Question

Name: Be sure to show your work for potential partial credit! 1. An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 138 lb and a standard deviation of 30.9 lb. a. If a pilot is randomly selected, find the probability that his weight is between 130 Ib and 191 lb. (Round to three decimal places as needed.) PC _ < X < __) = P(_ <Z< _) = b. Bonus: Is 130 or 191 a more extreme weight? Why? c. Find: 20.9 = — (Round to three decimal places) Relate this Z-score in to the pilot data by giving the 90th percentile for pilots' weights. (Hint: Remember units, round to 1 decimal place) d. If 30 different pilots are randomly selected, find the probability that their mean weight is between 130 lb and 191 lb. (Round to three decimal places as needed. Hint: Think about sampling distributions and the central limit theorem) Mz = – 0 = — P(_<i<_)=P( _<<< _) =

Answer 1

solution: e = 138 6 = 30.9 Z =x -all a) P (130 LX L191) =P ( 130 - 138 e 2c 191-138) 30.9 30.9 = P1 -0.26 CZC 1.72) = 0.5599 ab) plac130) = Plal-0.26) = 0.3974 P(z) VB 8) = Plz31-72) = 0.0422 Yes Ligiis considered as unusual because Plasisi) is less than o.of (C) Log = 1.282 Z = X - M ) 1.282 = X - 138 30.9 177.61 (d, dix= e = [138] - 30.9 5.6415 V3. P/130 27 (191) =P ( 130 - 13 8 1 10 1 1139 - P(-1.46 222 9.68 ) 0.9279