Tài liệu Statistics xac xuat

Statistics Technical Service Field Service Quality Assurance Statistical Sampling Variable • Figures Attribute • Yes or No Technical Service Field Service Quality Assurance Sampling Samples are divided into two groups. • Random Samples – Drawn from a lot giving every sample the same chance to get picked. • Aimed Samples – Drawn in direct connection to an event. NEVER MIX THE TWO! Technical Service Field Service Quality Assurance Sampling Aimed Samples, objective. • Used to monitor the influence of an event in a production lot. • Accumulated results can be used to improve specific production steps, operational and technical. Technical Service Field Service Quality Assurance Sampling Random Samples, objective. • Used to give an estimation of the average defect rate in a production lot. • Accumulated results can be used to estimate the average performance level of a plant. Technical Service Field Service Quality Assurance Sampling Random Sampling, • • • • The size of a production lot does not matter. Only the sample size determines the accuracy. Percentage is not a good measurement. Applying statistics is necessary due to limited amount of sampling (cost reasons). • For rare events we use the Poisson Distribution Technical Service Field Service Quality Assurance Sampling Defect rates detected at a certain sample size Technical Service Field Service Quality Assurance Sampling Calculating sample size depending on AQL 4.6 Technical Service Field Service Quality Assurance Sampling Calculating sample size depending on AQL Formula: n = 100 x z AQL (%) Example: AQL = 1:10,000 = 0.01% Defect detected (C) = 1 at a Probability of 99% In the chart follow the 99-line to the C=1 curve, then go to the y-axis to get the z-value. In this case 4.6. From the formula we get: 100 x 4.6 = 46,000 0.01 If we take 46.000 samples from a lot with a assumed defect rate of 1 : 10.000, we have 99% chance to find one defect. If we find < 1, we know with 99% probability that the actual defect rate is < 1 : 10.000 Technical Service Field Service Quality Assurance Sampling Calculating sample size depending on AQL 2.3 Technical Service Field Service Quality Assurance Sampling Calculating sample size depending on AQL Formula: n = 100 x z AQL (%) Example: AQL = 1:10,000 = 0.01% Defect detected (C) = 1 at a Probability of 90% In the chart follow the 90-line to the C=1 curve, then go to the y-axis to get the z-value. In this case 2.3. From the formula we get: 100 x 2.3 = 23,000 0.01 If we take 23.000 samples from a lot with a assumed defect rate of 1 : 10.000, we have 90% chance to find one defect. If we find < 1, we know with 90% probability that the actual defect rate is < 1 : 10.000 Technical Service Field Service Quality Assurance Sampling Probability of detecting defects (%) Defect level Sample size Technical Service Field Service Quality Assurance Designing a QC System • • • • • The AQL must be determined by the Management Definitions of Defectives must be made Limits for Raw Materials must be set (calculated) Determine Evaluation methods based on defect type and accuracy Determine Sampling plans based on confidence and verification frequency required Technical Service Field Service Quality Assurance The confidence level • The probability is an expression of the degree of likelihood that the conclusions drawn from the results obtained by testing a certain number (quantity) of material are correct • The risk is an expression of the degree of the likelihood that the conclusions drawn from the result of testing a certain sample are incorrect • % Probabilty + % Risk = 100% Probabilities are usually expressed as the ratio: % Probability/100 Technical Service Field Service Quality Assurance Simulation of uncertainty Throw a dice 30 times and register the number of sixes. On a average we get 5 sixes, but there is a considerable variation m =n x p = 30 x 1/6 = 5 Notation: n = number of times p = probability each time to get a six m = mean or average number of sixes among n: m=nxp Technical Service Field Service Quality Assurance Statistics • • • • Examine 2400 packages from a production. Assume defect rate is 0.1% (1 : 1000). How many defects will be found? Repeat. M = n x p = 2400 x 0.1% = 2400 * 0.001 = 2.4 On a average we will find 2.4 defects, but there is a considerable variation Estimate of defect rate if …defects are found: – 1 : 1/2400 = 0.04% – 2 : 2/2400 = 0.17% Technical Service Field Service Quality Assurance Poisson distribution • • • • Unsterility is a rare event It is an attribute p is small (less than 0.1) Poisson distribution: P (x=k) = mk x e-m k! • this formula allows us to calculate the chance of finding ‘k’ elements with a certain characteristic in a sample with size ‘n’ in a population that contains 100 p% elements with that characteristic Technical Service Field Service Quality Assurance Poisson distribution: examples Example 1: • 4% of the Dutch population is more than 70 years old. • What is the chance to find three persons in a group of 100 persons that are older than 70 years? – M = n x p = 100 x 0.04 = 4 – P(x=3) = 43 x e-4 = 64/6 x 0.0183 = 0.1952 = 19.52% 3! Example 2: • Assumed defect rate in production is 1:1000 = 0.1% (p=0.001) • QC takes 100 samples from each production • What is the chance to find 1 defect? – M = n x p = 100 x 0.001 = 0.1 – P(x=1) = 0.11 x e-0.1 = 0.1 x 0.90 = 0.09 = 9.1% 1! Technical Service Field Service Quality Assurance Poisson distribution: examples Example 3: • Assumed defect rate in production is 1:1000 = 0.1% (p=0.001) • QC takes 200 samples from each production • What is the chance to find 1 defect? – M = n x p = 200 x 0.001 = 0.2 – P(x=1) = 0.21 x e-0.2 = 0.2 x 0.82 = 0.164 = 16.4% 1! Example 4: • Accepted defect rate in production is 1:1000 = 0.1% (p=0.001) • QC takes 2303 samples from a commissioning run • What is the chance to find zero defects? – M = n x p = 2303 x 0.001 = 2.303 – P(x=0) = 2.3030 x e-2.303 = 1 x 0.0999 = 0.0999 = 9.99% (10%) 0! Technical Service Field Service Quality Assurance Probability Example 4 (cont’d) : • Accepted defect rate in production is 1:1000 = 0.1% (p=0.001) • QC takes 2303 samples from a commissioning run • What is the chance to find zero defects? – M = n x p = 2303 x 0.001 = 2.303 – P(x=0) = 2.3030 x e-2.303 = 0.1= 10% 0! • The chance to find 1 or more defects is 100 - 10 = 90% • If the outcome of the sterility test is that zero defects are found in 2303 samples, we know that with 90% probability the defect rate will be less than 1:1000 Technical Service Field Service Quality Assurance