Six Sigma Black Belt Study Guides

Transcription

1 Six Sigma Black Belt Study Guides 1 Powered by POeT Solvers Limited.

2 Analytical statistics helps to draw conclusions about a population by analyzing the data collected from a sample of such population. Accuracy of the conclusions about the population depends on the size of the sample studied. The bigger the sample size, the greater is the accuracy. Steps involved in analytical statistics: Constructing a hypothesis on the requirement Constructing a hypothesis on the requirement Collecting data Analyzing data Drawing inferences on the validity of the hypothesis Analytical statistics is used to: Estimate population parameters Determine the differences between two populations Determine the differences among a number of populations Determine the degree of relationship between two or more parameters 2 Powered by POeT Solvers Limited.

3 Probability Chance/possibility of some event occurring Expressed in terms of fraction (varies between 0 to 1) Example: The probability of getting a face value of 4 when you roll a fair die is 1/6. Terminology Experiment : Rolling a fair die Outcome: The result of an experiment. In the above experiment, there can be six outcomes, namely, obtaining die face values of 1, 2, 3, 4, 5, and 6. Sample space: The set of all possible outcomes of an experiment. In the above example, the sample space S = {1, 2, 3, 4, 5, 6}. Event: An event is a subset of the sample space. In the above example, the event A is face value of 4, or A = {4}. Union: The union of two events consisting of all the outcomes contained in either one of them. 3 Powered by POeT Solvers Limited.

4 Intersection: The intersection of two events is defined as the set of all the outcomes common to both the events. The intersection of two events A and B is denoted as A B. Complement: The complement of set A is defined by A c = S A. Hence, the complement of a set is the set of all outcomes of the sample space except those contained in the event itself. Worked example: In a group of employees of ABC Inc., 60% have passed GB certification, 45% have passed BB certification, and 25% have passed both. What is the percentage of employees who have passed neither? Solution: Let A and B denote the set of all employees in the group who have passed GB and BB certification respectively. Hence P (A) = 0.6, P (B) = 0.45, and P (A B) = We need to find out the percentage of employees who have passed neither of the exams, i.e. percentage of employees who failed in both the exams, i.e. P (A c B c ) To find out this, we need to know axioms of probability, De Morgan s Law, and addition rule. 4 Powered by POeT Solvers Limited.

5 Axioms of Probability: Consider a sample space S and an event A from that sample space. Then, P(S) = 1 0 P(A) 1 P(A c ) = 1 P(A) De Morgan s Law: A c B c = (AUB) c A c U B c = (A B) c Addition Rule: For any two events A and B from the sample space S, the probabilities of either one of the events occurring is P (AUB) = P (A) + P (B) P (A B) Remember Key words A or B Both A and B Only A Only B Notation AUB A B A (A B) B (A B) A B A B 5 Powered by POeT Solvers Limited.

6 Worked example (continued) To find out P (A c B c ) P(A c B c ) = P((AUB) c ) [using De Morgan s law] = 1 P (AUB) [using axioms of probability] = 1 {P (A) + P (B) P (A B)} [using the addition rule] = 1 { } = 0.2 Hence, 20% of the employees of the group have passed neither of the two exams. 6 Powered by POeT Solvers Limited.

7 Equally likely outcomes: When a sample space consists of N possible outcomes all of which are equally likely to occur, then the probability for occurrence of each outcome is 1/N. Example: We can take the example of rolling a fair die where the occurrence of the outcomes (face value 1, 2, 3, 4, 5, or 6) are equally likely. Hence, the probability of getting a face value 5 is 1/6. Mutually exclusive: Two events A and B are said to be mutually exclusive if A B = (null set or empty set, which contains no element), i.e. mutually exclusive events have no common outcomes. Example: If B = {2, 3, 5} and C = {1, 4, 7, 8}, then B C = C are mutually exclusive. ; hence B and Independency: Two events A and B are said to be independent if P (A B) = P (A) P (B). It can be generalized for more than two independent events A, B, C, P (A B C.) = P (A) P (B) P(C).. This rule is referred to as Multiplication rule. Example: What is the probability of getting tails all the three times on three consecutive flips of a coin? Solution: Since the probability of getting tails on flipping a fair coin once is 0.5, the probability of getting all tails on three consecutive flips is 0.5X0.5X0.5 = Powered by POeT Solvers Limited.

