《统计学基础(英文版·第7版)》课件les7e--ADA-0401.pptx

1、Elementary Statistics Seventh Edition Chapter 4 Discrete Probability DistributionsCopyright 2019,2015,2012,Pearson Education,Inc.Chapter Outline 4.1 Probability Distributions 4.2 Binomial Distributions 4.3 More Discrete Probability DistributionsSection 4.1Probability DistributionsSection 4.1 Objecti

2、ves How to distinguish between discrete random variables and continuous random variables How to construct a discrete probability distribution and its graph and how to determine if a distribution is a probability distribution How to find the mean,variance,and standard deviation of a discrete probabil

3、ity distribution How to find the expected value of a discrete probability distributionRandom Variables(1 of 3)Random Variable Represents a numerical value associated with each outcome of a probability distribution.Denoted by x Examples Numberof salescallsasalespersonmakesinone dayx.Hoursspentonsales

4、callsinonedayx.Random Variables(2 of 3)Discrete Random Variable Has a finite or countable number of possible outcomes that can be listed.Example Numberof salescallsasalespersonmakesinone dayx.Random Variables(3 of 3)Continuous Random Variable Has an uncountable number of possible outcomes,represente

5、d by an interval on the number line.Example Hoursspentonsalescallsinonedayx.Example:Discrete and Continuous Variables(1 of 2)Determine whether each random variable x is discrete or continuous.Explain your reasoning.1.Let x represent the number of Fortune 500 companies that lost money in the previous

6、 year.Solution:Discrete random variable(The number of companies that lost money in the previous year can be counted.)Example:Discrete and Continuous Variables(2 of 2)Determine whether each random variable x is discrete or continuous.Explain your reasoning.2.Let x represent the volume of gasoline in

7、a 21-gallon tank.Solution:Continuous random variable(The amount of gasoline in the tank can be any volume between 0 gallons and 21 gallons.)Discrete Probability DistributionsDiscrete probability distribution Lists each possible value the random variable can assume,together with its probability.Must

8、satisfy the following conditions:In WordsIn Symbols1.The probability of each value of the discrete random variable is between 0 and 1,inclusive.0P x2.The sum of all the probabilities is 1.1P xConstructing a Discrete Probability DistributionLet x be a discrete random variable with possible outcomes 1

9、2nx xx.1.Make a frequency distribution for the possible outcomes.2.Find the sum of the frequencies.3.Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.4.Check that each probability is between 0 and 1,inclusive,and that the sum of all the probabilit

10、ies is 1.Example:Constructing and Graphing a Discrete Probability DistributionAn industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees.Each individual was given a whole number score from 1 to 5,where 1 is extremely passive and 5 is extremely

11、 aggressive.A score of 3 indicated neither trait.The results are shown.Construct a probability distribution for the random variable x.Then graph the distribution using a histogram.Score,xFrequency,f124233342430521Solution:Constructing and Graphing a Discrete Probability Distribution(1 of 3)Divide th

12、e frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable.2410 16150P 3320 22150P 4230 28150P 3040 20150P 2150 14150P Discrete probability distribution:x12345P(x)0.160.220.280.200.14Solution:Constructing and Graphing a Dis

13、crete Probability Distribution(2 of 3)x12345P(x)0.160.220.280.200.14This is a valid discrete probability distribution since 1.Each probability is between 0 and 1,inclusive,0P x.2.The sum of the probabilities equals 1,0 160 220 280 200 14P x .Solution:Constructing and Graphing a Discrete Probability

14、Distribution(3 of 3)x12345P(x)0.160.220.280.200.14Because the width of each bar is one,the area of each bar is equal to the probability of a particular outcome.Also,the probability of an event corresponds to the sum of the areas of the outcomes included in the event.You can see that the distribution

15、 is approximately symmetric.Passive-Aggressive TraitsExample:Verifying a Probability DistributionVerify that the distribution for the three-day forecast and the number of days of rain is a probability distribution.Days of Rain,x0123Probability,P(x)0.2160.4320.2880.064Solution:Verifying a Probability

16、 DistributionSolutionIf the distribution is a probability distribution,then(1)each probability is between 0 and 1,inclusive,and(2)the sum of all the probabilities equals 1.1.Each probability is between 0 and 1.2.0 2160 4320 2880 064P x .Days of Rain,x0123Probability,P(x)0.2160.4320.2880.064Because b

17、oth conditions are met,the distribution is a probability distribution.Example:Identifying Probability Distributions(1 of 2)Determine whether each distribution is a probability distribution.Explain your reasoning.1.x5678P(x)0.280.210.430.15SolutionEach probability is between 0 and 1,but the sum of al

18、l the probabilities is 1.07,which is greater than 1.The sum of all the probabilities in a probability distribution always equals 1.So,this distribution is not a probability distribution.Example:Identifying Probability Distributions(2 of 2)Determine whether each distribution is a probability distribu

19、tion.Explain your reasoning.2.Solution The sum of all the probabilities is equal to 1,but P(3)and P(4)are not between 0 and 1.Probabilities can never be negative or greater than 1.So,this distribution is not a probability distribution.MeanMean of a discrete probability distribution xP x Each value o

20、f x is multiplied by its corresponding probability and the products are added.Example:Finding the MeanThe probability distribution for the personality inventory test for passive-aggressive traits is given.Find the mean score.Solution:2 942 9xP x Solution:Finding the MeanThe probability distribution

21、for the personality inventory test for passive-aggressive traits is given.Find the mean score.Solution:2 942 9xP x Recall that a score of 3 represents an individual who exhibits neither passive nor aggressive traits and the mean is slightly less than 3.So,the mean personality trait is neither extrem

22、ely passive nor extremely aggressive,but is slightly closer to passive.Variance and Standard DeviationVariance of a discrete probability distribution 22xP x Standard deviation of a discrete probability distribution 22xP xExample:Finding the Variance and Standard DeviationThe probability distribution

23、 for the personality inventory test for passive-aggressive traits is given.Find the variance and standard deviation.Score,xProbability,P(x)10.1620.2230.2840.2050.14Solution:Finding the Variance and Standard DeviationRecall 2.94Variance:21 6164xP x Standard Deviation:21 61641 3 Most of the data value

24、s differ from the mean by no more than 1.3.Expected ValueExpected value of a discrete random variable Equal to the mean of the random variable.E xxP x Example:Finding an Expected ValueAt a raffle,1500 tickets are sold at$2each for fourprizes of$500,$250,$150,and$75.You buy oneticket.Find the expecte

25、d value and interpret its meaning.Solution:Finding an Expected Value(1 of 2)To find the gain for each prize,subtract the price of the ticket from the prize:Your gain for the$500prize is$500$2$498 Your gain for the$250prize is$250$2$248 Your gain for the$150prize is$150$2$148 Your gain for the$75prize is$75$2$73 If you do not win a prize,your gain is$0$2$2 Solution:Finding an Expected Value(2 of 2)Probability distribution for the possible gains(outcomes)E xxP x11111496$498$248$148$73$215001500150015001500$1.35 You can expect to lose an average of$1 35for eachticket you buy.