Introduction to Probability and Statistics Chapter 12

Topics Types of Probability Fundamentals of Probability Statistical Independence and Dependence Expected Value The Normal Distribution

Sample Space and Event Probability is associated with performing an experiment whose outcomes occur randomly Sample space contains all the outcomes of an experiment An event is a subset of sample space Probability of an event is always greater than or equal to zero Probabilities of all the events must sum to one Events in an experiment are mutually exclusive if only one can occur at a time

Objective Probability Stated prior to the occurrence of the event Based on the logic of the process producing the outcomes Relative frequency is the more widely used definition of objective probability. Subjective Probability Based on personal belief, experience, or knowledge of a situation. Frequently used in making business decisions. Different people often arrive at different subjective probabilities.

Fundamentals of Probability Distributions Frequency Distribution organization of numerical data about the events Probability Distribution A list of corresponding probabilities for each event Mutually Exclusive Events If two or more events cannot occur at the same time Probability that one or more events will occur is found by summing the individual probabilities of the events: P(A or B) = P(A) + P(B)

Fundamentals of Probability A Frequency Distribution Example Grades for past four years.

P(M or F) = P(M) + P(F) - P(MF) Fundamentals of Probability Non-Mutually Exclusive Events & Joint Probability Probability that non-mutually exclusive events M and F or both will occur expressed as: P(M or F) = P(M) + P(F) - P(MF) A joint (intersection) probability, P(MF), is the probability that two or more events that are not mutually exclusive can occur simultaneously.

Fundamentals of Probability Cumulative Probability Distribution Determined by adding the probability of an event to the sum of all previously listed probabilities Probability that a student will get a grade of C or higher: P(A or B or C) = P(A) + P(B) + P(C) = .10 + .20 + .50 = .80

Statistical Independence and Dependence Independent Events Events that do not affect each other are independent. Computed by multiplying the probabilities of each event. P(AB) = P(A)  P(B) For coin tossed three consecutive times: Probability of getting head on first toss, tail on second, tail on third is: P(HTT) = P(H)  P(T)  P(T) = (.5)(.5)(.5) = .125

Statistical Independence and Dependence Independent Events – Bernoulli Process Definition Properties of a Bernoulli Process: Two possible outcomes for each trial. Probability of the outcome remains constant over time. Outcomes of the trials are independent. Number of trials is discrete and integer.

Binomial Distribution Used to determine the probability of a number of successes in n trials. where: p = probability of a success q = 1- p = probability of a failure n = number of trials r = number of successes in n trials Determine probability of getting exactly two tails in three tosses of a coin.

Example Microchips are inspected at the quality control station From every batch, four are selected and tested for defects Given defective rate of 20%, what is the probability that each batch contains exactly two defectives

Binomial Distribution Example – Quality Control What is probability that each batch will contain exactly two defectives? What is probability of getting two or more defectives? Probability of less than two defectives: P(r<2) = P(r=0) + P(r=1) = 1.0 - [P(r=2) + P(r=3) + P(r=4)] = 1.0 - .1808 = .8192

Dependent Events If the occurrence of one event affects the probability of the occurrence of another event, the events are dependent. Coin toss to select bucket, draw for blue ball. If tail occurs, 1/6 chance of drawing blue ball from bucket 2; if head results, no possibility of drawing blue ball from bucket 1. Probability of event “drawing a blue ball” dependent on event “flipping a coin.”

Dependent Events – Conditional Probabilities Unconditional: P(H) = .5; P(T) = .5, must sum to one. Conditional: P(RH) =.33, P(WH) = .67, P(RT) = .83, P(WT) = .17

P(AB) = P(AB)/P(B) or P(AB) = P(A|B).P(B) Math Formulation of Conditional Probabilities Given two dependent events A and B: P(AB) = P(AB)/P(B) or P(AB) = P(A|B).P(B) With data from previous example: P(RH) = P(RH)  P(H) = (.33)(.5) = .165 P(WH) = P(WH)  P(H) = (.67)(.5) = .335 P(RT) = P(RT)  P(T) = (.83)(.5) = .415 P(WT) = P(WT)  P(T) = (.17)(.5) = .085

Summary of Example Problem Probabilities

Bayesian Analysis In Bayesian analysis, additional information is used to alter (improve) the marginal probability of the occurrence of an event. Improved probability is called posterior probability A posterior probability is the altered marginal probability of an event based on additional information. Bayes’ Rule for two events, A and B, and third event, C, conditionally dependent on A and B:

Bayesian Analysis – Example (1 of 2) Machine setup; if correct 10% chance of defective part; if incorrect, 40%. 50% chance setup will be correct or incorrect. What is probability that machine setup is incorrect if sample part is defective? Solution: P(C) = .50, P(IC) = .50, P(DC) = .10, P(DIC) = .40 where C = correct, IC = incorrect, D = defective

Statistical Independence and Dependence Bayesian Analysis – Example (2 of 2) Previously, the manager knew that there was a 50% chance that the machine was set up incorrectly Now, after testing the part, he knows that if it is defective, there is 0.8 probability that the machine was set up incorrectly

Expected Value Random Variables When the values of variables occur in no particular order or sequence, the variables are referred to as random variables. Random variables are represented by a letter x, y, z, etc. Possible to assign a probability to the occurrence of possible values. Possible values of no. of heads are: Possible values of demand/week:

Expected Value Example (1 of 4) Machines break down 0, 1, 2, 3, or 4 times per month. Relative frequency of breakdowns , or a probability distribution:

