Question

Acceptance sampling is an important quality control technique, where a batch of data is tested to determine if the proportion of units having a particular attribute exceeds a given percentage. Suppose that 9% of produced items are known to be nonconforming. Every week a batch of items is evaluated and the production machines are adjusted if the proportion of nonconforming items exceeds 12%. [You may find it useful to reference the z table.) a. What is the probability that the production machines will be adjusted if the batch consists of 72 items? (Round "Z" value to 2 decimal places, and final answer to 4 decimal places.) Probability b. What is the probability that the production machines will be adjusted if the batch consists of 84 items? (Round "z" value to 2 decimal places, and final answer to 4 decimal places.) Probability

Answer 1

Pg No ① Answer:- Given that, Produced items = 9.1. proportion of nonconfirming = 12.4. a) what is the probability Batch consists of 72 items. adjusted if the and its exceeds The production machines will be batch consists of 72 items .12.4 is tapeeted value: the sampling distribution of (FLPP ELP) =p sample proportion is Standard Sampling deviation: distribution of Sample proportion is SDLP) =) PRIP) 72

pq no 2 024X0.91 79 V everyone 132 5x10? = 0.0332 3D (P) = 0.0337 Qe checks the normal Noor, approanimation condition , under the approximate by na72 Ř np>5 and n (1-P) >5 we will have to proceed assumption that we have to a normal distribution Now, we have to vegrily np = 7280.89 ... 6-48 n(1-P) - 72 x(1-0-07) = 72x091 265052

PINO to use ē np nul-e is not appropriate normal distribution approximation For because normat approximation which is both and are greater condition than 5 satisfied: production consists machines will of 72 items .: The probability that the be adjusted if the batch is and Exceeds 12.1. 17-P-ECB) I SOLO I 012-0-09 00337 = 0.03 0.03.2017 : : 0.890 Z: 089 we have to find the area subtract from 1 to get, P12%8:59) = 1-P(Z <0.89) 31-0181387 Now, are to the right

pg No G -0.18673 1. Plz>0.89) = 0.186731 The adjusted 18.45 probability that if the batch is 0:1867 the production consists of machine will be tą items exceeds b) what is the probability . The production machine the batch consists of 12.1. Expected value i ECP)=P batch consists 84 items. will be adjusted if 84 items and its exceeds - Omaq Standard deviation:- SOCB) EPC1-P) . .0911-08) V 84 ve 2619

Pg Nag approximation = 0.03122 SOLÔ) =0:03/ Now, we check the normal Condition. n=84 È np>5 and Nel-P)>5 we will have to proceed that we have to approximate distribution Now, we have to verify np = 84 X009 - 7-56 nll-P) - 8411-019) = 844091 : 76.44 under by the a assumption normal ł np n(1-P) is not normal distribution appropriate approximation to use contine Condition for because, normal approximation m which is both and are greater than 5 satisfied

pgno @ items and its . A batch Exceeds consists of 84 12.11 2=P-E(0) 5p10) 0.18-0.89 0.0312 = 0.03 - 0.0312 =0.961 IZ = 0.96 Now, we have to find the area to we subtract from 1 to get, PLZ >0.96) -l-P/Z<0.96) .:1-0.83147 the right - 016853 Pl7>0.96) - 0.16853 i. The probability that the production machine will be adjusted if batch consists of 84 items Gereeds 18.4 is 0.1685