【正文】 Z is Normally distributed with Mean = 0 and SD = 1 Z值平均值為 ―0‖ 標(biāo)準(zhǔn)偏差為 ―1‖的正態(tài)分布值 . LSL ? ?LSLXP ?Z score Z Scores Z值 55 ZLSL, ZUSL USL ? ?? ? ? ? ? ?? ? ? ?psxUSLZsxLSLPsxUSLsxXsxLSLPUSLXLSLP??????? ??????????? ????????? ?USLXP ?LSL ? ?LSLXP ?ZLSL ZUSL 多少 ? 過程 USL ? ?sxUSLscor eZ ??ZScore interpretation: How many standard deviations, s or shats, is the mean, xbar, from some specified value, x. Z值解釋 :是一個到平均值有幾個標(biāo)準(zhǔn)偏差的的特定值 xLet’s assume there is only an USL 讓我們假設(shè)只有只有規(guī)格下限 ?s ppm USL T ? A Six Sigma Process 25,000 ppm USL T ? A Two Sigma Process Basic Instructions for Minitab Computing Standard Normal Probabilities 計(jì)算標(biāo)準(zhǔn)正態(tài)概率的 MiniTab基本操作 57 MPT Confidential ? Select ―Calc‖, ―Probability Distributions‖ and ―Normal‖. ? Select ―Cumulative Probability‖, enter the ―Mean‖ and ―Standard Deviation‖, click on ―Input constant‖, enter the value and click on ―OK‖. Computing Percent Fallout Cumulative Distribution Function Normal with mean = and standard deviation = x P( X = x) ? ?? ?1???????USLXPUSLXP58 MPT Confidential Minitab Output USL = 12 T ? ?11 A One Sigma Process DPPM = () x 1,000,000 = 158700 ? ? ? ? 11 1112 ???????sxUSLsxxZsco reZUSLCumulative Distribution Function Normal with mean = and standard deviation = x P( X = x) 59 MPT Confidential Computing ZScore From Percent Fallout ? Select ―Inverse Cumulative probability‖, set the ―Mean‖ = 0 and ―Standard Deviation‖ =1, click on ―Input constant‖, enter the total area associated with fallout and click on ―OK‖. p = ? ? ? ???????scor eZZPzZPInverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = P( X = x) x 60 MPT Confidential ? ? 1 LSL =9 USL = 12 T ? ? 11 Determine the DPPM ZLSL, ZUSL and the Z score Exercise 61 MPT Confidential ? Select ―Calc‖, ―Probability Distributions‖ and ―Normal‖. ? Select ―Cumulative probability‖, enter the ―Mean‖ and ―Standard Deviation‖, click on ―Input constant‖, enter the value and click on ―OK‖. Cumulative Distribution Function Normal with mean = and standard deviation = x P( X = x) ? ? ? ? 11 1112 ??????sxUSLsxxZUSLSolution: Minitab 62 MPT Confidential ? Select ―Cumulative probability‖, enter the ―Mean‖ and ―Standard Deviation‖, click on ―Input constant‖, enter the value and click on ―OK‖. Cumulative Distribution Function Normal with mean = and standard deviation = x P( X = x) DPPM =( () +()) x 1,000,000 = 181500 Solution: Minitab ZLSL = (9 11) / 1 63 MPT Confidential ? Select ―Inverse Cumulative probability‖, set the ―Mean‖ = 0 and ―Standard Deviation‖ =1, click on ―Input constant‖, enter the total area associated with fallout and click on ―OK‖. p = 1(() +()) = = ? ? ? ???????scor eZZPzZPInverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = P( X = x) x Solution: Minitab 64 MPT Confidential Process Capability for NonNormal Data 非正態(tài)的過程能力 65 MPT Confidential Process Capability for NonNormal Data 非正態(tài)數(shù)據(jù)的過程能力 Not every measured characteristic is normally distributed. Some data follows distributions that are known, and these may be able to have their capability measured accurately using that knowledge 不是所有的測量參數(shù)都是正態(tài)分布的 ,一些數(shù)據(jù)分布已知 ,就可以準(zhǔn)確地應(yīng)用自己的過程能力分析方法去測量過程能力 Characteristic參數(shù) Distribution分布 Cycle Time循環(huán)時(shí)間 Weibull (Exponential)威布爾 (指數(shù) )分布 Reject Rate不合格率 Binomial二項(xiàng)式分布 Defect Rate缺陷率 Poisson泊松分布 66 MPT Confidential The Weibull Distribution is a general family of distribution with 威布爾分布為一個常見的分布 ,用下式表示 : where scale parameter ? is the value at which CDF=%, and shape parameter ? determines the shape of the PDF. ? 上式中 , 尺度參數(shù) θ為 CDF=%時(shí)的值 ? 形狀參數(shù) β確定了 PDF的形狀 Process Capability for Cycle Time 周期時(shí)間的過程能力 ? ? ??????? ??? ??????? x1 ex,。xWeibull67 MPT Confidential At ?=1, the Weibull Distribution is reduced to 當(dāng) β=1,威布爾分布可簡化為 :: For an Exponential Distribution, 對指數(shù)分布 ,有 The Exponential Distribution is thus a Weibull Distribution with ?=1. 指數(shù)分布為 β=1時(shí)的威布爾分布 . ????xe1Weibull (x。 ?=1, ?) ???xExponential (x。 ?) Process Capability for Cycle Time 周期時(shí)間的過程能力 68 MPT Confidential Example 4 A customer service manager wants to determine the process capability for his department. A primary performance index is the time taken to close a customer plaint. The goal for this index is to close a plaint within one calendar week. Performance over the last 400 plaints was reviewed. 一位客戶服務(wù)經(jīng)理想確定他的部門過程能力 ,主要評價(jià)指標(biāo)為處理客戶投訴的時(shí)間周期 ,目標(biāo)為在一周內(nèi)處理完一單客戶投訴 ,過去的 400%次投訴作為測量數(shù)據(jù) . 數(shù)據(jù)在此期 0103 Days 69 MPT Confidential Example 4 Stat ? Quality Tools ? Capability Analysis (Weibull) 70 MPT Confidential Example 4 2 52 01 51 050U S LP r o c e s s C a p a b i l i t y f o r C o m p l a i n t C l o s u r eC a l c u l a t i o n s B a s e d o n E x p o n e n t i a l D i s t r i b u t i o n M o d e lP P M T o t a lP P M U S LP P M L S LP P M T o t a lP P M U S LP P M L S LP p kP P LP P UP pS c a l eS h a p eS a m p l e NM e a nL S LT a r g e tU S L1 2 2 9 7 0 . 8 01 2 2 9 7 0 . 8 0 * 7 5 0 0 0 . 0 0 7 5 0 0 0 . 0 0 *0 . 3 9 *0 . 3 9 *3 . 3 41 . 0 04 0 03 . 3 4 * *7 . 0 0E x p e c t e d L T P e r f o r m a n c eO b s e r v e d L T P e r f o r m a