practice_prel1_sol - Engineering Statistics ENGRD270 Prelim...

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Engineering Statistics ENGRD270 Prelim No.1 Solutions Spring 2008 1. (a) n = 24 + 9 + 27 = 60, Median = 1 2 ( X (30) + X (31) ) = 154 , Q 1 = 1 2 ( X (15) + X (16) = 138 , Q 3 = 1 2 ( X (45) + X (46) ) = 168 . 5 (b) IQR = 168 . 5 - 138 = 30 . 5 . 168 . 5 + 1 . 5 × 30 . 5 = 214 . 75 so 219 is an outlier on upper side. 138 - 1 . 5 × 30 . 5 = 92 . 25 so no outliers on lower side. Boxplot: Box: median line at 154, box ends at 138 and 168.5. Whisker ends at 96 and 210 (not 92.25, 214.75). Outlier at 219. (c) 90th percentile. ( i - 1 2 ) /n = 0 . 9 so i = 54 . 5. q . 9 = 1 2 ( X (54) + X (55) ) = 189 . 5 2. Note: Pr(incorrect answer) = 1 2 irrespective of whether correct answer is “true” or “false”. Pr[Pass Prelim] = ± 1 2 6 + 6 × 1 2 × ± 1 2 5 = 7 × 2 - 6 = 0 . 1094 Pr[Pass Final] = ± 1 2 10 + 10 × 1 2 × ± 1 2 9 = 11 × 2 - 10 = 0 . 01074 Pr[Pass Course] = Pr[Pass Prelim] 2 × Pr[Pass Final] = (7 × 2 - 6 ) 2 × 11 × 2 - 10 = 49 × 11 × 2 - 22 = 0 . 000129 . 3. Let A be the event that the morning lecture is bad and B that the afternoon lecture
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practice_prel1_sol - Engineering Statistics ENGRD270 Prelim...

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