Statistics for tayo
Describe Poisson Probability Distribution?
A Poisson distributed random variable is often useful in estimating the number of occurrences over a specified interval of time or space. It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2, . . . ) Examples of Poisson distributed random variables: number of knotholes in 14 linear feet of pine board number of vehicles arriving at a toll booth in one hour Bell Labs used the Poisson distribution to model the arrival of phone calls. Two Properties of a Poisson Experiment The probability of an occurrence is the same for any two intervals of equal length. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval. Poisson Probability Function 𝑓(𝑥)=(𝜇^𝑥 𝑒^(−𝜇))/𝑥! where: x = the number of occurrences in an interval f(x) = the probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828 x! = x(x - 1)(x - 2) . . . (2)(1) Poisson Probability Distribution Poisson Probability Function- Since there is no stated upper limit for the number of occurrences, the probability function f(x) is applicable for values x = 0, 1, 2, ... without limit. In practical applications, x will eventually become large enough so that f(x) is approximately zero and the probability of any larger values of x becomes negligible. Poisson Probability Distribution Example: Mercy Hospital Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening? Poisson Probability Distribution Example: Mercy Hospital Using the probability function: = 6/hour = 3/half-hour, x = 4 𝑓(4)=(3^4 (2.71828)^(−3))/4! = .1680 Poisson Probability Distribution A property of the Poisson distribution is that the mean and variance are equal. m = s 2 Example: Mercy Hospital Variance for Number of Arrivals during 30-Minute periods m = s 2 = 3
What is a random experiment ?
A Random experiment is a process that generates well-defined experimental outcomes.
Explain Random Experiment and Its Sample Space?
A Random experiment is a process that generates well-defined experimental outcomes. The sample space for an experiment is the set of all experimental outcomes. An experimental outcome is also called a sample point.
Explain Decision Trees?
A decision tree provides a graphical representation showing the sequential nature of the decision-making process. Each decision tree has two types of nodes: round nodes correspond to the states of nature square nodes correspond to the decision alternatives The branches leaving each round node represent the different states of nature while the branches leaving each square node represent the different decision alternatives. At the end of each limb of a tree are the payoffs attained from the series of branches making up that limb. Once we have defined the decision alternatives and states of nature for the chance events, we focus on determining probabilities for the states of nature. The classical method, relative frequency method, or subjective method of assigning probabilities may be used. Because one and only one of the N states of nature can occur, the probabilities must satisfy two conditions: P(sj) > 0 for all states of nature ∑_(𝑗=1)^𝑁▒〖𝑃(𝑠_𝑗 )=𝑃(𝑠_1 )+𝑃(𝑠_2 )+...+𝑃(𝑠_𝑁 )=1〗 Then we use the expected value approach to identify the best or recommended decision alternative. The expected value of each decision alternative is calculated (explained on the next slide). The decision alternative yielding the best expected value is chosen.
What is Random Variables?
A random variable is a numerical description of the outcome of an experiment. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals.
Explain Uniform Probability Distribution?
A random variable is uniformly distributed whenever the probability is proportional to the interval's length. The uniform probability density function is: f (x) = 1/(b - a) for a < x < b = 0 elsewhere where: a = smallest value the variable can assume b = largest value the variable can assume Expected Value of x E(x) = (a + b)/2 Variance of x Var(x) = (b - a)2/12 Example: Slater's Buffet Slater's customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Uniform Probability Density Function f(x) = 1/10 for 5 < x < 15 = 0 elsewhere where: x = salad plate filling weight Expected Value of x E(x) = (a + b)/2 = (5 + 15)/2 = 10 Variance of x Var(x) = (b - a)2/12 = (15 - 5)2/12 = 8.33
Explain Events and Their Probabilities?
An event is a collection of sample points. The probability of any event is equal to the sum of the probabilities of the sample points in the event. If we can identify all the sample points of an experiment and assign a probability to each, we can compute the probability of an event. Example: Bradley Investments Event M = Markley Oil Profitable M = {(10, 8), (10, -2), (5, 8), (5, -2)} P(M) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) = .20 + .08 + .16 + .26 = .70 Example: Bradley Investments Event C = Collins Mining Profitable C = {(10, 8), (5, 8), (0, 8), (-20, 8)} P(C) = P(10, 8) + P(5, 8) + P(0, 8) + P(-20, 8) = .20 + .16 + .10 + .02 = .48
What is Assigning Probabilities?
