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Unformatted text preview: ST 314 HW #3 Solutions 1. Text, p. 133, exercise 3.40. The r.v. Y is distributed exponential with = 0.01. a. P(Y < 30) = 1 e  (30) = 1 e.01(30) = 1 e .30 = .2592 b. P(Y > 15) = 1 [1 e  (15) ] = e.15 = .8607 c. P(Y = 100) = 0 d. P(50 Y 150) = [1 e  (150) ] [1 e  (50) ] = e1.5 + e.5 = .3834 2. Text, p. 141, exercise 3.47. Let Y = thickness of bolts. Then Y is normally distributed with mean = 10.0 mm and standard deviation = 1.6 mm. a. P(9.2 Y 10.8) = P( (9.210)/1.6 (Y )/ (10.810)/1.6 ) = P (.5 Z .5) = P(Z < .5) P(Z < .5) = .6915 .3085 = .3830 b. P(Y 9.2) = P ((Y )/ (9.210)/1.6) = P(Z .5) = .3085 3. Text, p. 142, exercise 3.53. Let Y = volume percent. Then Y is normally distributed with = 70 and = 2. a. P(Y > 75) = P((Y )/ > (7570)/2) = P(Z > 2.5) = 1  .9938 = .0062 b. P(67 < Y < 73) = P((6770)/2 < Z < (7370)/2) = P(1.5 < Z < 1.5) = P(Z < 1.5) P(67 < Y < 73) = P((6770)/2 < Z < (7370)/2) = P(1....
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This note was uploaded on 05/27/2008 for the course ST 314 taught by Professor Arthur during the Spring '08 term at Oregon State.
 Spring '08
 Arthur

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