...Binomial Distribution This is a discrete random variable, where the process of obtaining the Binomial distribution is called “Bernoulli “ process. An experiment that often consists of repeated trials, each with two possible outcomes, which could be labeled as “success” or “failure”. This experiment is known as binomial experiment. A binomial experiment is one that possesses the following properties: 1. The experiment consists of n repeated trials. 2. Each trial has only 2 possible outcomes that can be classified as “Success” or “Failure”. 3. The probability of a success and failure , denoted by pand q, remains constant from trial to trial. 4. The repeated, trials are independent. Formula for success in the Bernoulli process. The probability of r success from n trial is : [pic] where; r - the number of success n – the number of trial p – the probability of success from one trial and q – the probability of failure from one trial. q = 1 - p [pic] with mean, [pic] and variance, [pic] It is written as X ~ B(n,p) and read as X is Binomial distribution with parameters n success and the probability of success, p. Example 7.4 One coin is tossed 5 times. If X is random variable for number of heads . Find the probability of getting (a) no heads (b) one head (c) 3 heads (d) at least 3 heads Solution X is...
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...Binomial Distributions How to recognize binomial random variable: 1. The sample size n is fixed. 2. The n observations (or trials) are independent. 3. There are only 2 possible outcomes for each observation. They are labeled “Success” and “Failure” 4. The probability of success is the same for each trial. Let p = success probability and 1 – p = failure probability 5. The binomial random variable is . . . X = the number of successes out of n observations. Binomial distributions are identified specifically by two parameters: n and p. X takes on values 0, 1, 2, . . . , n Notation: [pic] Mean: [pic] Variance: [pic] Standard deviation: [pic] The Binomial Probability Formula: [pic] Suppose [pic], then the probability of observing 16 successes out of 20 observations is [pic] Find the mean, variance, and standard deviation of X. Mean: [pic] Variance: [pic] Standard deviation: [pic] Table 3 in Appendix II contains probability distributions for a variety of binomial distributions. Look up the following probabilities: For X ~ B(20, 0.7), find P(X=16). Answer: P(X=16) = 0.130 For X ~ B(8, 0.45), find P(X > 5). Answer: P(X > 5) = P(6) + P(7) + P(8) = 0.070 + 0.016 + 0.002 = 0.088 More examples: 1. A research team found that 10% of...
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...DISTRIBUSI BINOMIAL Distribusi binomial berasal dari percobaan binomial yaitu suatu proses Bernoulli yang diulang sebanyak n kali dan saling bebas. Distribusi Binomial merupakan distribusi peubah acak diskrit. Secara langsung, percobaan binomial memiliki ciri-ciri sebagai berikut: – – – – percobaan tersebut dilakukan berulang-ulang sebanyak n kali setiap percobaan menghasilkan keluaran yang dapat dikategorikan sebagai gagal dan sukses probabilitas sukses p tetap konstan dari satu percobaan ke percobaan lain percobaan yang berulang adalah saling bebas Ruang sampel A untuk percobaan E yang terdiri dari himpunan tak hingga tetapi masih terhitung dari titik – titik sampel: Jika S = Sukses dan G = Gagal E1 : S (sukses pada percobaan pertama) E2 : GS (gagal pada percobaan pertama dan sukses pada percobaan kedua) E3 : SG (sukses pada percobaan pertama, gagal pada percobaan kedua) E4 : GGS (gagal pada percobaan 1 dan 2, sukses pada percobaan ketiga) E5 : GSG (gagal pada percobaan 1 dan 3, sukses pada percobaan kedua) E6 : SGG (gagal pada percobaan 2 dan 3, sukses pada percobaan pertama) En : SSS ...S GGG...G (sukses sebanyak x kali, gagal sebanyak n – x kali) x n x Jika peluang sukses dinotasikan dengan p maka peluang gagal adalah q = 1 – p. Peubah acak X menyatakan banyaknya sukses dari n percobaan yang saling bebas. Maka peluang X pada masing – masing percobaan E adalah: P(X) p P(X) qp pq P(X) pq 2 2 untuk E1 untuk E2 untuk E3 P(X) q p pq untuk E4...
