This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: R2 = 0.89 + (0.80 x 0.11) = 0.978 R3 = 0.95 + (0.80 x 0.05) = 0.99 Now we can calculate the reliability of the entire system: R = R1 x R2 x R3 = 0.98 x 0.978 x 0.99 = 0.9488556 Problem # 8 R 5 = 0.998 R = 0.998 (1/5) = 0.9995996 Problem # 10 R = 0.90 x (0.85 + (0.85x0.15)) x ((0.90 + (0.90 x 0.1)) x 0.95 = = 0.90 x 0.9775 x 0.99 x 0.95 = 0.8274048...
View
Full
Document
This note was uploaded on 05/30/2011 for the course BUS 515 taught by Professor Blossom during the Spring '10 term at Strayer.
 Spring '10
 BLOSSOM
 Management

Click to edit the document details