8 Conditional probability: The conditional probability of an event A, given that another event B has already occurred, is defined by P (AIB) = P (A B) or P (A B) = P (B) P (AIB) P (B) Worked example: A box contains 130 components, of which 7 are defective. One component is randomly taken from the box and placed in a machine. A second component is then taken from the box. What is the probability that the second component is defective? Solution: Let A denote the event the first component is defective and B denote the event the first component is good. Hence, P (A) = 7/130 and P (B) = (130-7)/130. Now let C denote the event the second component is defective. The required probability is P(C). P(C) = P(C (AUB)) = P ((C A) U (C B)) = P (C A) + P (C B) P (C A) = P (A) P (CIA) = 7/130 X 6/129 = P (C B) = P (B) P (CIB) = 123/130 X 7/129 = P(C) = = So, the probability that the second component is defective is Powered by POeT Solvers Limited.

9 Combination: Combination is related to a simple selection of elements from a set without any order. It is the number of ways of choosing k objects from a set of n objects ignoring the order of choosing. It is mathematically denoted as n C k = n! (n k)! k! Example: In how many different ways can you select 2 books out of 5 books (ignore the order of selection)? Let us number the books as 1, 2, 3, 4, and 5. Now, we can select two books as: (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), and (4, 5). Thus, this problem can be easily solved by using the formula for combination, i.e. 9 Powered by POeT Solvers Limited.

10 Permutation: Permutation is related to the orderly arrangement of objects. A permutation is the number of ways of selecting k objects from a set of n objects while the order of selection is also taken into account. This can be mathematically denoted as: n P k = n! (n k)! Consider the last example but with the order of selection also taken into account: The permutations are: (1,2) (2,3) (3,5) (4,1) (5,2) (1,3) (2,4) (4,5) (5,1) (4,3) (1,4) (2,5) (2,1) (3,2) (5,3) (1,5) (3,4) (3,1) (4,2) (5,4) The problem can easily be solved by using the formula for permutation: 10 Powered by POeT Solvers Limited.

11 Probability distribution Distribution of a set of data is typically expressed by its shape, average (central tendency), or standard deviation (spread). It describes the nature of the data location, i.e. whether the data is concentrated near the mean (symmetry) or the concentration is on either side (skewed). The shape of the distribution is determined by the attributes like symmetry, skewness, and kurtosis Powered by POeT Solvers Limited.

12 Important Continuous Distributions Normal Bivariate normal Lognormal Exponential Weibull Chi-square Student s t F Important Discrete Distributions Binomial Poisson Hypergeometric 12 Powered by POeT Solvers Limited.

13 Discrete Distributions Binomial distribution: Bin (n, p) Bernoulli Trial: It is an experiment that has only two possible outcomes - success (s) and failure (f). Now consider a sequence of n Bernoulli trials having the following properties: 1) The trials are independent of one another, i.e. the outcome of one trial does not depend on the outcome of any other trial. 2) The trials are identical, i.e. P (success) = constant for all trials. Let us denote this by p. Such a sequence is called a sequence of n independent and identical Bernoulli (p) trials Powered by POeT Solvers Limited.

14 Conditions for Use of Binomial Distribution Events are independent. Events are mutually exclusive with 2 outcomes (Success/Failure, Good/Defective, Yes/No, etc.). Probability (p) of each outcome should remain constant during trials. Number of trials are finite (n) Powered by POeT Solvers Limited.

15 Application in Six Sigma Projects: Binomial distribution is used to determine the probability of obtaining a certain number of defectives in a sample of known size derived from a population with known defectives percentage. Formula: P (X = x) = n C x p x (1 p) (n-x) x = 0, 1, 2, 3,., n; 0< p < 1 Where n = sample size, x = number of defectives, p = defective rate in the population Bin (n, p) Sample Size Defective Percentage Remember 15 Powered by POeT Solvers Limited.

16 Worked example: A random sample of size 8 is selected from a batch with 12.54% defectives. Find the probability that the sample has exactly two defectives? Solution: Substitute n = 8, x = 2, and p = in the formula P (X = x) = n C x p x (1- p) (n-x) P(X = 2) = 8 C 2 (0.1254) 2 ( ) (8-2) = The probability that the sample contains exactly two defectives is Powered by POeT Solvers Limited.

