书签 分享 收藏 举报 版权申诉 / 44
上传文档赚钱

类型戴维商务统计学第7版英文版教学课件BSFC7e-CH04.ppt

  • 上传人(卖家):晟晟文业
  • 文档编号:3996358
  • 上传时间:2022-11-02
  • 格式:PPT
  • 页数:44
  • 大小:979.22KB
  • 【下载声明】
    1. 本站全部试题类文档,若标题没写含答案,则无答案;标题注明含答案的文档,主观题也可能无答案。请谨慎下单,一旦售出,不予退换。
    2. 本站全部PPT文档均不含视频和音频,PPT中出现的音频或视频标识(或文字)仅表示流程,实际无音频或视频文件。请谨慎下单,一旦售出,不予退换。
    3. 本页资料《戴维商务统计学第7版英文版教学课件BSFC7e-CH04.ppt》由用户(晟晟文业)主动上传,其收益全归该用户。163文库仅提供信息存储空间,仅对该用户上传内容的表现方式做保护处理,对上传内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知163文库(点击联系客服),我们立即给予删除!
    4. 请根据预览情况,自愿下载本文。本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
    5. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007及以上版本和PDF阅读器,压缩文件请下载最新的WinRAR软件解压。
    配套讲稿:

    如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。

    特殊限制:

    部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。

    关 键  词:
    戴维 商务 统计学 英文 教学 课件 BSFC7e_CH04
    资源描述:

    1、Basic ProbabilityChapter 4ObjectivesThe objectives for this chapter are:nTo understand basic probability concepts.nTo understand conditional probability nTo be able to use Bayes Theorem to revise probabilitiesnTo learn various counting rulesBasic Probability ConceptsnProbability the chance that an u

    2、ncertain event will occur(always between 0 and 1)nImpossible Event an event that has no chance of occurring(probability=0)nCertain Event an event that is sure to occur(probability=1)Assessing ProbabilityThere are three approaches to assessing the probability of an uncertain event:1.a priori -based o

    3、n prior knowledge of the process2.empirical probability3.subjective probabilityoutcomespossibleofnumbertotaloccurseventthein which waysofnumberTX based on a combination of an individuals past experience,personal opinion,and analysis of a particular situation outcomespossibleofnumbertotaloccurseventt

    4、hein which waysofnumber Assuming all outcomes are equally likelyprobability of occurrenceprobability of occurrenceExample of a priori probabilityWhen randomly selecting a day from the year 2015 what is the probability the day is in January?2015in days ofnumber totalJanuaryin days ofnumber January In

    5、 Day ofy ProbabilitTX365312015in days 365Januaryin days 31 TXExample of empirical probabilityTaking StatsNot Taking StatsTotalMale 84145229Female 76134210Total160279439Find the probability of selecting a male taking statistics from the population described in the following table:191.043984people ofn

    6、umber totalstats takingmales ofnumber Probability of male taking statsSubjective probabilitynSubjective probability may differ from person to personnA media development team assigns a 60%probability of success to its new ad campaign.nThe chief media officer of the company is less optimistic and assi

    7、gns a 40%of success to the same campaignnThe assignment of a subjective probability is based on a persons experiences,opinions,and analysis of a particular situationnSubjective probability is useful in situations when an empirical or a priori probability cannot be computedEventsEach possible outcome

    8、 of a variable is an event.nSimple eventnAn event described by a single characteristicne.g.,A day in January from all days in 2015nJoint eventnAn event described by two or more characteristicsne.g.A day in January that is also a Wednesday from all days in 2015nComplement of an event A (denoted A)nAl

    9、l events that are not part of event Ane.g.,All days from 2015 that are not in JanuarySample SpaceThe Sample Space is the collection of all possible eventse.g.All 6 faces of a die:e.g.All 52 cards of a bridge deck:Organizing&Visualizing EventsnVenn Diagram For All Days In 2015Sample Space(All Days In

