but in the west it has become a standard practice. Previously studying analytics involved an in depth knowledge of statistics and programming languages. But widespread availability of statistical package software has changed the reality to some extent. Now more emphasis is given on the application of the techniques to solve the business problems. So there is a need to understand the meaning of the statistical procedures. This book has been written to cater that need. In this book, all the necessary concepts have been explained keeping the business problem in mind. Also, to remove the apathy for statistics, use of mathematical expressions have been limited. That doesn‘t imply that we don‘t have to study the mathematics part. The intention is to put the substance over matter. As the students get accustomed to these statistical concepts, they can go for further investigations using various mathematical and statistical techniques. A list of suggested books and links have been given in the appendix. This book is directly related to the instructor‘s presentation. So it is highly advised that students should go through this material at the end of each class. As for general ...
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