《《统计学基础(英文版·第7版)》教学课件les7e04 02.pptx》由会员分享，可在线阅读，更多相关《《统计学基础(英文版·第7版)》教学课件les7e04 02.pptx（37页珍藏版）》请在三一办公上搜索。
1、统计学基础(英文版第7版)教学课件les7e_ppt_04_02-(1),统计学基础(英文版第7版)教学课件les7e_ppt_,Chapter Outline,4.1 Probability Distributions4.2 Binomial Distributions4.3 More Discrete Probability Distributions,.,Chapter Outline4.1 Probability,Section 4.2,Binomial Distributions,.,Section 4.2Binomial Distributi,Section 4.2 Objecti
2、ves,How to determine whether a probability experiment is a binomial experimentHow to find binomial probabilities using the binomial probability formulaHow to find binomial probabilities using technology, formulas, and a binomial probability tableHow to construct and graph a binomial distributionHow
3、to find the mean, variance, and standard deviation of a binomial probability distribution,.,Section 4.2 ObjectivesHow to d,Binomial Experiments,The experiment is repeated for a fixed number of trials, where each trial is independent of other trials.There are only two possible outcomes of interest fo
4、r each trial. The outcomes can be classified as a success (S) or as a failure (F).The probability of a success, P(S), is the same for each trial.The random variable x counts the number of successful trials.,.,Binomial ExperimentsThe experi,Notation for Binomial Experiments,.,Notation for Binomial Ex
5、perime,Example: Identifying and Understanding Binomial Experiments,Decide 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.,A certain surgical procedure has an 85% chance of
6、 success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries.,.,Example: Identifying and Under,Solution: Identifying and Understanding Binomial Experiments,Binomial ExperimentEach surgery represents a trial. There are eight surgeries,
7、 and each one is independent of the others.There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F).The probability of a success, P(S), is 0.85 for each surgery.The random variable x counts the number of successful surgeries.,.,Solution: Identifying and Unde,
8、Solution: Identifying and Understanding Binomial Experiments,Binomial Experimentn = 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),.,Solution: Identifying and Unde,Example: Identifyin
9、g and Understanding Binomial Experiments,Decide 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.,A jar contains five red marbles, nine blue marbles, and six green marbles.
10、You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles.,.,Example: Identifying and Under,Solution: Identifying and Understanding Binomial Experiments,Not a Binomial ExperimentThe probability of selecting a red marble on the first
11、 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 not independent and the probability of a success is not the same for each trial.,.,Solution: Identifying and Unde,Binomial Probability Formula,Binomial Probabil
12、ity FormulaThe probability of exactly x successes in n trials is,n = number of trialsp = probability of successq = 1 p probability of failurex = number of successes in n trialsNote: number of failures is n x,.,Binomial Probability FormulaBi,Example: Finding a Binomial Probability,Rotator cuff surger
13、y 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. Louis),.,Example: Finding a Binomial Pr,Solution: Finding a Binomial Probability,Method 1: Draw a tree diagra
14、m and use the Multiplication Rule.,.,Solution: Finding a Binomial P,Solution: Finding a Binomial Probability,Method 2: Use the binomial probability formula.,.,Solution: Finding a Binomial P,Binomial Probability Distribution,Binomial Probability DistributionList the possible values of x with the corr
15、esponding probability of each.Example: Binomial probability distribution for Microfacture knee surgery: n = 3, p = Use binomial probability formula to find probabilities.,.,Binomial Probability Distribut,Example: Constructing a Binomial Distribution,In a survey, U.S. adults were asked to identify wh
16、ich 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 of adults who respond yes. (Source: Pe
17、w Research),.,Example: Constructing a Binomi,Solution: Constructing a Binomial Distribution,p = 0.68 and q = 0.32n = 6, possible values for x are 0, 1, 2, 3, 4, 5 and 6,.,Solution: Constructing a Binom,Solution: Constructing a Binomial Distribution,Notice in the table that all the probabilities are
18、between 0 and 1 and that the sum of the probabilities is 1.,.,Solution: Constructing a Binom,Example: Finding a Binomial Probabilities Using Technology,A survey found that 26% of U.S. adults believe there is no difference between secured and unsecured wireless networks. (A secured network uses barri
19、ers, such as firewalls and passwords, to protect information; an unsecured network does not.) You randomly select 100 adults. What is the probability that exactly 35 adults believe there is no difference between secured and unsecured networks? Use technology to find the probability. (Source: Univers
20、ity of Phoenix),.,Example: Finding a Binomial Pr,Solution: Finding a Binomial Probabilities Using Technology,.,SolutionMinitab, Excel, StatCrunch, and the TI-84 Plus each have features that allow you to find binomial probabilities. Try using these technologies. You should obtain results similar to t
21、hese displays.,Solution: Finding a Binomial P,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 between secured and unsecured networks is about 0.012. Because 0.012 is less
22、 than 0.05, this can be considered an unusual event.,Solution: Finding a Binomial P,Example: Finding Binomial Probabilities Using Formulas,A survey found that 17% of U.S. adults say that Google News is a major source of news for them. You randomly select four adults and ask them whether Google News
23、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),.,Example: Finding Binomial Prob,Solution: Finding Binomial Probabilities Using Formulas
24、,.,Solution: Finding Binomial Pro,Solution: Finding Binomial Probabilities Using Formulas,SolutionTo 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)
25、= 4C2(0.17)2(0.83)2 = 6(0.17)2(0.83)2P(3) = 4C3(0.17)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 Pro,Solution: Finding Binomial Probabilities Using Formulas,.,Solution: Finding Binomial Pro,Solution: Finding Binomial Probabilities Using Formulas,
26、.,Solution: Finding Binomial Pro,Example: Finding a Binomial Probability Using a Table,About 10% of workers (ages 16 years and older) in the United States commute to their jobs by carpooling. You randomly select eight workers. What is the probability that exactly four of them carpool to work? Use a
27、table to find the probability. (Source: American Community Survey),Solution:Binomial with n = 8, p = 0.1, x = 4,.,Example: Finding a Binomial Pr,Solution: Finding Binomial Probabilities Using a Table,A portion of Table 2 is shown,According to the table, the probability is 0.005.,.,Solution: Finding
28、Binomial Pro,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 considered an unusual event.,.,Solution: Finding Binomial P
29、ro,Example: Graphing a Binomial Distribution,Sixty-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 distribution.
30、 (Source: National Cancer Institute),Solution: n = 6, p = 0.62, q = 0.38Find the probability for each value of x,.,Example: Graphing a Binomial D,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
31、.,Solution: Graphing a Binomial,Solution: Graphing a Binomial Distribution,Histogram:,.,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.,Solution: Graphing a Binomial,Mean, Varian
32、ce, and Standard Deviation,Mean: = npVariance: 2 = npqStandard Deviation:,.,Mean, Variance, and Standard D,Example: Mean, Variance, and Standard Deviation,In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy
33、 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 = 0.44,Mean: = np = 300.56 = 16.8Variance: 2 = npq = 300.560.44 7.4Standard Deviation:,.,Example: Mean, Variance, and S,Solution: Mean, Varian
34、ce, 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.,.,Solution: Mean, Variance, and,感谢聆听,感谢聆听,
