251y0224 10 30 02 ECO 251 QUANTITATIVE BUSINESS ANALYSIS I SECOND EXAM OCTOBER 29 2002 NAME KEY SECTION ENROLLED MWF TR 10 11 12 30 Circle both days and time separately or just write down days and time Part I Multiple Choice 40 Points Remember that At least one means one or two or three or four or It does not mean exactly one 1 If A and B are complementary events a P A B 0 b P A B 1 c P A 1 P B d All of the above e None of the above 2 The following table shows the probabilities of the Whizbang Corporation s stock movement Economic Conditions Good Fair Poor Total Stock Price Rises 30 20 Stock Price Falls 04 20 44 Total Complete the table and then find the probability that the stock price goes up given that economic conditions are poor a 0600 b 2308 c 7600 d 1456 e 1075 Solution Here is the completed table Economic Stock Price Stock Price Total Conditions Rises Falls Good 30 04 34 Fair 20 20 40 Poor 06 20 26 Total 56 44 1 00 Using the notation in the next problem P U Po P U Po 06 2308 P Po 26 251y0224 10 30 02 Let us name the events in the previous problem as follows G Economic Conditions Good F Economic Conditions Fair Po Economic Conditions Poor U Stock price goes up and D Stock price goes down Then the probability that conditions are poor if the stock price is going down is a P D Po 3 b P D Po c P Po D d P D Po e None of the above 4 In the two problems above the joint probability of a rising stock price and poor economic conditions is a P U Po 06 b P U Po 76 c P U Po 23 d P U Po 06 e P U Po 76 5 If the probability that a child is a boy or a girl is equally likely and I have 4 children what is the probability that at least one of the first three is a girl a 375 b 875 c 625 d 667 e 750 Solution Forget about the fourth child The probability of at least one girl in 3 tries could be the sum of the probabilities of 1 2 and 3 girls To see how to do it this way look at the examples where we found the probabilities of 1 2 or three heads To do it right find P BBB 1 P BBB 1 5 5 5 1 125 875 6 A factory has three veeblefetzers Veeblefetzer A produces defective items 4 percent of the time is only used for 10 of the output Veeblefetzer B produces 20 of the output and 3 of its output is defective Because only 2 of the output of veeblefetzer C is defective 70 of the output is produced on veeblefetzer C If a defective item is found the probability that it comes from veeblefetzer C is closest to a 70 b 60 c 50 d 40 e 30 2 251y0224 10 30 02 Solution Here is the completed table A 0 4 D 9 6 B C 0 6 1 4 Total 2 4 19 4 68 6 97 6 D Total 10 20 70 100 To get the total row take 100 and multiply it by the percents given for production by each machine To get the D row multiply the total by the percent defective For example 4 of 10 is 0 4 Get the total defective by adding the row Then the fraction of defective items that come from C is 0 7 out of 1 4 or 50 Formally you were given the following P A 10 P B 20 P C 70 P D A 04 P D B 03 and P D C 02 You were asked for P C D By Bayes rule P C D P D C P C P D 02 70 024 014 0 5833 where we found P D using 024 P D P D A P D B P D C P D A P A P D B P B P D C P C 04 10 03 20 01 70 003 004 007 014 7 In problem 6 the proportion of output that is defective is closest to a 2 b 3 c 3 5 d 4 e 1 5 8 The event A has a probability of events are independent p the event B also has a probability of p If the two a P A B 2 p p 2 b P A B 2 p c d e P A B 2 p p 2 P A B 0 P A B p 2 Solution From the addition rule P A B P A P B P A B P A p P B p and since A and B are independent P A B P A P B p 2 So P A B p p p 2 2 p p 2 3 251y0224 10 30 02 If the events A and B each have probabilities above zero and below 1 A and B are not independent P A B 75 and P B A 6 then the following must be true a P A B is 1 b P A B is below 6 and above zero c P A B is between 6 and 75 d Either b or c could be true e There is not enough information to answer this question Solution From the multiplication rule P A B P A B P B and 9 P A B P B A P B A P A Since P A and P B are both less than 1 P A B must be less than both P A B and P B A 10 The following table shows the number of days absent among your employees in a month and the probability Number of days absent Probability 0 50 1 30 2 11 3 05 4 04 5 0 What is the expected value mean of the number of days absent a 1 00 b 0 83 c 2 00 d 0 166 e 10 0 Solution Here is the completed table for both the definitional and the computational formula as generated by Minitab There is no reason to bother with the definitional method x x P x x 2 P x row x P x xP x x 2 P x 1 0 0 50 0 00 0 00 0 83 0 4150 0 344450 2 1 0 30 0 30 0 30 0 17 0 0510 0 008670 3 2 0 11 0 22 0 44 1 17 0 1287 0 150579 4 3 0 05 0 15 0 45 2 17 0 1085 0 235445 5 4 0 04 0 16 0 64 3 17 0 1268 0 401956 1 00 0 83 1 83 0 0000 1 1411 We have a valid distribution since all the probabilities are between one and zero …
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