12 13 00 251y0043 ECO251 QBA1 Name key FINAL EXAM Class DECEMBER 15 2000 Hint Though you should never give 2 answers to a problem never cross out the answer to a problem that you have given up on I may see something worth partial credit in the answer Part I Do all the Following 20 Points Make Diagrams A z N 0 1 1 P z 2 93 P z 0 P 0 z 2 93 5 4983 0017 Diagram These are expected and will be provided in class for version 2 Half of you didn t read this question You assumed that it said P z because that was on last year s exam 2 P 3 42 z 0 19 P 3 42 z 0 P 0 19 z 0 4997 0753 4244 3 P 1 07 z 1 07 2 P 0 z 1 07 2 3577 7154 4 P 1 07 z 2 57 P 0 z 2 57 P 0 z 1 07 4949 3577 1372 5 The 93rd percentile of the distribution z 07 is the point with a probability of 07 above it or 93 below it Since 93 is below this point and 50 is below zero from the diagram P 0 z z 11 4300 The closest we can come to this probability using the Normal table is P 0 z 1 48 4306 So z 07 1 48 Note that probabilities cannot be above 1 or below 0 1 12 13 00 251y0043 x As usual people made diagrams of x with zero in the middle Make up your mind If you are diagramming x put the mean in the middle if you are diagramming z put zero in the middle B x N 1 2 5 Remember z 1 P x 2 93 2 93 1 2 P z P z 0 35 P z 0 P 0 z 0 35 5 1368 3632 5 0 19 1 2 3 42 1 2 z P 0 92 z 0 28 2 P 3 42 x 0 19 P 5 5 P 0 92 z 0 P 0 28 z 0 3212 1103 2109 1 07 1 2 1 07 1 2 z P 0 45 z 0 03 3 P 1 07 x 1 07 P 5 5 P 0 45 z 0 P 0 03 z 0 1736 0120 1616 2 57 1 2 1 07 1 2 z P 0 03 z 0 27 4 P 1 07 x 2 57 P 5 5 P 0 z 0 27 P 0 z 0 03 1064 0120 1184 5 The 93rd percentile of the distribution x 07 is the point with a probability of 07 above it or 93 below it Because it is above the 50th percentile which is the mean for the Normal distribution the point we want will be above 1 2 On the previous page we found z 07 1 48 So x 97 z 97 1 2 1 48 5 8 60 To check to see if this is correct P x 8 60 8 60 1 2 P z P z 1 48 P z 0 P 0 z 1 48 5 4306 9306 93 5 2 12 13 00 251y0043 II 4 points 2 point penalty for not trying Show your work We are investigating the cost of a business trip for people in the financial industry The data for a sample of 6 are below x 1433 1225 1433 1573 1333 941 Compute the sample standard deviation Solution s x 1 2 3 4 5 6 x 1433 1225 1433 1573 1333 941 7938 Show your work 4 x2 2053489 1500625 2053489 2474329 1776889 885481 10744302 x 7938 1323 s x2 n 6 x 2 nx 2 n 1 10744302 6 1323 2 242328 48465 6 5 5 s x 48465 6 220 1490 How about those jokers who are still trying to compute x x 2 by computing x x 2 It didn t work last term and it won t work this term I don t encourage using definitional formulas but if you insist on using them I will be happy to give you a personal tutorial on how to do it They are generally harder to use than computational formulas 3 12 13 00 251y0043 III Do at least 4 of the following 6 Problems at least 12 each or do sections adding to at least 48 points Anything extra you do helps and grades wrap around Show your work Please indicate clearly what sections of the problem you are answering If you are following a rule like E ax aE x please state it If you are using a formula state it If you answer a yes or no question explain why If you are using the Poisson or Binomial table state things like n p or the mean Avoid crossing out answers that you think are inappropriate you might get partial credit 1 a Find z 085 2 b If x N 18 5 find a symmetrical interval around the mean with a probability of 83 2 c If x N 18 5 find F 22 6 2 d If computer parts last an average of eighteen months with a standard deviation of 5 what is the probability that 15 of them will all last less than 22 6 months 2 e If a sample mean is taken from a sample of 15 from a normal population with a mean of 18 and a standard deviation of 5 what is the standard deviation of the sample mean 2 f What is the probability that the sample mean in 5 above is less than 22 6 2 g If sample mean of 14 3 comes from a sample of 15 from a normal population with a population standard deviation of 5 do an 83 confidence interval for the population mean 3 Solution From the problem statement x N 18 5 i e 18 and 5 a z 085 is the point with a probability of 085 above it or 915 below it It is the 91 5 percentile of z From the diagram P 0 z z 085 4150 the closest we can come to this probability using the Normal table is P 0 z 1 37 4147 So z 085 1 37 Make a diagram b A symmetrical interval around the mean with a probability of 83 We can split the 83 into two halves of 41 5 on either side of zero Make a diagram From the diagram we want two points z 085 and z 915 z 085 so that P z 915 z z 085 8500 And since we found z 085 1 37 in part a our interval for z is 1 37 to 1 37 The interval for x can then be written x z 075 18 1 37 5 18 6 85 or 11 15 to 24 85 24 85 18 11 15 18 z To check this P 11 15 x 24 85 P 5 5 P 1 37 z 1 37 2 P 0 z 1 …
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