书签 分享 收藏 举报 版权申诉 / 37
上传文档赚钱

类型《统计学基础(英文版·第7版)》教学课件les7e--04-02-.pptx

  • 上传人(卖家):晟晟文业
  • 文档编号:4287501
  • 上传时间:2022-11-26
  • 格式:PPTX
  • 页数:37
  • 大小:2.75MB
  • 【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《《统计学基础(英文版·第7版)》教学课件les7e--04-02-.pptx》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:
    统计学基础英文版·第7版 统计学 基础 英文 教学 课件 les7e_04_02
    资源描述:

    1、统计学基础(英文版第7版)教学课件les7e_ppt_04_02-(1)Chapter Outline 4.1 Probability Distributions 4.2 Binomial Distributions 4.3 More Discrete Probability Distributions.Section 4.2Binomial Distributions.Section 4.2 Objectives How to determine whether a probability experiment is a binomial experiment How to find bin

    2、omial probabilities using the binomial probability formula How to find binomial probabilities using technology,formulas,and a binomial probability table How to construct and graph a binomial distribution How to find the mean,variance,and standard deviation of a binomial probability distribution.Bino

    3、mial Experiments1.The experiment is repeated for a fixed number of trials,where each trial is independent of other trials.2.There are only two possible outcomes of interest for each trial.The outcomes can be classified as a success(S)or as a failure(F).3.The probability of a success,P(S),is the same

    4、 for each trial.4.The random variable x counts the number of successful trials.Notation for Binomial ExperimentsSymbolDescriptionnThe number of times a trial is repeatedpThe probability of success in a single trialqThe probability of failure in a single trial(q=1 p)xThe random variable represents a

    5、count of the number of successes in n trials:x=0,1,2,3,n.Example:Identifying and Understanding Binomial ExperimentsDecide whether each experiment is a binomial experiment.If it is,specify the values of n,p,and q,and list the possible values of the random variable x.If it is not,explain why.1.A certa

    6、in surgical procedure has an 85%chance of success.A doctor performs the procedure on eight patients.The random variable represents the number of successful surgeries.Solution:Identifying and Understanding Binomial ExperimentsBinomial Experiment1.Each surgery represents a trial.There are eight surger

    7、ies,and each one is independent of the others.2.There are only two possible outcomes of interest for each surgery:a success(S)or a failure(F).3.The probability of a success,P(S),is 0.85 for each surgery.4.The random variable x counts the number of successful surgeries.Solution:Identifying and Unders

    8、tanding Binomial ExperimentsBinomial Experiment n=8(number of trials)p=0.85(probability of success)q=1 p=1 0.85=0.15(probability of failure)x=0,1,2,3,4,5,6,7,8(number of successful surgeries).Example:Identifying and Understanding Binomial ExperimentsDecide whether each experiment is a binomial exper

    9、iment.If it is,specify the values of n,p,and q,and list the possible values of the random variable x.If it is not,explain why.2.A jar contains five red marbles,nine blue marbles,and six green marbles.You randomly select three marbles from the jar,without replacement.The random variable represents th

    10、e number of red marbles.Solution:Identifying and Understanding Binomial ExperimentsNot a Binomial Experiment The probability of selecting a red marble on the first trial is 5/20.Because the marble is not replaced,the probability of success(red)for subsequent trials is no longer 5/20.The trials are n

    11、ot independent and the probability of a success is not the same for each trial.Binomial Probability FormulaBinomial Probability Formula The probability of exactly x successes in n trials is!()()!xn xxn xnxnP xC p qp qnxxn=number of trialsp=probability of successq=1 p probability of failurex=number o

    12、f successes in n trialsNote:number of failures is n x.Example:Finding a Binomial ProbabilityRotator cuff surgery has a 90%chance of success.The surgery is performed on three patients.Find the probability of the surgery being successful on exactly two patients.(Source:The Orthopedic Center of St.Loui

    13、s).Solution:Finding a Binomial ProbabilityMethod 1:Draw a tree diagram and use the Multiplication Rule.81310000.243 Solution:Finding a Binomial ProbabilityMethod 2:Use the binomial probability formula.213!91(2)(32)!2!101081131001081310000.243P Binomial Probability DistributionBinomial Probability Di

    14、stribution List the possible values of x with the corresponding probability of each.Example:Binomial probability distribution for Microfacture knee surgery:n=3,p=Use binomial probability formula to find probabilities.x0123P(x)0.0160.1410.4220.42234.Example:Constructing a Binomial DistributionIn a su

    15、rvey,U.S.adults were asked to identify which social media platforms they use.The results are shown in the figure.Six adults who participated in the survey are randomly selected and asked whether they use the social media platform Facebook.Construct a binomial probability distribution for the number

    16、of adults who respond yes.(Source:Pew Research).Solution:Constructing a Binomial Distributionp=0.68 and q=0.32n=6,possible values for x are 0,1,2,3,4,5 and 6.060660(0)0.680.321 0.680.320.001PC 151561(1)0.680.326 0.680.320.014PC 242462(2)0.680.3215 0.680.320.073PC 333363(3)0.680.3220 0.680.320.206PC