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...Solutions Manual for Statistical Inference, Second Edition George Casella University of Florida Roger L. Berger North Carolina State University Damaris Santana University of Florida 0-2 Solutions Manual for Statistical Inference “When I hear you give your reasons,” I remarked, “the thing always appears to me to be so ridiculously simple that I could easily do it myself, though at each successive instance of your reasoning I am baffled until you explain your process.” Dr. Watson to Sherlock Holmes A Scandal in Bohemia 0.1 Description This solutions manual contains solutions for all odd numbered problems plus a large number of solutions for even numbered problems. Of the 624 exercises in Statistical Inference, Second Edition, this manual gives solutions for 484 (78%) of them. There is an obtuse pattern as to which solutions were included in this manual. We assembled all of the solutions that we had from the first edition, and filled in so that all odd-numbered problems were done. In the passage from the first to the second edition, problems were shuffled with no attention paid to numbering (hence no attention paid to minimize the new effort), but rather we tried to put the problems in logical order. A major change from the first edition is the use of the computer, both symbolically through Mathematicatm and numerically using R. Some solutions are given as code in either of these languages. Mathematicatm can be purchased from Wolfram Research, and R is a...
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...Lin20522_appb_726-743.indd Page 726 28/10/13 11:55 AM user-f-w-198 /201/MH02018/Lin20522_disk1of1/0078020522/Lin20522_pagefiles A P P E N D I X B : TA B L E S B.1 Binomial Probability Distribution n51 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 0.950 0.050 0.900 0.100 0.800 0.200 0.700 0.300 0.600 0.400 0.500 0.500 0.400 0.600 0.300 0.700 0.200 0.800 0.100 0.900 0.050 0.950 n52 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 0.903 0.095 0.003 0.810 0.180 0.010 0.640 0.320 0.040 0.490 0.420 0.090 0.360 0.480 0.160 0.250 0.500 0.250 0.160 0.480 0.360 0.090 0.420 0.490 0.040 0.320 0.640 0.010 0.180 0.810 0.003 0.095 0.903 n53 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 3 0.857 0.135 0.007 0.000 0.729 0.243 0.027 0.001 0.512 0.384 0.096 0.008 0.343 0.441 0.189 0.027 0.216 0.432 0.288 0.064 0.125 0.375 0.375 0.125 0.064 0.288 0.432 0.216 0.027 0.189 0.441 0.343 0.008 0.096 0.384 0.512 0.001 0.027 0.243 0.729 0.000 0.007 0.135 0.857 n54 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 3 4 0.815 0.171 0.014 0.000 0.000 0...
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.... . . . . . . . . . . . . . . . . . . . 11 2.3 Five Number Summaries and Box and Whisker Displays . . . . . 12 3 Probability 13 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2 Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2.2 Expected Value . . . . . . . . . . . . . . . . . . . . . . . . 17 3.2.3 Variance and Standard Deviation . . . . . . . . . . . . . . 17 3.2.4 “Shortcuts” for Binomial Random Variables . . . . . . . . 18 1 4 Probability Distributions 19 4.1 Binomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . 19 4.2 Poisson Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.2 As an Approximation to the Binomial . . . . . . . . . . . 22 Normal Distributions . ....
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...The Al Gore Effect: An Inconvenient Truth and Voluntary Carbon Offsets∗ [Job Market Paper] Grant Jacobsen University of California-Santa Barbara 2120 North Hall University of California Santa Barbara, 93106-9210 jacobsen@econ.ucsb.edu Phone: (717) 315-5503 Fax: (805) 893-8830 I thank Matthew Kotchen, Robert Deacon, Olivier Deschenes, and Charles Kolstad for helpful comments. I also thank participants at a UCSB seminar, the Western Economics International Association’s Conference, and the University of Colorado Environmental and Resource Economics Workshop. ∗ 1 The Al Gore Effect: An Inconvenient Truth and Voluntary Carbon Offsets Abstract This paper examines the relationship between climate change awareness and household behavior by testing whether Al Gore’s documentary An Inconvenient Truth caused an increase in the purchase of voluntary carbon offsets. I find that in the two months following the film’s release, zip codes within a 10-mile radius of a zip code where the film was shown experienced a 50 percent relative increase in the purchase of voluntary carbon offsets. During other times, offset purchasing patterns for zip codes inside the 10-mile radius were similar to the patterns of zip codes outside the 10-mile radius. There is, however, little evidence that individuals who purchased an offset due to the film renewed them again a year later. This research has implications for how information campaigns, which are commonly used by policy-makers to...
