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From Wikipedia
Acceptance sampling uses statistical sampling to determine whether to accept or reject a production lot of material. It has been a common quality control technique used in industry and particularly the military for contracts and procurement.
A wide variety of acceptance sampling plans are available.
History
Acceptance sampling procedures became common during WWII. Sampling plans, such as MILSTD105, were developed by Harold F. Dodge and others and became frequently used as standards.
More recently, quality assurance broadened the scope beyond final inspection to include all aspects of manufacturing. Broader quality management systems include methodologies such as statistical process control, HACCP, six sigma, and ISO 9000. Some use of acceptance sampling still remains.
Rationale
Sampling provides one rational means of verification that a production lot conforms with the requirements of technical specifications. 100% inspection does not guarantee 100% compliance and is too time consuming and costly. Rather than evaluating all items, a specified sample is taken, inspected or tested, and a decision is made about accepting or rejecting the entire production lot.
Plans have known risks: an acceptable quality limit and a rejectable quality level (LTDP) are part of the operating characteristic curve of the sampling plan. These are primarily statistical risks and do not necessarily imply that defective product is intentionally being made or accepted. Plans can have a known average outgoing quality limit (AOQL).
Attribute plans
MILSTD105 was a United States defense standard that provided procedures and tables for sampling by attributes (pass or fail characteristic). MILSTD105E was cancelled in 1995 but is available in related documents such as ANSI/ASQ Z1.4, "Sampling Procedures and Tables for Inspection by Attributes". Several levels of inspection are provided and can be indexed to several AQLs. The sample size is specified and the basis for acceptance or rejection (number of defects) is provided.
Variables plans
When a measured characteristic produces a number, other sampling plans such as those based on MILSTD414 are often used. Compared with attriute sampling plans, these often use a smaller sample size for the same indexed AQL.
The sample mean or empirical mean and the sample covariance are statistics computed from a collection of data.
Sample mean and covariance
Given a random sample \textstyle \mathbf{x}_{1},\ldots,\mathbf{x}_{N} from an \textstyle ndimensional random variable \textstyle \mathbf{X} (i.e., realizations of \textstyle N independent random variables with the same distribution as \textstyle \mathbf{X}), the sample mean is
 \mathbf{\bar{x}}=\frac{1}{N}\sum_{k=1}^{N}\mathbf{x}_k.
In coordinates, writing the vectors as columns,
 \mathbf{x}_{k}=\left[ \begin{array} [c]{c}x_{1k}\\ \vdots\\ x_{nk}\end{array} \right] ,\quad\mathbf{\bar{x}}=\left[ \begin{array} [c]{c}\bar{x}_1 \\ \vdots\\ \bar{x}_n \end{array} \right] ,
the entries of the sample mean are
 \bar{x}_{i}=\frac{1}{N}\sum_{k=1}^{N}x_{ik},\quad i=1,\ldots,n.
The sample covariance of \textstyle \mathbf{x}_{1},\ldots,\mathbf{x}_{N} is the nbynmatrix \textstyle \mathbf{Q}=\left[ q_{ij}\right] with the entries given by
 q_{ij}=\frac{1}{N1}\sum_{k=1}^{N}\left( x_{ik}\bar{x}_i \right) \left( x_{jk}\bar{x}_j \right) .
The sample mean and the sample covariance matrix are unbiased estimates of the mean and the covariance matrix of the random variable \textstyle \mathbf{X}. The reason why the sample covariance matrix has \textstyle N1 in the denominator rather than \textstyle N is essentially that the population mean E(X) is not known and is replaced by the sample mean \textstyle\bar{x}. If the population mean E(X) is known, the analogous unbiased estimate
 q_{ij}=\frac{1}{N}\sum_{k=1}^N \left( x_{ik}E(X_i)\right) \left( x_{jk}E(X_j)\right)
with the population mean indeed does have \textstyle N. This is an example why in probability and statistics it is essential to distinguish between upper case letters (random variables) and lower case letters (realizations of the random variables).
The maximum likelihoodestimate of the covariance
 q_{ij}=\frac{1}{N}\sum_{k=1}^N \left( x_{ik}\bar{x}_i \right) \left( x_{jk}\bar{x}_j \right)
for the Gaussian distribution case has N as well. The ratio of 1/N to 1/(N − 1) approaches 1 for large N, so the maximum likelihood estimate approximately equals the unbiased estimate when the sample is large.
Weighted samples
In a weighted sample, each vector \textstyle \textbf{x}_{k} is assigned a weight \textstyle w_k \geq0. Without loss of generality, assume that the weights are normalized:
 \sum_{k=1}^{N}w_k = 1.
