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Unformatted text preview: 1. The width of a slot of a duralumin forging is (in inches) normally distributed with m = 0.9000 and s= 0.0030. The specification limits were given as 0.9000±0.0050. (a) What percentage of forgings will be defective? (b) What is the maximum allowable value of σ that will permit no more than 1 in 100 defectives when the widths are normally distributed with µ=0.9000 and σ? 1. The weight of anodized reciprocating pistons produced by Brown Company follows a normal distribution with mean 10 lb and standard deviation 0.2lb. a. The heaviest 2.5% of the pistons produced are rejected as overweight. What weight, in pounds, determines the overweight classification? Give your arguments. Let Y= weight (in lb) Find the z such that 0.975 = Prob ( Z < (Y10)/0.2) . This z is 1.96 (from the table). So the weight is Y= 1.96x0.2 + 10 = 10.392. So the weight that determines the overweight classification is 10.392 lbs. Prob(Y > 10.392) 10 10.392 Weight Prob(z > 1.96) 1.96 z b. Suppose Brown Company can sell only those pistons weighing between 9.8 and 10.4 lbs. What proportion of the pistons is lost? Prob( 9.8 < Y < 10.4) = Prob( 1 < z < 2) = Prob(z <2) – Prob(z < 1) = 0.9772 0.1587= 0.8185. This is the proportion of pistons that can be sold. The proportion of pistons that are lost is 10.8185 = 0.1815. 0.8185 9.8 10.4 weight 0.8185 1 2 z c. If a sample of pistons is available, how might your answer (a) and (b) without the normal population model? For part (a), find the 97.5th percentile of the sample. For part (b), find the proportion of sample values between 9.8 and 10.4. 2. A paint company knows from experience that the area in square feet, that 1 gallon of its premium paint will cover follows an N(250, 25) distribution model. The company wishes to advertise that if a gallon of this paint fails to cover t square feet, it will refund the purchase price plus 10%. What is the largest value of t that will ensure that on average no more than 0.5% of all purchases will be given the refund? Let Y = area covered. Find the z such that 0.005 = Prob(Z < (t250)/5) = Prob(Y < t). This is z = 2.57. The t = 2.57x25+250= 185.75. So if a gallon of this paint fails to cover 185.75 square feet, the company will refund the purchase price plus 10%. 3. An analog signal received at a detector (measured in microvolts) is normally distributed with a mean of 100 and a variance of 256; What is the probability that the signal will be less than 120 microvolts given that it is larger than 110 microvolts. Let X=signal in microvolts P ( X < 120  X > 110) = P ( X < 120 and X > 110) P ( X > 110) 120 − 100 ⎞ ⎛ 110 − 100 P⎜ <Z< ⎟ ⎝ P (110 < X < 120) 256 256 ⎠ = = 110 − 100 ⎞ P ( X > 110) ⎛ 1 − P⎜ Z < ⎟ ⎝ 256 ⎠ = P ( Z < 1.25) − P ( Z < 0.625) 0.8944 − 0.7324 = = 0.6053 1 − P ( Z < 0.625) 1 − 0.7324 ...
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This document was uploaded on 03/20/2011.
 Winter '09

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