This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (f) 9 6 ( ;9,0.7) k b k = = 1  B(5; 9, 0.7) = 1(0.2703) = 0.7297 4.16 (a) b(1;12, 0.05) = B(1;12, 0.05) – B(0;12,0.05) = (0.8816)(0.5404) = 0.3412 (b) B(2; 12, 0.05) = 0.9804 (c) 1 B(1;12, 0.05) = 1(0.8816)=0.1184 4.18 (a) The probability that 1 or more components in a sample of 15 is defective when the true probability of being good is 0.95 , is 1  b(0;15, 0.05) = 1 15 (0.95) = 0.5367 (b) The probability that 0 are defective when the true probability is 0.90 is b(0; 15, 0.10) = 0.2059 (c) When the true probability is 0.90, we have b(0; 15, 0.20)= 0.0352...
View
Full
Document
This homework help was uploaded on 01/31/2008 for the course AMS 310.01 taught by Professor Mendell during the Fall '03 term at SUNY Stony Brook.
 Fall '03
 Mendell

Click to edit the document details