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...your students while teaching or reviewing concepts including: – Probability and Independence – The Binomial distribution – The Normal approximation to the binomial – Hypothesis testing Semantics • Remove all ACEs from the deck • Place one red ACE and one black ACE face up • Ask each student, in order, to guess the color of a card which they cannot see • Place card onto appropriate colored ACE stack • After 10 (or so) guesses, place opposite color ACE on each stack • Repeat second step above • Reveal to students their hidden psychic ability! Probability • Define “success” as guessing right • What is the probability of “success”? – Bayesian? – Frequentist? • What is the probability of getting all n right? Independence • What is the probability of getting all n right? • Does the guess of one student affect the next student? – Series of reds often forces the next guess to be black – Does the probability of “success” change? Binomial Distribution • Recall requirements of – Dichotomous event – Independence – N trials – Constant probability of success (50%) • What is the probability of getting all n right? • What is the probability of getting n-1 out of n right (modification)? Hypothesis Testing • p-value: If the true probability of “success” is 50%, what is the chance you get all n right? – Binomial calculation – Normal approximation to the binomial • Only need n=10 until np and n(1-p) = 5 • In the end, students don’t need to understand how the...
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...Acme Electronics Case Date: March 6, 2012 To: Jetson, on behalf of Acme Electronics From: Team 4 Consulting Firm Re: Legal and statistical evaluation of problems facing Acme Per your request, we have assembled a report with a legal and statistical evaluation of the problems facing Acme. If you have any questions, feel free to contact us at any time. Group 4: Acme Electronics Case Executive Summary This report is dealing with the case of ACME Electronics vs. Otto Gunter. Gunter purchased a computer from ACME Electronics in 2002. In 2004, the hard drive crashed and he brought it in to ACME Electronics to have it replaced, as well as request for his old hard drive back to retrieve his information. ACME replaced the hard drive and gave Gunter his supposed old hard drive. Gunter spent $800 retrieving information from the hard drive only to find out it was mixed with another customer’s. The other customer disposed of Gunter’s old hard drive, so he cannot retrieve it. Gunter is suing ACME for $800 for retrieving information from the wrong hard drive supplied by ACME, $5,000 for time spent reconstructing the lost information, and $10,000 for punitive damages. The point of this report is to analyze the legal ramifications and statistics based on the situation at hand and to deduce the appropriate course of action to take based off the analysis. We analyzed a survey using statistics...
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...Overview and examples from Finite Mathematics Using Microsoft Excel® Revathi Narasimhan Saint Peter's College An electronic supplement to Finite Mathematics and Its Applications, 6th Ed. , by Goldstein, Schneider, and Siegel, Prentice Hall, 1997 Introduction In any introductory mathematics course designed for non-mathematics majors, it is important for the student to understand and apply mathematical ideas in a variety of contexts. With the increased use of advanced software in all fields, it is also important for the student to effectively interact with the new technology. Our goal is to integrate these two objectives in a supplement for the text Finite Mathematics and Its Applications, by Goldstein, Schneider, and Siegel. The package consists of interactive tutorials and projects in an Excel workbook format. The software platform used is the Microsoft Excel 5.0 spreadsheet. It was chosen for the following reasons: • • • suited to applications encountered in a finite math course widespread use outside of academia ease of creating reports with a professional look Use of Excel 5.0 was put into effect in the author's sections of the Finite Mathematics II course in the Spring 1996 semester. It was expanded to cover the Finite Mathematics I course for the Fall semester of 1996. Using a combination of specially designed projects and tutorials, students are able to analyze data, draw conclusions, and present their analysis in a professional format. The mathematical...
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...student, University of Alabama in Huntsville Abstract. In their article published in the April 2013 edition of Control Engineering Practice, Silvia Joekes and Emanuel Pimentel Barbosa present the use of a correction factor, the Cornish-Fisher quantile correction, to improve the fraction nonconforming control chart (p-chart) when dealing with high quality processes. Introduction. The use of the standard Shewhart control chart for the fraction of nonconforming parts will produce a type I error–signaling that the process is out of control when, in fact, it is still in control–with probability 0 = 0.0027. When dealing with a process with both high quality and a large yield, the control limits for the Shewhart p-chart based on the normal distribution will become tighter, leaving a greater chance of a false alarm signal occurring. This journal entry proposes a modified version of the p-chart based on the CornishFisher expansion, which allows monitoring processes with very low values of p. This new corrected p-chart is shown using one and two terms of correction. Shewhart p-chart basics. Shewhart control charts use 3 limits for both upper and lower control limits, i.e. 3 . The use of Shewhart control charts is commonplace due to their ease of use: the process will only ever signal in control or out of control. This type of pass/fail or conforming/nonconforming relationship is used for attributes data: data that is not easily assigned some numerical value for the characteristic...
