### FINAL EXAM - STATISTICS 3093 - 1996 DECEMBER SOLUTION NOTES FOR VERSION ONE

 ```QUESTION ONE [24/80] 0.0 0.1 1.4 2.2 2.8 3.0 3.2 4.0 4.4 4.7 6.4 QUARTILES === === === MEAN (AVERAGE) 2.93 STANDARD DEVIATION 1.95 95% CONFIDENCE INTERVALS FOR MEDIAN X(2) to X(10) 0.1 TO 4.7 FOR MEAN 2.93 +- (2.228*1.95/sqrt(11)) 2.93 +- 1.31 ``` ```QUESTION TWO [8/80] a) The expected number of rejections 0.1*16 = 1.6 b) Probability that nobody rejects it 0.9^16 = 0.1853 c) Probability that (exactly) one person rejects it. 0.3294 This is a binomial distribution computation. ``` ```QUESTION THREE [8/80] This is an open question. Various approaches are correct. These are only sample answers: A1 1.3 2.4 3.3 2.1. 2.6 1.7 1.3 2.2 A2 1.3 2.4 3.3 2.1. 2.6 100 1.3 2.2 B1 Plot of X 0 1 2 3 4 5 Y 10 22 29 43 51 59 B2 Plot of X 0 1 2 3 4 5 Y 10 22 29 99 57 59 ``` ```QUESTION FOUR [24/80] X 1 2 3 4 5 sum of squares of residuals Y 18 18 19 16 34 = 126 FITTED LINE 15 18 21 24 27 s = sqrt(126/(5-2)) RESIDUAL 3 0 -2 -8 7 = 6.48 ``` ```QUESTION FIVE [16/80] X 1 2 3 TOTAL Probability 0.3 0.5 0.2 1 60/X 60.0 30.0 20.0 P*60/X 18.0 15.0 4.0 37 37 = E{60/X} D = (60/X)-37 23.0 -7.0 -17.0 D*D 529.0 49.0 289.0 P*D*D 158.7 24.5 57.8 241 241 = V{60/X} Alternate approach: The population of 60/X is .... 60 60 60 30 30 30 30 30 20 20 The mean of this is ............................. 37 The variance (divide by 10, not 9) is ........... 247 ``` ``` a) X and Y are independent E(60/X 30/Y) = 55.5 V(60/X 30/Y) = V(60/X) + V(30/Y) Independence = V(60/X) + V(30/X) X & Y have same distribution = V(60/X)+0.25V(60/X) Rules for constants = 241 + 0.25*241 Substitution b) 60/X and 30/X are not independent E(60/X 30/X) = 55.5 V(60/X 30/Y) = V(90/X) Algebra = 2.25 V(60/X) Rules for constants = 2.25*241 Substitution ```