Basic Requirements for Assigning Probabilities 1. The probability assigned to each experimental outcome must be between 0 and 1, inclusively. 0 < P(Ei) < 1 for all i where: Ei is the i th experimental outcome and P(Ei) is its probability Basic Requirements for Assigning Probabilities 2. The sum of the probabilities for all experimental outcomes must equal 1. P(E1) + P(E2) + . . . + P(En) = 1 where: n is the number of experimental outcomes Classical Method Assigning probabilities based on the assumption of equally likely outcomes Relative Frequency Method Assigning probabilities based on experimentation or historical data Subjective Method Assigning probabilities based on judgment
Describe Binomial Probabilities and Cumulative Probabilities ?
Binomial Probabilities and Cumulative Probabilities Statisticians have developed tables that give probabilities and cumulative probabilities for a binomial experiment random variable . These tables can be found in some statistics textbooks. With modern calculators and the capability of statistical software packages, such tables are almost unnecessary.
How to solving the Solving for the Reorder Point?
By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about .20 to .05. This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer's desire to make a purchase.
What is Standard Normal Probability Distribution?
Characteristics A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. The letter z is used to designate the standard normal random variable. Converting to the Standard Normal Distribution z = (𝑥−𝜇)/𝜎 We can think of z as a measure of the number of standard deviations x is from . Example: Pep Zone Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for a replenishment order. It has been determined that demand during replenishment lead-tThe manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 gallons? me is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. P(x > 20) = ? Standard Normal Probability Distribution Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. z = (x - )/ = (20 - 15)/6 = .83 Step 2: Find the area under the standard normal curve to the left of z = .83. z = (x - )/ = (20 - 15)/6 = .83 Standard Normal Probability Distribution Solving for the Stockout Probability Step 3: Compute the area under the standard normal curve to the right of z = .83. P(z > .83) = 1 - P(z < .83) = 1- .7967 = .2033 If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than .05, what should the reorder point be? (Hint: Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.) Standard Normal Probability Distribution Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of x x = + z.05 = 15 + 1.645(6) = 24.87 or 25 A reorder point of 25 gallons will place the probability of a stockout during lead time at (slightly less than) .05
What is compliments of Events?
Complement of an Event The complement of event A is defined to be the event consisting of all sample points that are not in A. The complement of A is denoted by Ac. Event A Ac Venn Diagram
Describe Computing Branch Probabilities with Using Bayes' Theorem?
Computing Branch Probabilities Using Bayes' Theorem Bayes' Theorem can be used to compute branch probabilities for decision trees. For the computations we need to know: the initial (prior) probabilities for the states of nature, the conditional probabilities for the outcomes or indicators of the sample information given each state of nature. A tabular approach is a convenient method for carrying out the computations. For each state of nature, multiply the prior probability by its conditional probability for the indicator. This gives the joint probabilities for the states and indicator. Step 2 Sum these joint probabilities over all states. This gives the marginal probability for the indicator. Step 3 For each state, divide its joint probability by the marginal probability for the indicator. This gives the posterior probability distribution.
What is Mutual Exclusiveness and Independence?
Do not confuse the notion of mutually exclusive events with that of independent events. Two events with nonzero probabilities cannot be both mutually exclusive and independent. If one mutually exclusive event is known to occur, the other cannot occur.; thus, the probability of the other event occurring is reduced to zero (and they are therefore dependent). Two events that are not mutually exclusive, might or might not be independent.
Describe Discrete Random Variable with a Finite Number of Values?
Example: JSL Appliances Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4) We can count the TVs sold, and there is a finite upper limit on the number that might be sold (which is the number of TVs in stock). Example: JSL Appliances Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive.
What is Posterior Probabilities?
Example: L. S. Clothiers Given the planning board's recommendation not to approve the zoning change, we revise the prior probabilities as follows: 𝑃(𝐴_1│𝐵)= (𝑃(𝐴_1 )𝑃(𝐵|𝐴_1))/(𝑃(𝐴_1 )𝑃(𝐵│𝐴_1 )+𝑃(𝐴_2 )𝑃(𝐵│𝐴_2 ) ) = ((.7)(.2))/((.7).2)+(.3).9)) = .34 Example: L. S. Clothiers The planning board's recommendation is good news for L. S. Clothiers. The posterior probability of the town council approving the zoning change is .34 compared to a prior probability of .70.
