Y n150 25 b the probability that the linkage is

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Y N.150; :25/ (b) The probability that the linkage is larger than 150.2 is P.Y > 150:2/ D 1 ˆOE.150:2 150/=:5 ç D :3446 (c) The probability that the linkage is within tolerance is P.149:83 Y 150:21/ D ˆOE.150:21 150/=:5 ç ˆOE.149:83 150/=:5 ç D :2958
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4 40. Let X be the circumference of battery posts. Then X follows the Weibull distribution with lo- cation parameter D 3:25 , shape parameter ˛ D 1=3 and scale parameter D 1=:005 . (The BCNN text denotes the shape parameter by ˇ and the scale parameter by 1=˛ .) The cumulative distribution function is F.x/ D 1 exp OE ..x 3:25/=:005/ 1=3 ç ; x 3:25 (a) The probability of a post having a circumference greater than 3.40 is P.X > 3:40/ D 1 F.3:40/ D :0447 (b) The probability of a post not meeting tolerance is 1 P.3:3 < X < 3:5/ D 1 F.3:5/ C F.3:3/ D :9091 52. The time window should be 6:00 to 10:00 a.m. (a) By calculating areas we get ƒ.t/ D Z t 0 .s/ ds D ( 30t; 0 t < 1 45t 15; 1 t < 2 35 C 20t; 2 t 4 . (b) The number of arrivals between 6:30 and 8:30 a.m. is a Poisson random variable with rate (mean) ƒ.2:5/ ƒ.0:5/ D 70 . (c) The answer is P 59 k D 0 e 70 70 k =k ä .
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  • Spring '08
  • ALEXOPOULOS
  • Probability theory, Exponential distribution, Qj, Weibull, 000 hours, 10,000 hours

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