17 Hypergeometric Distribution Finite population equivalent of the binomial distribution Let the objects be selected at random one after another without replacement from a finite population consisting of k successes and N - k failures [N = population size]. Here, the trails are not independent since the selection is made without replacement. The mean is given by nk/n (note that in the Binomial (n, p) distribution, the mean is given by np). The binomial p = k/n provides a good approximation to the hypergeometric distribution in many cases Powered by POeT Solvers Limited.

18 Formula: P(X = x) = k C x (N-k) C (n-x) x = max (0, n + k n) to min (k,n) N C n N = finite population size n = sample size k = number of successes in the population (in Six Sigma terms, this is the number of defective units in the population) x = number of successes in the sample (in Six Sigma terms, this is the number of defective units in the sample) 18 Powered by POeT Solvers Limited.

19 Worked example From among the 60 persons applying for the Quality Analyst position in an organization, only 32 have GB certification. If 6 are randomly selected from this group, what is the probability that none of them possesses the certification? Solution: Substitute N = 60 (number of persons applying for the position), n = 6 (sample size), k = 32 (number of GB certified persons. Note: Here, GB certification is the success) and x = 0 in the formula P(X = 0) = This is the probability that none among the group selected is GB certified Powered by POeT Solvers Limited.

20 Poisson Distribution Poi (λ) It is the number of events that occur in a specified time interval with the following assumptions: Events occur singly (one after another) at a constant rate, say λ per unit time The number of events that occur in non-overlapping time intervals are independent of one another. The events are described as occurring as a Poisson process with rate λ. Poisson distribution describes the number of events that occur in a specified interval of time Powered by POeT Solvers Limited.

21 Applications of Poisson Distribution in Six Sigma Projects Used to describe the number of defects per unit time Used as a basis for the λ for c and u control charts Other Applications Poisson distribution is also used to analyze: The number of transactions monitored per day in a BPO The number of alpha particles emitted by a radio active element per unit time 21 Powered by POeT Solvers Limited.

22 Poisson distribution as a very good approximation to the Binomial If a sequence of binomial (n, p) distributions be such that the mean np is held constant as λ, the two limiting conditions n and p 0 being satisfied simultaneously (typically n 100 and p 0.05), then the limit leads to the Poisson ( λ) distribution. Formula: P(X = x) = (λ x e -λ ) / x! x = 0, 1, 3,.; λ > 0 λ = mean number of counts in an interval (λ > 0) x = number of defects / counts in the sample / interval e = a constant which is approximately equal to P(X = x) = probability of x occurrences in the sample / interval 22 Powered by POeT Solvers Limited.

23 Worked Example: ABC Inc. is a BPO company working in three shifts and the defects per each shift follow a Poisson distribution with mean values as given below: Shift Mean Find the probability that the third shift produces less than three defects. Solution: The required probability is P(X < 3) P(X < 3) = P(X = 0, 1, 2) = P(X = 0) + P(X = 1) + P(X = 2) Note that here = 4.7 (mean value for the third shift is 4.7 and is the mean of the Poisson distribution) The required probability P(X < 3) = = Powered by POeT Solvers Limited.

24 Remember Defect Use Poisson (λ) λ = average number of defects Defective Use Binomial (n,p) n = sample size, p = defectives percentage 24 Powered by POeT Solvers Limited.

25 Continuous Distributions Probability Density Function (PDF) The probability density function (PDF) is a mathematical function [generally denoted as f(x)] that models the probability density reflected in a histogram. PDF suggests the shape of the probability distribution. It is used in probability distributions for continuous variables. PDF uses integrals summation of all the areas under the curve such as the one shaded in the following diagram: 25 Powered by POeT Solvers Limited.

26 Uniform Distribution: U (a, b) The probability density function (PDF) of the uniform distribution is 1 f (x) = b a for a x b 0 for x < a and X > b The plot of a uniform distribution has a horizontal line as the probability of occurrence is constant for all values Powered by POeT Solvers Limited.

27 Normal Distribution: N (µ, σ) The PDF is given by f (x) = 1 exp (x - µ) 2 2π σ 2σ 2 - <x<, - <µ<, 0<σ< Properties of a normal curve It is bell shaped (i.e. most of the data points are concentrated near the mean) Symmetric about the mean Mean = median = mode Area under the standard normal curve = 1 Unimodal (i.e. has one mode) The standard normal distribution has mean = 0 and standard deviation = Powered by POeT Solvers Limited.