What is Prior Probabilities?
Example: L. S. Clothiers Let: A1 = town council approves the zoning change A2 = town council disapproves the change Using subjective judgment: P(A1) = .7, P(A2) = .3
Explain Conditional Probabilities?
Example: L. S. Clothiers Past history with the planning board and the town council indicates the following: P(B|A1) = .2 and P(B|A2) = .9 Hence P(BC|A1) = .8 and P(BC|A2) = .1
What is Classical Method?
Example: Rolling a Die. If an experiment has n possible outcomes, the classical method would assign a probability of 1/n to each outcome. Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a 1/6 chance of occurring
Explain the Using Excel to Compute Normal Probabilities?
Excel has two functions for computing cumulative probabilities and x values for any normal distribution: NORM.DIST is used to compute the cumulative probability given an x value. NORM.INV is used to compute the x value given a cumulative probability.
Describe Expected Value?
Expected Value The expected value, or mean, of a random variable is a measure of its central location. E(x) = = ∑xf(x) The expected value is a weighted average of the values the random variable may assume. The weights are the probabilities. The expected value does not have to be a value the random variable can assume.
What is Binomial Probability Distribution?
Four Properties of a Binomial Experiment 1. The experiment consists of a sequence of n identical trials. 2. Two outcomes, success and failure, are possible on each trial. 3. The probability of a success, denoted by p, does not change from trial to trial. (This is referred to as the stationarity assumption.) 4. The trials are independent.
Explain the Expected Value with information?
Frequently, information is available that can improve the probability estimates for the states of nature. The expected value of perfect information (EVPI) is the increase in the expected profit that would result if one knew with certainty which state of nature would occur. The EVPI provides an upper bound on the expected value of any sample or survey information. Expected Value of Perfect Information The expected value of perfect information is defined as EVPI = |EVwPI - EVwoPI| where EVPI = expected value of perfect information EVwPI = expected value with perfect information about the states of nature EVwoPI = expected value without perfect information about the states of nature Expected Value of Perfect Information EVPI Calculation Step 1: Determine the optimal return corresponding to each state of nature. Step 2: Compute the expected value of these optimal returns. Step 3: Subtract the EV of the optimal decision from the amount determined in step (2). Expected Value of Perfect Information Calculate the expected value for the optimum payoff for each state of nature and subtract the EV of the optimal decision. EVPI = .4(10,000) + .2(18,000) + .4(21,000) - 14,000 = $2,000
Explain the Counting Rule for Multiple Step Experiments?
If an experiment consists of a sequence of k steps in which there are n1 possible results for the first step, n2 possible results for the second step, and so on, then the total number of experimental outcomes is given by (n1)(n2) . . . (nk). A helpful graphical representation of a multiple-step experiment is a tree diagram.
What is Independent Events?
If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent. Two events A and B are independent if: P(A|B) = P(A) or P(B|A) = P(B)
What is statistics Experiment?
In statistics, the notion of an experiment differs somewhat from that of an experiment in the physical sciences. In statistical experiments, probability determines outcomes. Even though the experiment is repeated in exactly the same way, an entirely different outcome may occur. For this reason, statistical experiments are sometimes called random experiments.
Introduction of Probability?
Introduction to Probability Random Experiments, Counting Rules, and Assigning Probabilities Events and Their Probability Some Basic Relationships of Probability Conditional Probability Bayes' Theorem
Describe Decision Analysis With Sample Information?
Knowledge of sample (survey) information can be used to revise the probability estimates for the states of nature. Prior to obtaining this information, the probability estimates for the states of nature are called prior probabilities. With knowledge of conditional probabilities for the outcomes or indicators of the sample or survey information, these prior probabilities can be revised by employing Bayes' Theorem. The outcomes of this analysis are called posterior probabilities or branch probabilities for decision trees. Decision Analysis With Sample Information Decision Strategy A decision strategy is a sequence of decisions and chance outcomes. The decisions chosen depend on the yet to be determined outcomes of chance events. The approach used to determine the optimal decision strategy is based on a backward pass through the decision tree. Backward Pass Through the Decision Tree At Chance Nodes: Compute the expected value by multiplying the payoff at the end of each branch by the corresponding branch probability. At Decision Nodes: Select the decision branch that leads to the best expected value. This expected value becomes the expected value at the decision node. Decision Analysis With Sample Information Example: Burger Prince Burger Prince must decide whether to purchase a marketing survey from Stanton Marketing for $1,000. The results of the survey are "favorable" or "unfavorable". The conditional probabilities are: P(favorable | 80 customers per hour) = .2 P(favorable | 100 customers per hour) = .5 P(favorable | 120 customers per hour) = .9
What is uncertainties?
