【正文】 e items, and the actual fraction of defective items. This derivation appears below. The number of different samples of size n that can be selected from a finite population N is termed a mathematical bination and is puted as: ()where a factorial, n! is n*(n1)*(n2)...(1) and zero factorial (0!) is one by convention. The number of possible samples with exactly x defectives is the bination associated with obtaining x defectives from m possible defective items and nx good items from Nm good items:()Given these possible numbers of samples, the probability of having exactly x defective items in the sample is given by the ratio as the hypergeometric series: ()With this function, we can calculate the probability of obtaining different numbers of defectives in a sample of a given size. Suppose that the actual fraction of defectives in the lot is p and the actual fraction of nondefectives is q, then p plus q is one, resulting in m = Np, and N m = Nq. Then, a function g(p) representing the probability of having r or less defective items in a sample of size n is obtained by substituting m and N into Eq. () and summing over the acceptable defective number of items: ()If the number of items in the lot, N, is large in parison with the sample size n, then the function g(p) can be approximated by the binomial distribution:()or ()The function g(p) indicates the probability of accepting a lot, given the sample size n and the number of allowable defective items in the sample r. The function g(p) can be represented graphical for each bination of sample size n and number of allowable defective items r, as shown in Figure 131. Each curve is referred to as the operating characteristic curve (OC curve) in this graph. For the special case of a single sample (n=1), the function g(p) can be simplified:()so that the probability of accepting a lot is equal to the fraction of acceptable items in the lot. For example, there is a probability of that the lot may be accepted from a single sample test even if fifty percent of the lot is defective. Figure 131 Example Operating Characteristic Curves Indicating Probability of Lot AcceptanceFor any bination of n and r, we can read off the value of g(p) for a given p from the corresponding OC curve. For example, n = 15 is specified in Figure 131. Then, for various values of r, we find: r=0r=0r=1r=1p=24%p=4%p=24%p=4%g(p) 2%g(p) 54%g(p) 10%g(p) 88%The producer39。s and consumer39。s risk can be related to various points on an operating characteristic curve. Producer39。s risk is the chance that otherwise acceptable lots fail the sampling plan (ie. have more than the allowable number of defective items in the sample) solely due to random fluctuations in the selection of the sample. In contrast, consumer39。s risk is the chance that an unacceptable lot is acceptable (ie. has less than the allowable number of defective items in the sample) due to a better than average quality in the sample. For example, suppose that a sample size of 15 is chosen with a trigger level for rejection of one item. With a four percent acceptable level and a greater than four percent defective fraction, the consumer39。s risk is at most eightyeight percent. In contrast, with a four percent acceptable level and a four percent defective fraction, the producer39。s risk is at most 1 = or twelve percent.In specifying the sampling plan implicit in the operating characteristic curve, the supplier and consumer of materials or work must agree on the levels of risk acceptable to themselves. If the lot is of acceptable quality, the supplier would like to minimize the chance or risk that a lot is rejected solely on the basis of a lower than average quality sample. Similarly, the consumer would like to minimize the risk of accepting under the sampling plan a deficient lot. In addition, both parties presumably would like to minimize the costs and delays associated with testing. Devising an acceptable sampling plan requires trade off the objectives of risk minimization among the parties involved and the cost of testing. Example 133: Acceptance probability calculation Suppose that the sample size is five (n=5) from a lot of one hundred items (N=100). The lot of materials is to be rejected if any of the five samples is defective (r = 0). In this case, the probability of acceptance as a function of the actual number of defective items can be puted by noting that for r = 0, only one term (x = 0) need be considered in Eq. (). Thus, for N = 100 and n = 5: For a two percent defective fraction (p = ), the resulting acceptance value is:Using the binomial approximation in Eq. (), the parable calculation would be:which is a difference of , or percent from the actual value of found above.If the acceptable defective proportion was two percent (so p1 = p2 = ), then the chance of an incorrect rejection (or producer39。s risk) is 1 g() = 1 = or ten percent. Note that a prudent producer should insure better than minimum quality products to reduce the probability or chance of rejection under this sampling plan. If the actual proportion of defectives was one percent, then the producer39。s risk would be only five percent with this sampling plan. Example 134: Designing a Sampling Plan Suppose that an owner (or product consumer in the terminology of quality control) wishes to have zero defective items in a facility with 5,000 items of a particular kind. What would be the different amounts of consumer39。s risk for different sampling plans? With an acceptable quality level of no defective items (so p1 = 0), the allowable defective items in the sample is zero (so r = 0) in the sampling plan. Using the binomial approximation, the probability of accepting the 5,000 items as a function of the fraction of actual defective items and the sample size is:To insure a ninety percent chance of rejecting a lot with an actual p