exam1-sol

exam1-sol - 1.a: [15pt] 145 1.645 8 35 = 145 2.22 =...

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1.a: [15pt] 145 ± 1 . 645 × 8 35 = 145 ± 2 . 22 = [142 . 78 , 147 . 22] 1.b: [15pt] H 0 : μ = 142 vs. H a : μ > 142. Z = 145 - 142 8 / 35 = 2 . 22. p -value is 0.0132. Reject H 0 at 5% significance level. 1.c:[15pt] n = ( 1 . 96 × 8 1 . 5 ) 2 = 109 . 3 110 2.a: [5pt] 26 3 × 10 3 = 17 , 576 , 000 2.b: [5pt] 26 × 25 × 24 × 10 × 9 × 8 / 17 , 576 , 000 = 0 . 6390533 3.a: [15pt] For each passenger’s baggage, the probability of being overweight is P ( X 100) = P ( Z 100 - 95 35 ) = P ( Z 0 . 14) = 0 . 4443. Hence, the probability that all three passengers have overweight baggages is 0 . 4443 3 = 0 . 0877 (by multiplication rule for independent events). 3.b: [15pt] Since n = 200 > 30, by CLT (of sample sum), we have 200 X i =1 x i ˙ N (95 × 200 , 200 × 35). Hence, the probability of exceed the 20,000-pound limit is
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This note was uploaded on 12/28/2011 for the course EXST 7003 taught by Professor Moser,e during the Fall '08 term at LSU.

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