Managers often base their decisions on an analysis of uncertainties such as the following: What are the chances that sales will decrease if we increase prices? What is the likelihood a new assembly method will increase productivity? What are the odds that a new investment will be profitable?
What is Mutually Exclusive Events?
Mutually Exclusive Events Two events are said to be mutually exclusive if the events have no sample points in common. Two events are mutually exclusive if, when one event occurs, the other cannot occur. If events A and B are mutually exclusive, P(A B) = 0. The addition law for mutually exclusive events is: P(A B) = P(A) + P(B)
Explain Counting rules for Combinations?
Number of Combinations of N Objects Taken n at a Time A second useful counting rule enables us to count the number of experimental outcomes when n objects are to be selected from a set of N objects.
Explain Counting Rule for Permutations?
Number of Permutations of N Objects Taken n at a Time. A third useful counting rule enables us to count the number of experimental outcomes when n objects are to be selected from a set of N objects, where the order of selection is important. 𝑃■8(𝑁@𝑛)=𝑛!(■8(𝑁@𝑛)) = 𝑁!/(𝑁−𝑛)! where: N! = N(N - 1)(N - 2) . . . (2)(1) n! = n(n - 1)(n - 2) . . . (2)(1) 0! = 1
What is Bayes' Theorem?
Often we begin probability analysis with initial or prior probabilities. Then, from a sample, special report, or a product test we obtain some additional information. Given this information, we calculate revised or posterior probabilities. Bayes' theorem provides the means for revising the prior probabilities. Example: L. S. Clothiers A proposed shopping center will provide strong competition for downtown businesses like L. S. Clothiers. If the shopping center is built, the owner of L. S. Clothiers feels it would be best to relocate to the shopping center. The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council. To find the posterior probability that event Ai will occur given that event B has occurred, we apply Bayes' theorem. 𝑃(𝐴_𝑖│𝐵)= (𝑃(𝐴_𝑖 )𝑃(𝐵|𝐴_𝑖))/(𝑃(𝐴_1 )𝑃(𝐵│𝐴_1 )+𝑃(𝐴_2 )𝑃(𝐵│𝐴_2 )+...+𝑃(𝐴_𝑛 )𝑃(𝐵|𝐴_𝑛)) Bayes' theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space. Bayes' Theorem: Tabular Approach Example: L. S. Clothiers Step 1 Prepare the following three columns: Column 1 - The mutually exclusive events for which posterior probabilities are desired. Column 2 - The prior probabilities for the events. Column 3 - The conditional probabilities of the new information given each event. Bayes' Theorem: Tabular Approach Example: L. S. Clothiers Step 2 Prepare the fourth column Column 4 Compute the joint probabilities for each event and the new information B by using the multiplication law. Multiply the prior probabilities in column 2 by the corresponding conditional probabilities in column 3. That is, P(Ai B) = P(Ai) P(B|Ai). Bayes' Theorem: Tabular Approach Example: L. S. Clothiers Step 3 Sum the joint probabilities in Column 4. The sum is the probability of the new information, P(B). The sum .14 + .27 shows an overall probability of .41 of a negative recommendation by the planning board. Example: L. S. Clothiers Step 3 Sum the joint probabilities in Column 4. The sum is the probability of the new information, P(B). The sum .14 + .27 shows an overall probability of .41 of a negative recommendation by the planning board. Bayes' Theorem: Tabular Approach Example: L. S. Clothiers Step 4 Prepare the fifth column: Column 5 Compute the posterior probabilities using the basic relationship of conditional probability. 𝑃(𝐴𝑖│𝐵)= (𝑃(𝐴𝑖∩𝐵))/(𝑃(𝐵)) The joint probabilities P(Ai I B) are in column 4 and the probability P(B) is the sum of column 4.
What is Binomial Probability Distribution?