Expected Value Example (2 of 4) Computed by multiplying each possible value of the variable by its probability and summing these products. The weighted average, or mean, of the probability distribution of the random variable. Expected value of number of breakdowns per month: E(x) = (0)(.10) + (1)(.20) + (2)(.30) + (3)(.25) + (4)(.15) = 0 + .20 + .60 + .75 + .60 = 2.15 breakdowns

Expected Value Example (3 of 4) Variance is a measure of the dispersion of random variable values about the mean. Variance computed as follows: Square the difference between each value and the expected value. Multiply resulting amounts by the probability of each value. Sum the values compiled in step 2. General formula: 2 = [xi - E(xi)] 2 P(xi)

Expected Value Example (4 of 4) Standard deviation computed by taking the square root of the variance. For example data: 2 = 1.425 breakdowns per month standard deviation =  = sqrt(1.425) = 1.19 breakdowns per month

Poisson Distribution Based on the number of outcomes occurring during a given time interval or in a specified regions Examples # of accidents that occur on a given highway during a 1-week period # of customers coming to a bank during a 1-hour interval # of TVs sold at a department store during a given week # of breakdowns of a washing machine per month

Conditions Consider the # of breakdowns of a washing machine per month example Each breakdown is called an occurrence Occurrences are random that they do not follow any pattern (unpredictable) Occurrence is always considered with respect to an interval (one month)

The Probability Mass Distribution X = number of counts in the interval Poisson random variable with  > 0 PMF f(x)= x=0,1,2, Mean and Variance E[X] =  , V (X) = 

Example If a bank gets on average  = 6 bad checks per day, what are the probabilities that it will receive four bad checks on any given day?10 bad checks on any two consecutive days? Solution x = 4 and  = 6, then f(4) = = 0.135  = 12 and x = 10, then f(10) = = 0.105

Example The number of failures of a testing instrument from contamination particle on the product is a Poisson random variable with a mean of 0.02 failure per hour. What is the probability that the instrument does not fail in an 8-hour shift? What is the probability of at least one failure in one 24-hour day?

Solution Let X denote the failure in 8 hours. Then, X has a Poisson distribution with =0.16 P(X=0)=0.8521 Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with =0.48 P(Y1) = 1-P(Y = 0) =0.3812

The Normal Distribution Continuous Random Variables Continuous random variable can take on an infinite number of values within some interval. Continuous random variables have values that are not countable Cannot assign a unique probability to each value

The Normal Distribution Definition The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean. The center of a normal distribution is its mean . The area under the normal curve represents probability, and total area under the curve sums to one.

The Normal Distribution Example (1 of 5) Mean weekly carpet sales of 4,200 yards, with standard deviation of 1,400 yards. What is probability of sales exceeding 6,000 yards?  = 4,200 yd;  = 1,400 yd; probability that number of yards of carpet will be equal to or greater than 6,000 expressed as: P(x6,000).

The Normal Distribution Example (2 of 5) -

The Normal Distribution Standard Normal Curve (1 of 2) The area or probability under a normal curve is measured by determining the number of standard deviations from the mean. Number of standard deviations a value is from the mean designated as Z. Z = (x - )/

The Normal Distribution Standard Normal Curve (2 of 2)

The Normal Distribution Example (3 of 5) Z = (x - )/  = (6,000 - 4,200)/1,400 = 1.29 standard deviations P(x 6,000) = .5000 - .4015 = .0985

The Normal Distribution Example (4 of 5) Determine probability that demand will be 5,000 yards or less. Z = (x - )/ = (5,000 - 4,200)/1,400 = .57 standard deviations P(x 5,000) = .5000 + .2157 = .7157

The Normal Distribution Example (5 of 5) Determine probability that demand will be between 3,000 yards and 5,000 yards. Z = (3,000 - 4,200)/1,400 = -1,200/1,400 = -.86 P(3,000  x  5,000) = .2157 + .3051= .5208

Different Table P(3,000  x  5,000)= P((3,000 - 4,200)/1,400)  z  ((5,000 - 4,200)/1,400) P(-0.86 z  0.57)= P( z  0.57)- P( z  -0.86)= P( z  0.57)- P( z ≥0.86)= P( z  0.57)- [1-P( z  0.86)]= (0.7157)-[1-0.8051]=0.5208

The Normal Distribution Sample Mean and Variance The population mean and variance are for the entire set of data being analyzed. The sample mean and variance are derived from a subset of the population data and are used to make inferences about the population.

The Normal Distribution Computing the Sample Mean and Variance

The Normal Distribution Example Problem Re-Done Sample mean = 42,000/10 = 4,200 yd Sample variance = [(190,060,000) - (1,764,000,000/10)]/9 = 1,517,777 Sample std. dev. = sqrt(1,517,777) = 1,232 yd

The Normal Distribution Chi-Square Test for Normality (1 of 2) It can never be simply assumed that data are normally distributed. A statistical test must be performed to determine the exact distribution. The Chi-square test is used to determine if a set of data fit a particular distribution. It compares an observed frequency distribution with a theoretical frequency distribution that would be expected to occur if the data followed a particular distribution (testing the goodness-of-fit).

The Normal Distribution Chi-Square Test for Normality (2 of 2) In the test, the actual number of frequencies in each range of frequency distribution is compared to the theoretical frequencies that should occur in each range if the data follow a particular distribution. A Chi-square statistic is then calculated and compared to a number, called a critical value, from a chi-square table. If the test statistic is greater than the critical value, the distribution does not follow the distribution being tested; if it is less, the distribution does exist. Chi-square test is a form of hypothesis testing.

Statistical Analysis with Excel (1 of 3)

Statistical Analysis with Excel (2 of 3)

Statistical Analysis with Excel (3 of 3)