28 Calculate the area under the standard normal curve between standard deviations and standard deviations. ᶲ(z) 0 z P (1.24<X<1.65) =P(X < 1.65) P (X < 1.24) =Area to the left of 1.65 Area to the left of 1.24 = [From the standard normal table] = Powered by POeT Solvers Limited.

29 Hence, the area under the curve between the two values is Since the total area under the curve is 1, the area under the curve between the two vertical lines is 5.802% of the total area under the curve. In most of the applications, area under the curve represents the population under consideration. So, in this context, it can be said that 5.802% of the population lies between 1.24 and Powered by POeT Solvers Limited.

30 Bivariate Normal Distribution This is used in bivariate cases (X, Y) involving two variables each of which is normally distributed. The resulting distribution is called bivariate normal distribution Powered by POeT Solvers Limited.

31 Density Lognormal Distribution Analytical Statistics The PDF is given by f (x) = 1 exp (ln (x) - µ) 2 2 where x > 0. xσ 2π σ 2σ It is a right skewed distribution with most of the data residing in the left tail. For this distribution, natural logarithms of the original data follow a normal distribution. Applications The lognormal distribution can be used to model various situations involving response time, time to failure data, time to repair data, etc Distribution Plot Lognormal, Loc=0, Scale=1, Thresh= X 31 Powered by POeT Solvers Limited.

32 Density Analytical Statistics Exponential (λ) Distribution The PDF is given by f (x) = λe-2 -λx, where x > Distribution Plot Lognormal, Loc=0, Scale=1, Thresh= X The exponential distribution and poisson distribution are closely related. If X follows a poisson distribution, then the reciprocal Y = 1/X is exponentially distributed. Example: Suppose average failures are 0.60 per hour (discrete data - poisson distribution), then the mean time between failures (MTBF) is 1/0.6 = 1.67 hours (continuous data exponential distribution). Applications This is used to model lifetimes of certain elements. To determine the average time between failures or average time between a number of occurrences. Frequently used to analyze reliability 32 Powered by POeT Solvers Limited.

33 Weibull Distribution The PDF is given by f (x) = αβ(x y) β 1 -α(x y)β e where α = scale parameter, β = shape parameter, y = location parameter Depending on the value of β, the weibull function takes many shapes as indicated below (for β = 1, it is exponential and for β = 3.5, it is approximately normal). Applications Widely used in reliability Can be used in similar applications as in lognormal distribution, for example, to measure time to fail, time to repair, material strength, etc Powered by POeT Solvers Limited.

34 Student s t Distribution Commonly used to determine the confidence interval of the population mean Used to test hypothesis when the means of sample populations are compared (This will be discussed in detail in the subsequent chapters). The shape of the t distribution approaches the normal distribution as the sample size increases Typically at n 31, t and normal distribution will be the same. t distribution has one parameter n, which is a positive integer that specifies the number of degrees of freedom (degree of freedom is defined as an unrestricted variable in a distribution) The shape of the distribution takes various forms as determined by its number of degrees of freedom n 34 Powered by POeT Solvers Limited.

35 Density Analytical Statistics Chi-Square (x 2 ) Distribution It is used in the testing of hypothesis to test population variance against its known or assumed value. It is a right skewed distribution, or, in other words, the curve has its long tail toward the higher values of the distribution. It has one parameter k, which is a positive integer that specifies the number of degrees of freedom Distribution Plot Chi-Square, df = X 35 Powered by POeT Solvers Limited.

36 Density Analytical Statistics F Distribution F distribution is used in testing of hypothesis for equality of variances. It is very important in ANOVA, which is a technique used in design of experiments (DOE), for testing significant differences in variance within and between test runs. F distribution has 2 parameters n1 and n2 called degrees of freedom. Shape of the distribution curve is non-symmetrical and varies with the degrees of freedom Distribution Plot F, df1 = 10, df2 = X 36 Powered by POeT Solvers Limited.

37 In this chapter, we have learned about: Introduction to Analytical Statistics Probability Continuous and discrete distributions 37 Powered by POeT Solvers Limited.