Our interest is in the number of successes occurring in the n trials. Let x denote the number of successes occurring in the n trials. Binomial Probability Function 𝑓(𝑥)=𝑛!/𝑥!(𝑛−𝑥)! 𝑝^𝑥 (1−𝑝)^((𝑛−𝑥)) where: x = the number of successes p = the probability of a success on one trial n = the number of trials f(x) = the probability of x successes in n trials n! = n(n - 1)(n - 2) ..... (2)(1) Binomial Probability Function Example: Evans Electronics Evans Electronics is concerned about a low retention rate for its employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Example: Evans Electronics The probability of the first employee leaving and the second and third employees staying, denoted (S, F, F), is given by p(1 - p)(1 - p) With a .10 probability of an employee leaving on any one trial, the probability of an employee leaving on the first trial and not on the second and third trials is given by (.10)(.90)(.90) = (.10)(.90)2 = .081 Example: Evans Electronics Two other experimental outcomes result in one success and two failures. The probabilities for all three experimental outcomes involving one success follow. Experimental Outcome (S, F, F) (F, S, F) (F, F, S) Probability of Experimental Outcome p(1 - p)(1 - p) = (.1)(.9)(.9) = .081 (1 - p)p(1 - p) = (.9)(.1)(.9) = .081 (1 - p)(1 - p)p = (.9)(.9)(.1) = .081 Total = .243 Example: Evans Electronics Using the probability function: Let: p = .10, n = 3, x = 1 𝑓(𝑥)=𝑛!/𝑥!(𝑛−𝑥)! 𝑝^𝑥 (1−𝑝)^((𝑛−𝑥)) 𝑓(1)=3!/1!(3−1)! (0.1)^1 (0.9)^2 = .243 Expected Value E(x) = = np Variance Var(x) = s 2 = np(1 - p) Standard Deviation 𝜎=√(𝑛𝑝(1−𝑝)) Binomial Probability Distribution Example: Evans Electronics Expected Value E(x) = np = 3(.1) = .3 employees out of 3 Variance Var(x) = np(1 - p) = 3(.1)(.9) = .27 Standard Deviation 𝜎=√(3(.1).9))= .52 employees
What is Probability?
Probability is a numerical measure of the likelihood that an event will occur. Probability values are always assigned on a scale from 0 to 1. A probability near zero indicates an event is quite unlikely to occur. A probability near one indicates an event is almost certain to occur.
Describe Decision Analysis?
Problem Formulation Decision Making with Probabilities Decision Analysis with Sample Information Computing Branch Probabilities using Bayes' Theorem
Describe the Discrete Probability Distributions
Random Variables Developing Discrete Probability Distributions Expected Value and Variance Binomial Probability Distribution Poisson Probability Distribution Hypergeometric Probability Distribution The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable We can describe a discrete probability distribution with a table, graph, or formula. The probability distribution is defined by a probability function, denoted by f(x), that provides the probability for each value of the random variable. The required conditions for a discrete probability function are: f(x) > 0 and ∑f(x) = 1 There are three methods for assigning probabilities to random variables: classical method, subjective method, and relative frequency method. The use of the relative frequency method to develop discrete probability distributions leads to what is called an empirical discrete distribution. Example: JSL Appliances Using past data on TV sales, a tabular representation of the probability distribution for sales was developed. Number x f(x) 0 .40 = 80/200 1 .25 2 .20 3 .05 4 .10 1.00 Units Sold of Days 0 80 1 50 2 40 3 10 4 20 20 Discrete Probability Distributions In addition to tables and graphs, a formula that gives the probability function, f(x), for every value of x is often used to describe the probability distributions. Several discrete probability distributions specified by formulas are the discrete-uniform, binomial, Poisson, and hypergeometric distributions. Discrete Probability Distributions The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula. The discrete uniform probability function is f(x) = 1/n where: n = the number of values the random variable may assume The values of the random variable are equally likely.
What is Relative Frequency Method?
Relative Frequency Method Example: Lucas Tool Rental Lucas Tool Rental would like to assign probabilities to the number of car polishers it rents each day. Office records show the following frequencies of daily rentals for the last 40 days.
Explain Addition Law?
The addition law provides a way to compute the probability of event A, or B, or both A and B occurring. The law is written as: P(A B) = P(A) + P(B) - P(A B) Example: Bradley Investments Event M = Markley Oil Profitable Event C = Collins Mining Profitable We know: P(M) = .70, P(C) = .48, P(M C) = .36 Thus: P(M C) = P(M) + P(C) - P(M C) = .70 + .48 - .36 = .82 (This result is the same as that obtained earlier using the definition of the probability of an event.)
