104CHAPTER 49.Letnbe the required number of measurements. LetXbe the average of thenmeasurements.Then the true value isμX, and the standard deviation isσX1n.NowP μX0 25XμX0 250 95. In any normal population, 95% of the population is within 1.96standard deviations of the mean.Therefore 1 96σX0 25. SinceσX1n,n61 47. The smallest value ofnis thereforen62.11. (a) LetXrepresent the number of defective parts in a shipment.ThenXBin 400 0 20 , soXis approximately normal with meanμX400 0 2080 and standard deviationσX400 0 20 88.To findP X90 , use the continuity correction and find thez-score of 90.5.Thez-score of 90.5 is 90 58081 31.The area to the right ofz1 31 is 10 90490 0951.P X900 0951.(b) LetYrepresent the number of shipments out of 500 that are returned.From part (a) the probability that a shipment is returned is 0.0951, soYBin 500 0 0951 .It follows thatYis approximately normal with meanμY500 0 095147 55 and standard deviationσY500 0 09510 90496 5596.To findP Y60 , use the continuity correction and find thez-score of 59.5.Thez-score of 59.5 is 59 547 556 55961 82.The area to the right of
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