Assignment Week 2 Bus308

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Assignment week 2 Milena I Barry Statistics for Managers Instructor Nicholas Bergan BUS 308 June 18, 2012 4.4 Suppose a couple will have three children. Letting B denote a boy and G denote a girl: a Draw a tree diagram depicting the sample space outcome for this experiment b List the sample space outcome that correspond to each of the following events: 1 All three children will have the same gender {BBB,GGG} 2 Exactly two of the children will be girls {BGG,GGB,GBG} 3 Exactly one of the three children will be a girl {BBG,BGB,GBB} 4 None of the children would be a girl {BBB} c Assuming that all sample space outcomes are equally likely, find the probability of each of the events given in part b 1= 2/8=1/4 2=3/8 3=3/8 4=1/8 4.20 John and Jane are married. The probability that John watches a certain television show is .4. The probability that Jane watches the show is .5. The probability that John watches the show, given that Jane does is .7. a Find the probability that both John and Jane watches the show P(John | Jane) * P(Jane)=P(John and Jane)/P(Jane) =(0.7)×(0.5)=0.35 b Find the probability that Jane watches the show, given that John does. P(Jane | John)=P(Jane and John)/P(John)=(0.35)/(0.4)=0.875 c Do John and Jane watch the show independently of each other? Justify jour answer. P(Jane) * P(John) = (0.5) * (0.4) = 0.20 P(Jane and John) ≠ P(Jane) * P(John) = 0.35 ≠ 0.20 This means that they don’t watch the show independently of each other. 5.12 Suppose the probability distribution of a random variable x can be described by the formula p(x) = X/15. for each of the values x - 1, 2,3,4 and 5. Example P(x+2) = p(2) =p(2) =2/15. a. Write out the probability distribution of x x | P(x)=x/15 | X*p(x) | (x^2) * P(x) | 1 | 0.0667 | 0.0667 | 0.0667 | 2 | 0.1333 | 0.2667 | 0.5333 | 3 | 0.2000 | 0.6000 | 1.8000 | 4 | 0.2667 |

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