    10、 2015)January DaysWednesdaysDays That Are In January and Are WednesdaysOrganizing&Visualizing EventsnContingency Tables -For All Days in 2015nDecision TreesAll Days In 2015Not Jan.Jan.Not Wed.Wed.Wed.Not Wed.Sample SpaceTotalNumberOfSampleSpaceOutcomesNot Wed.27 286 313 Wed.4 48 52Total 31 334 365 J

    11、an.Not Jan.Total 4 27 48286(continued)Definition:Simple ProbabilitynSimple Probability refers to the probability of a simple event.nex.P(Jan.)nex.P(Wed.)P(Jan.)=31/365P(Wed.)=52/365Not Wed.27 286 313 Wed.4 48 52Total 31 334 365 Jan.Not Jan.TotalDefinition:Joint ProbabilitynJoint Probability refers t

    12、o the probability of an occurrence of two or more events(joint event).nex.P(Jan.and Wed.)nex.P(Not Jan.and Not Wed.)P(Jan.and Wed.)=4/365P(Not Jan.and Not Wed.)=286/365Not Wed.27 286 313 Wed.4 48 52Total 31 334 365 Jan.Not Jan.TotalnMutually exclusive eventsnEvents that cannot occur simultaneouslyEx

    13、ample:Randomly choosing a day from 2015 A=day in January;B=day in FebruarynEvents A and B are mutually exclusiveMutually Exclusive EventsCollectively Exhaustive EventsnCollectively exhaustive eventsnOne of the events must occur nThe set of events covers the entire sample spaceExample:Randomly choose

    14、 a day from 2015 A=Weekday;B=Weekend;C=January;D=Spring;nEvents A,B,C and D are collectively exhaustive(but not mutually exclusive a weekday can be in January or in Spring)nEvents A and B are collectively exhaustive and also mutually exclusiveComputing Joint and Marginal ProbabilitiesnThe probabilit

    15、y of a joint event,A and B:nComputing a marginal(or simple)probability:nWhere B1,B2,Bk are k mutually exclusive and collectively exhaustive eventsoutcomeselementaryofnumbertotalBandAsatisfyingoutcomesofnumber)BandA(P)BdanP(A)BandP(A)BandP(AP(A)k21Joint Probability ExampleP(Jan.and Wed.)36542015in da

    16、ys ofnumber total Wed.are and Jan.in are that days ofnumber Not Wed.27 286 313 Wed.4 48 52Total 31 334 365 Jan.Not Jan.TotalMarginal Probability ExampleP(Wed.)36552365483654)Wed.andJan.P(Not Wed.)andJan.(PNot Wed.27 286 313 Wed.4 48 52Total 31 334 365 Jan.Not Jan.Total P(A1 and B2)P(A1)TotalEventMar

    17、ginal&Joint Probabilities In A Contingency TableP(A2 and B1)P(A1 and B1)EventTotal1Joint ProbabilitiesMarginal(Simple)Probabilities A1 A2B1B2 P(B1)P(B2)P(A2 and B2)P(A2)Probability Summary So FarnProbability is the numerical measure of the likelihood that an event will occurnThe probability of any e

    18、vent must be between 0 and 1,inclusivelynThe sum of the probabilities of all mutually exclusive and collectively exhaustive events is 1CertainImpossible0.5100 P(A)1 For any event A1P(C)P(B)P(A)If A,B,and C are mutually exclusive and collectively exhaustiveGeneral Addition RuleP(A or B)=P(A)+P(B)-P(A

    19、 and B)General Addition Rule:If A and B are mutually exclusive,then P(A and B)=0,so the rule can be simplified:P(A or B)=P(A)+P(B)For mutually exclusive events A and BGeneral Addition Rule ExampleP(Jan.or Wed.)=P(Jan.)+P(Wed.)-P(Jan.and Wed.)=31/365+52/365-4/365 =79/365Dont count the four Wednesdays