(If they are not, divide the weights by their sum.) Then the weighted mean \textstyle \mathbf{\bar{x}} and the weighted covariance matrix \textstyle \mathbf{Q}=\left[ q_{ij}\right] are given by
 \mathbf{\bar{x}}=\sum_{k=1}^N w_k \mathbf{x}_k
and
 q_{ij}=\frac{\sum_{k=1}^N w_k \left( x_{ki}\bar{x}_i \right) \left( x_{kj}\bar{x}_j \right) }{1\sum_{k=1}^{N}w_k^2}.
If all weights are the same, \textstyle w_{k}=1/N, the weighted mean and covariance reduce to the sample mean and covariance above.
Criticism
The sample mean and sample covariance are widely used in statistics and applications, and are extremely common measures of location and dispersion, respectively, likely the most common: they are easily calculated and possess desirable characteristics.
However, they suffer from certain drawbacks; notably, they are not robust statistics, meaning that they are thrown off by outliers. As robustness is often a desired trait, particularly in realworld applications, robust alternatives may prove desirable, notably quantilebased statistics such the sample median for location, and interquartile range (IQR) for dispersion. Other alternatives include trimming and Winsorising, as in the trimmed mean and the Winsorized mean.
A sample is a subject chosen from a population for investigation. A random sample is one chosen by a method involving an unpredictable component. Random sampling can also refer to taking a number of independent observations from the same probability distribution, without involving any real population. The sample usually is not a representative of the population from which it was drawn— this random variation in the results is known as sampling error. In the case of random samples, mathematical theory is available to assess the sampling error. Thus, estimates obtained from random samples can be accompanied by measures of the uncertainty associated with the estimate. This can take the form of a standard error, or if the sample is large enough for the central limit theorem to take effect, confidence intervals may be calculated.
Types of random sample
 A simple random sample is hi selected so that all samples of the same size have an equal chance of being selected from the population.
 A selfweighting sample, also known as an EPSEM (Equal Probability of Selection Method) sample, is one in which every individual, or object, in the population of interest has an equal opportunity of being selected for the sample. Simple random samples are selfweighting.
 Stratified sampling involves selecting independent samples from a number of subpopulations, group or strata within the population. Great gains in efficiency are sometimes possible from judicious stratification.
 Cluster sampling involves selecting the sample units in groups. For example, a sample of telephone calls may be collected by first taking a collection of telephone lines and collecting all the calls on the sampled lines. The analysis of cluster samples must take into account the intracluster correlation which reflects the fact that units in the same cluster are likely to be more similar than two units picked at random.
Methods of producing random samples
 Random number table
 Mathematical algorithms for pseudorandom number generators
 Physical randomization devices such as coins, playing cards or sophisticated devices such as ERNIE
From Yahoo Answers
Answers:Your question is incomplete, but what I think you are saying is that if we sample 80 parts and find 2 or more defective parts we will not accept the shipment. Let X be the number of defective parts. X has the binomial distribution with n = 80 trials and success probability p = 0.03 In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(nx)!) * p^x * (1p)^(nx) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n  x failures. Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials. X ~ Binomial( n = 80 , p = 0.03 ) the mean of the binomial distribution is n * p = 2.4 the variance of the binomial distribution is n * p * (1  p) = 2.328 the standard deviation is the square root of the variance = ( n * p * (1  p)) = 1.525778 The first few elements of the Probability Mass Function, PMF, f(X) = P(X = x) is: P( X = 0 ) = 0.08744576 P( X = 1 ) = 0.2163606 P( X = 2 ) = 0.2643169 P( X = 3 ) = 0.2125435 we know that P( X 2) = 1  P(X < 2) = 1  (P(X = 0) + P(X = 1)) = 1  (0.08744576 + 0.2163606) = 0.6961936 P(X 2) = 0.6961936 this is the probability we reject the lot. the probability we accept is: P(X < 2) = 0.08744576 + 0.2163606 = 0.3038064
Answers:This is just the binomial probability of finding (x=) 0 defects in (n=) 4 attempts, given that the probability of a defect is (p=) 100/5000. Just use the binomial formula. (PS  this is barely acceptance sampling)
Answers:you mean past paper for IB exams? if not then i dunno..but if yes then i have some~ for math HL & SL, physics HL , psychology HL& SL, english A2 which subjects do you want?
Answers:These are straightforward binomial distributions, using the formula (a + b)^n where a = probability of a good item, b = probability of a defective item, n = number in sample. In (a), probability of a good item = 49/50, or 0.98, probability of a defective item = 1/50, or 0.02. Number in sample = 5, therefore the formula is (0.98 + 0.02)^5 and the probability of zero defective in the sample is the first term in the expansion, which is 0.98^5 = 0.9039 The others are exactly similar : (b) (0.96 + 0.04)^10 gives probability of zero defective = 0.96^10 = 0.6648 (c) (0.98 + 0.02)^20 gives a probability of 0.98^20 = 0.6676
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