Downloaded from https://www.everyspec.com AMCP 706-1" Example No. 24 (Cont'd) .1 Write the equations for the 4. The accept line it accept and reject lines. Use A= -1.3208 + 0.03 2 50n Eqs. 3-13a and 3-13b. The reject line 18 = + 1.320t + 0.03 2 50n. b. Compute the 5 nominal prob- 5. Po(1) = 1 ab)laties of r.oceptanoe from P,(O. 99) 1 - 10% - 90% Eq. 3-14. e,(o 967) = 2. 1072 lPa(0. 9675) a2. 1972 297250% P,(o. 95) = 10% Pa(0) = 0%. 2.1972 6. Conrpute the average number 6. ASN 1 . 0 = 0.047 = 40.64 of items tested before a de- cteton, for R values in Eq. (0. 90 X 2. 1972) + (0.10 X 2.1972) 3-15. ASN°'89 (0.01 x 1.6094) + (0.99 x 0.08407) -31.56 2.1972 2 1972 0.05407 1,6094 (0.10 x 2.
1972) + (0.90 x 2. 1972) A8N 9 (0.05 X 1. 6094) + (•.95 0.05407) - 16. 67 2.1!972 ASN° = 1.6094 - 1.37 - 2. No trunuatlon lines are calculated fov this example.