    20、 in January twice!Not Wed.27 286 313 Wed.4 48 52Total 31 334 365 Jan.Not Jan.TotalComputing Conditional ProbabilitiesnA conditional probability is the probability of one event,given that another event has occurred:P(B)B)andP(AB)|P(AP(A)B)andP(AA)|P(BWhere P(A and B)=joint probability of A and B P(A)

    21、=marginal or simple probability of AP(B)=marginal or simple probability of BThe conditional probability of A given that B has occurredThe conditional probability of B given that A has occurrednWhat is the probability that a car has a GPS,given that it has AC?i.e.,we want to find P(GPS|AC)Conditional

    22、 Probability ExamplenOf the cars on a used car lot,70%have air conditioning(AC)and 40%have a GPS.20%of the cars have both.Conditional Probability ExampleNo GPSGPSTotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0nOf the cars on a used car lot,70%have air conditioning(AC)and 40%have a GPS and 20%of the ca

    23、rs have both.0.28570.70.2P(AC)AC)andP(GPSAC)|P(GPS (continued)Conditional Probability ExampleNo GPSGPSTotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0nGiven AC,we only consider the top row(70%of the cars).Of these,20%have a GPS.20%of 70%is about 28.57%.0.28570.70.2P(AC)AC)andP(GPSAC)|P(GPS (continued)U

    24、sing Decision TreesHas ACDoes not have ACHas GPSDoes not have GPSHas GPSDoes not have GPSP(AC)=0.7P(AC)=0.3P(AC and GPS)=0.2P(AC and GPS)=0.5P(AC and GPS)=0.1P(AC and GPS)=0.27.5.3.2.3.1.AllCars7.2.Given AC or no AC:ConditionalProbabilitiesUsing Decision TreesHas GPSDoes not have GPSHas ACDoes not h

    25、ave ACHas ACDoes not have ACP(GPS)=0.4P(GPS)=0.6P(GPS and AC)=0.2P(GPS and AC)=0.2P(GPS and AC)=0.1P(GPS and AC)=0.54.2.6.5.6.1.AllCars4.2.Given GPS or no GPS:(continued)ConditionalProbabilitiesIndependencenTwo events are independent if and only if:nEvents A and B are independent when the probabilit

    26、y of one event is not affected by the fact that the other event has occurredP(A)B)|P(AMultiplication RulesnMultiplication rule for two events A and B:P(B)B)|P(AB)andP(A P(A)B)|P(ANote:If A and B are independent,thenand the multiplication rule simplifies toP(B)P(A)B)andP(A Marginal ProbabilitynMargin

    27、al probability for event A:nWhere B1,B2,Bk are k mutually exclusive and collectively exhaustive events)P(B)B|P(A)P(B)B|P(A)P(B)B|P(A P(A)kk2211Bayes TheoremnBayes Theorem is used to revise previously calculated probabilities based on new information.nDeveloped by Thomas Bayes in the 18th Century.nIt

    28、 is an extension of conditional probability.Bayes Theoremnwhere:Bi=ith event of k mutually exclusive and collectively exhaustive eventsA=new event that might impact P(Bi)P(BB|P(A)P(BB|P(A)P(BB|P(A)P(BB|P(AA)|P(Bk k 2 2 1 1 i i i Bayes Theorem ExamplenA drilling company has estimated a 40%chance of s

    29、triking oil for their new well.nA detailed test has been scheduled for more information.Historically,60%of successful wells have had detailed tests,and 20%of unsuccessful wells have had detailed tests.nGiven that this well has been scheduled for a detailed test,what is the probability that the well

    30、will be successful?nLet S=successful well U=unsuccessful wellnP(S)=0.4,P(U)=0.6 (prior probabilities)nDefine the detailed test event as DnConditional probabilities:P(D|S)=0.6 P(D|U)=0.2nGoal is to find P(S|D)Bayes Theorem Example(continued)0.6670.120.240.24(0.2)(0.6)(0.6)(0.4)(0.6)(0.4)U)P(U)|P(DS)P