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...event of one person testing positive is independent of any other person testing positive. 2. Every person testing positive for the disease is equally likely. Solution • Let N be the total population, r be the number of infected people in the population. Hence, N-r represents the number of people in the population without the disease. • Let n denote the number of samples in a group. Therefore, [pic] be the number of groups. Let ‘p’ represent the probability that a group tests negative for the disease, that is, none of the group members carry the antigen. Probability Distribution In a population of N, there are r people who carry the disease and N-r people who are free from the disease. Samples of size n are chosen from this population. Let X denote the number of people who test positive for the disease in the sample n. Then, p follows hypergeometric distribution. [pic] For the problem at hand, k = 0. [pic] Therefore, 1-p will represent the probability that the group tests positive. Number of tests to be performed through sampling = Number of groups + Testing each person in the groups testing positive If the proposed technique is to require lesser number of chemical analyses than those required by the existing procedure [pic] [pic] [pic] f(x) was plotted on MS Excel for different values of N, n and r, as shown here: |N...
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...Descriptive Statistics Descriptive statistics involves organizing, summarizing and illustrating statistical data. The objective is to show important characteristics of the data without drawing any conclusions. Inferential statistics involves using a representative subset of data (a sample) in order to draw conclusions about unknown characteristics of an entire set of data (a population). Population: The entire set of elements of interest (i.e. all humans, all working-age people in Canada, all IT companies). A population parameter is a characteristic used to describe a population. For example, Population mean ( Population standard deviation ( Population median ( The values of the population parameters would be preferable for use in decision-making but seldom will these values be known since collecting all the population elements (a census) is usually too expensive and/or time consuming. Sample: A representative subset of the entire set of elements of interest that is used to gain insight about the population. A sample statistic is a characteristic used to describe a sample. For example, Sample mean [pic] Sample standard deviation s Sample median Md It is cheaper, less time-consuming and more practical to use sample statistics as estimates for population parameters in making business decisions. How well the sample represents the population depends on...
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...Solution to selected exercises from chapter 3 3-14. f X (0) P( X 0) 1/ 6 1/ 6 1/ 3 f X (1.5) P( X 1.5) 1/ 3 f X (2) 1/ 6 f X (3) 1/ 6 a) P(X = 2) = 1/6 b) P(0.6 < X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P(0 X 2) P( X 0) P( X 1.5) 1/ 3 1/ 3 2 / 3 e) P(X = 0 or X = 2) = 1/3 + 1/6 = ½ --------------------------------------------------------------------------------------------------------------------3-15. All probabilities are greater than or equal to zero and sum to one. a) P(X 1) = 1/8 + 2/8 + 2/8 + 2/8 = 7/8 b) P(X > 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(1 X 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X 1 or X = 2) = 1/8 + 2/8 +1/8 = 4/8 = ½ --------------------------------------------------------------------------------------------------------------------3-20. P(X = 0) = 0.053 = 1.25 104 P(X = 1) = 3[0.95(0.05)(0.05)] = 7.125 103 P(X = 2) = 3[0.95(0.95)(0.05)] = 0.1354 P(X = 3) = 0.953 = 0.8574 --------------------------------------------------------------------------------------------------------------------- 3-25. P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = 0.5 million) = 0.1 --------------------------------------------------------------------------------------------------------------------- 3-32. x0 0, 1/ 3 0 x 1.5 F ( x) 2 / 3 1.5 x 2 where 5 / 6 2 x 3 3 x 1 f X (0) P( X 0) 1/ 6 1/ 6 1/ 3 f X (1.5) P( X ...
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...statistics (proportions-population) - past events, Probability (sample) - future events. 1) A U B, A or B P(A or B) = P(A) + P(B) - P(A and B) Mutually exclusive => P(A and B)=0 2) P(B/A) Independent=P(B). Dependent=P(B/A)= P(A and B)/ P(A) 3) Intersection A ∩ B ///Dependent :P(A and B) = P(A) x P(B|A) Independent Events: P(A and B) = P(A) x P(B) 4) A stands for “not A” ; Complement Rule: P(A) + P(not A) = 1; P(A) = 1- P(not A); P(At least one) = 1 – P(none) ……………………………………………………………………………………………………………………………………………………………………… ……………………………………………………………………………. 1)UNIFORM DISTRIBUTION: The area under the uniform distribution: P( Mx1 ) z1= 1 - CI% x% (confidence interval dat) of the observation fall below X 3) BINOMIAL DISTRIBUTION: calculez : p(success); n= total no; x = number of successes in sample- ni se da in intrebare; BINOMIAL FORMULA: p(x)= [n!/ (X! (n-x)!)]*px(1-p)n-x ; P(x=x) =P(1)+ P(2)+………………..+P(n) BINOMIAL...
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