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Unformatted text preview: Statistics 427: Homework # 7 Solutions to Suggested Problems 4.29 a. .9838 is found in the 2.1 row and the .04 column of the standard normal table so c = 2.14. b. P(0 Z c) = .291 (c) = .7910 c = .81 c. P(c Z) = .121 1 P(c Z) = P(Z < c) = (c) = 1 .121 = .8790 c = 1.17 d. P(c Z c) = (c) (c) = (c) (1 (c)) = 2 (c) 1 (c) = .9920 c = .97 e. P( c  Z  ) = .016 1 .016 = .9840 = 1 P(c  Z  ) = P(  Z  < c ) = P(c < Z < c) = (c) (c) = 2 (c) 1 (c) = .9920 c = 2.41 4.33 a. P(X 18) = 25 . 1 15 18 Z P = P(Z 2.4) = (2.4) = .9452 b. P(10 X 12) = P(4.00 Z 2.40) P(Z 2.40) = (2.40) = .0082 c. P( X 15 1.5(1.25) ) = P( Z 1.5) = P(1.5 Z 1.5) = (1.5) (1.5) = .8664 4.39 a. + (91 st percentile from std normal) = 30 + 5(1.34) = 36.7 b. 30 + 5( 1.555) = 22.225 c. = 3.000 m; = 0.140. We desire the 90 th percentile: 30 + 1.28(0.14) = 3.179 4.44 a. P( 1.5 X + 1.5 ) = P(1.5 ) = P(1....
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This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.
 Spring '08
 Staff
 Statistics

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