SAMPLING INSPECTION TABLES 25 



TABLE A 



Values of x and y for Given Values of c 



Used in equation (18) for determining p^ when N, n and c are given, or in equation 



(22) for determining n when N, c and p/^ are given 



The value of c that minimizes / (equation (8'), using p = p), is given directly 

 by Fig. 9, which uses coordinates of M = pN and k = p/pL- The curves bound- 

 ing the c zones on Fig. 9 were obtained directly from relations between equations 

 (18) and (8'), using p = p, that define values of M and k such that / is the same 

 for c and c + \. 



The value of n, corresponding to the value of c given on Fig. 9, is determined 

 from equation (18), expressed as 



yN 



PlN + y 



(22) 



Example: Given: N = 750, pL = 0.01, p = 0.004. 

 To Find: n and c. 



Solution: M = pN = (0.004) (750) = 3; ^ = ^ = ^^^ = 0.4. 



_ pL 0.01 



Consulting Fig. 9, for M =3 and k = 0.4, read c = 1. 



From Table A, for c = 1, read y = 0.8408. 



F .• (11^ (0.8408) (750) 



From equation (22), n = (q.OI) (750) + 0.8408 = ^'•'- 



Sampling Plan: n = 76, c = 1. 



Double Sampling 



Given: Lot Size (N), AOQL {pi), process average fraction defective {p). 

 To find: Values of «i, n^, ci and C2 that will minimize I. 

 The average quality after inspection {Pa) is found by substituting in equation 

 (12), the value of / given in equation (9). 



h n OT=c I / m=C 2—c \—\ 



pA = ~\ {N - th) ^ Pni.pnx + {N - »i - rh) { Pei + l.pni ^ Pm,pn^_ 



L »»=0 \ '"=0 



+ ••• + -Pc2,prn-Po,;)n2 ) • (23) 



Differentiating equation (23) with respect to p and equating to 0, in accordance 

 with equation (14), and .solving for p, gives the value of ^ = pi that makes p^ 

 a maximum; i.e., pA = />/,. The resulting equation is not reproduced here since 