    31、(S)|P(DS)P(S)|P(DD)|P(SBayes Theorem Example(continued)Apply Bayes Theorem:So the revised probability of success,given that this well has been scheduled for a detailed test,is 0.667nGiven the detailed test,the revised probability of a successful well has risen to 0.667 from the original estimate of

    32、0.4Bayes Theorem ExampleEventPriorProb.Conditional Prob.JointProb.RevisedProb.S(successful)0.40.6(0.4)(0.6)=0.240.24/0.36=0.667U(unsuccessful)0.60.2(0.6)(0.2)=0.120.12/0.36=0.333Sum=0.36(continued)Counting Rules Are Often Useful In Computing ProbabilitiesnIn many cases,there are a large number of po

    33、ssible outcomes.nCounting rules can be used in these cases to help compute probabilities.Counting RulesnRules for counting the number of possible outcomesnCounting Rule 1:nIf any one of k different mutually exclusive and collectively exhaustive events can occur on each of n trials,the number of poss

    34、ible outcomes is equal tonExamplenIf you roll a fair die 3 times then there are 63=216 possible outcomesknCounting RulesnCounting Rule 2:nIf there are k1 events on the first trial,k2 events on the second trial,and kn events on the nth trial,the number of possible outcomes isnExample:nYou want to go

    35、to a park,eat at a restaurant,and see a movie.There are 3 parks,4 restaurants,and 6 movie choices.How many different possible combinations are there?nAnswer:(3)(4)(6)=72 different possibilities(k1)(k2)(kn)(continued)Counting RulesnCounting Rule 3:nThe number of ways that n items can be arranged in o

    36、rder isnExample:nYou have five books to put on a bookshelf.How many different ways can these books be placed on the shelf?nAnswer:5!=(5)(4)(3)(2)(1)=120 different possibilitiesn!=(n)(n 1)(1)(continued)Counting RulesnCounting Rule 4:nPermutations:The number of ways of arranging X objects selected fro

    37、m n objects in order isnExample:nYou have five books and are going to put three on a bookshelf.How many different ways can the books be ordered on the bookshelf?nAnswer:different possibilities(continued)X)!(nn!Pxn6021203)!(55!X)!(nn!PxnCounting RulesnCounting Rule 5:nCombinations:The number of ways

    38、of selecting X objects from n objects,irrespective of order,isnExample:nYou have five books and are going to select three are to read.How many different combinations are there,ignoring the order in which they are selected?nAnswer:different possibilities(continued)X)!(nX!n!Cxn10(6)(2)1203)!(53!5!X)!(nX!n!CxnChapter SummaryIn this chapter we covered:nUnderstanding basic probability concepts.nUnderstanding conditional probability nUsing Bayes Theorem to revise probabilitiesnVarious counting rules

    展开阅读全文
    提示  163文库所有资源均是用户自行上传分享,仅供网友学习交流,未经上传用户书面授权,请勿作他用。
    关于本文
    本文标题:戴维商务统计学第7版英文版教学课件BSFC7e-CH04.ppt
    链接地址:https://www.163wenku.com/p-3996358.html

    Copyright@ 2017-2037 Www.163WenKu.Com  网站版权所有  |  资源地图   
    IPC备案号:蜀ICP备2021032737号  | 川公网安备 51099002000191号


    侵权投诉QQ:3464097650  资料上传QQ:3464097650
       


    【声明】本站为“文档C2C交易模式”,即用户上传的文档直接卖给(下载)用户,本站只是网络空间服务平台,本站所有原创文档下载所得归上传人所有,如您发现上传作品侵犯了您的版权,请立刻联系我们并提供证据,我们将在3个工作日内予以改正。

    163文库