A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2. a. ?What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05? .10? .20? .25? b. ?Let ?p ?denote the actual proportion of defectives in the batch. A graph of ?P?(batch is accepted) as a function of ?p, ?with ?p ?on the horizontal axis and ?P?(batch is accepted) on the vertical axis, is called the ?operating characteristic curve ?for the acceptance sampling plan. Use the results of part (a) to sketch this curve for 0 ? p ? 1. c.? ?Repeat parts (a) and (b) with “1” replacing “2” in the acceptance sampling plan. d.? ?Repeat parts (a) and (b) with “15” replacing “10” in the acceptance sampling plan. e. ?Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

Answer Step 1 of 6 Given n=10 Let x=no.of defective Here X~B(X,P) The pmf of binomial distribution is P(x)=c x)(p )(q ), x=0,1,2……..,n Step 2 of 6 When p=0.01, q=0.99 P(accepting the batch)=P(X2) =P(x=0)+P(x=1)+P(x=2) =0.9044+0.0913+0.0042 =0.9999 When p=0.05, q=0.95 P(accepting the batch)=P(X2) =P(x=0)+P(x=1)+P(x=2) =0.5987+0.3151+0.0746 =0.9984 When p=0.1, q=0.9 P(accepting the batch)=P(X2) =P(x=0)+P(x=1)+P(x=2) =0.3487+0.3874+0.1937 =0.9298 When p=0.2, q=0.8 P(accepting the batch)=P(X2) =P(x=0)+P(x=1)+P(x=2) =0.1074+0.2684+0.302 =0.6778 When p=0.25, q=0.75 P(accepting the batch)=P(X2) =P(x=0)+P(x=1)+P(x=2) =0.0563+0.1877+0.2816=0.5256 x 0.01 0.05 0.1 0.2 0.25 p(x) 0.9999 0.9984 0.9298 0.6778 0.5256 b)Step 3 of 6 Step 4 of 6 c) Repeat parts (a) and (b) with “1” replacing “2” in the acceptance sampling plan. Here X~B(X,P)=B(10,p) When p=0.01, P(accepting the batch)=P(X1)=0.996 When p=0.05, P(accepting the batch)=P(X1)=0.914 When p=0.1, P(accepting the batch)=P(X1)=0.736 When p=0.2, P(accepting the batch)=P(X1)=0.3758 When p=0.25 P(accepting the batch)=P(X1)=0.244 x 0.01 0.05 0.1 0.2 0.25 p(x) 0.996 0.914 0.736 0.3758 0.244 Step 5 of 6 d ) Repeat parts (a) and (b) with“15” replacing “10” in the acceptance sampling plan. Here X~B(X,P)=B(15,p) When p=0.01, P(accepting the batch)=P(X2)=0.9995 When p=0.05, P(accepting the batch)=P(X2)=0.964 When p=0.1, P(accepting the batch)=P(X2)=0.816 When p=0.2, P(accepting the batch)=P(X2)=0.398 When p=0.25 P(accepting the batch)=P(X2)=0.236 x 0.01 0.05 0.1 0.2 0.25 p(x) 0.9995 0.964 0.816 0.398 0.236 Step 6 of 6 e) Plan (d) is the most satisfactory plan Because P(accept) is high for p 0.1 and low for p >0.1