    17、424264(4)0.680.3215 0.680.320.328PC 515165(5)0.680.326 0.680.320.279PC 606066(6)0.680.321 0.680.320.099PCSolution:Constructing a Binomial DistributionNotice in the table that all the probabilities are between 0 and 1 and that the sum of the probabilities is 1.Example:Finding a Binomial Probabilities

    18、 Using TechnologyA survey found that 26%of U.S.adults believe there is no difference between secured and unsecured wireless networks.(A secured network uses barriers,such as firewalls and passwords,to protect information;an unsecured network does not.)You randomly select 100 adults.What is the proba

    19、bility that exactly 35 adults believe there is no difference between secured and unsecured networks?Use technology to find the probability.(Source:University of Phoenix).Solution:Finding a Binomial Probabilities Using Technology.SolutionMinitab,Excel,StatCrunch,and the TI-84 Plus each have features

    20、that allow you to find binomial probabilities.Try using these technologies.You should obtain results similar to these displays.Solution:Finding a Binomial Probabilities Using Technology.SolutionFrom these displays,you can see that the probability that exactly 35 adults believe there is no difference

    21、 between secured and unsecured networks is about 0.012.Because 0.012 is less than 0.05,this can be considered an unusual event.Example:Finding Binomial Probabilities Using FormulasA survey found that 17%of U.S.adults say that Google News is a major source of news for them.You randomly select four ad

    22、ults and ask them whether Google News is a major source of news for them.Find the probability that(1)exactly two of them respond yes,(2)at least two of them respond yes,and(3)fewer than two of them respond yes.(Source:Ipsos Public Affairs).Solution:Finding Binomial Probabilities Using Formulas.Solut

    23、ion:Finding Binomial Probabilities Using FormulasSolution2.To find the probability that at least two adults will respond yes,find the sum of P(2),P(3),and P(4).Begin by using the binomial probability formula to write an expression for each probability.P(2)=4C2(0.17)2(0.83)2=6(0.17)2(0.83)2P(3)=4C3(0

    24、.17)3(0.83)1=4(0.17)3(0.83)1P(4)=4C4(0.17)4(0.83)0=1(0.17)4(0.83)0.Solution:Finding Binomial Probabilities Using Formulas.Solution:Finding Binomial Probabilities Using Formulas.Example:Finding a Binomial Probability Using a TableAbout 10%of workers(ages 16 years and older)in the United States commut

    25、e to their jobs by carpooling.You randomly select eight workers.What is the probability that exactly four of them carpool to work?Use a table to find the probability.(Source:American Community Survey)Solution:Binomial with n=8,p=0.1,x=4.Solution:Finding Binomial Probabilities Using a Table A portion

    26、 of Table 2 is shownAccording to the table,the probability is 0.005.Solution:Finding Binomial Probabilities Using a Table You can check the result using technology.So,the probability that exactly four of the eight workers carpool to work is 0.005.Because 0.005 is less than 0.05,this can be considere

    27、d an unusual event.Example:Graphing a Binomial DistributionSixty-two percent of cancer survivors are ages 65 years or older.You randomly select six cancer survivors and ask them whether they are 65 years of age or older.Construct a probability distribution for the random variable x.Then graph the di

    28、stribution.(Source:National Cancer Institute)Solution:n=6,p=0.62,q=0.38 Find the probability for each value of x.Solution:Graphing a Binomial Distribution.Notice in the table that all the probabilities are between 0 and 1 and that the sum of the probabilities is 1.Solution:Graphing a Binomial Distri

    29、butionHistogram:.From the histogram,you can see that it would be unusual for none or only one of the survivors to be age 65 years or older because both probabilities are less than 0.05.Mean,Variance,and Standard Deviation Mean:=np Variance:2=npq Standard Deviation:npq.Example:Mean,Variance,and Stand

    30、ard DeviationIn Pittsburgh,Pennsylvania,about 56%of the days in a year are cloudy.Find the mean,variance,and standard deviation for the number of cloudy days during the month of June.Interpret the results and determine any unusual values.(Source:National Climatic Data Center)Solution:n=30,p=0.56,q=0

    31、.44Mean:=np=300.56=16.8Variance:2=npq=300.560.44 7.4Standard Deviation:30 0.56 0.442.7npq.Solution:Mean,Variance,and Standard Deviation=16.8 2 7.4 2.7 On average,there are 16.8 cloudy days during the month of June.The standard deviation is about 2.7 days.Values that are more than two standard deviations from the mean are considered unusual.16.8 2(2.7)=11.4;A June with 11 cloudy days or less would be unusual.16.8+2(2.7)=22.2;A June with 23 cloudy days or more would also be unusual.

    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:《统计学基础(英文版·第7版)》教学课件les7e--04-02-.pptx
    链接地址:https://www.163wenku.com/p-4287501.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