Describe Area as a Measure of Probability?
The area under the graph of f(x) and probability are identical. This is valid for all continuous random variables. The probability that x takes on a value between some lower value x1 and some higher value x2 can be found by computing the area under the graph of f(x) over the interval from x1 to x2.
What is Payoff Tables?
The consequence resulting from a specific combination of a decision alternative and a state of nature is a payoff. A table showing payoffs for all combinations of decision alternatives and states of nature is a payoff table. Payoffs can be expressed in terms of profit, cost, time, distance or any other appropriate measure.
Explain Expected Value Approach?
The expected value of a decision alternative is the sum of weighted payoffs for the decision alternative. The expected value (EV) of decision alternative di is defined as 𝐸𝑉(𝑑_𝑖 )=∑_(𝑗=1)^𝑁▒〖𝑃(𝑠_𝑗)𝑉_𝑖𝑗 〗 where: N = the number of states of nature P(sj ) = the probability of state of nature sj Vij = the payoff corresponding to decision alternative and state of nature. Expected Value Approach Example: Burger Prince Burger Prince Restaurant is considering opening a new restaurant on Main Street. It has three different restaurant layout models (A, B, and C), each with a different seating capacity. Burger Prince estimates that the average number of customers served per hour will be 80, 100, or 120. The payoff table for the three models is on the next slide. Expected Value Approach Calculate the expected value for each decision. The decision tree on the next slide can assist in this calculation. Here d1, d2, d3a represent the decision alternatives of models A, B, and C. And s1, s2, s3 represent the states of nature of 80, 100, and 120 customers per hour.
Describe Expected Value of Sample Information?
The expected value of sample information (EVSI) is the additional expected profit possible through knowledge of the sample or survey information. EVSI = |EVwSI - EVwoSI| where: EVSI = expected value of sample information EVwSI = expected value with sample information about the states of nature EVwoSI = expected value without sample information about the states of nature Expected Value of Sample Information EVwSI Calculation Step 1: Determine the optimal decision and its expected return for the possible outcomes of the sample using the posterior probabilities for the states of nature. Step 2: Compute the expected value of these optimal returns. If the outcome of the survey is "favorable", choose Model C. If the outcome of the survey is "unfavorable", choose Model A. EVwSI = .54($17,855) + .46($11,433) = $14,900.88 Expected Value of Sample Information Subtract the EVwoSI (the value of the optimal decision obtained without using the sample information) from the EVwSI. EVSI = .54($17,855) + .46($11,433) - $14,000 = $900.88 Conclusion Because the EVSI is less than the cost of the survey, the survey should not be purchased
what is Exponential Probability Distribution?
The exponential probability distribution is useful in describing the time it takes to complete a task. The exponential random variables can be used to describe: Time between vehicle arrivals at a toll booth Time required to complete a questionnaire Distance between major defects in a highway In waiting line applications, the exponential distribution is often used for service time. A property of the exponential distribution is that the mean and standard deviation are equal. The exponential distribution is skewed to the right. Its skewness measure is 2. Example: Al's Full-Service Pump The time between arrivals of cars at Al's full-service gas pump follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less.
Explain Problem Formulation?
The first step in the decision analysis process is problem formulation. We begin with a verbal statement of the problem. Then we identify: the decision alternatives the states of nature (uncertain future events) the payoffs (consequences) associated with each specific combination of: decision alternative state of nature A decision problem is characterized by decision alternatives, states of nature, and resulting payoffs. The decision alternatives are the different possible strategies the decision maker can employ. The states of nature refer to future events, not under the control of the decision maker, which may occur. States of nature should be defined so that they are mutually exclusive and collectively exhaustive.
What is a sample space
The sample space for an experiment is the set of all experimental outcomes
What is Hypergeometric Probability Distribution?
The hypergeometric distribution is closely related to the binomial distribution. However, for the hypergeometric distribution: the trials are not independent, and the probability of success changes from trial to trial. Hypergeometric Probability Distribution Hypergeometric Probability Function 𝑓(𝑥)=(■8(𝑟@𝑥))(■8(𝑁−𝑟@𝑛−𝑥))/((■8(𝑁@𝑛)) ) where: x = number of successes n = number of trials f(x) = probability of x successes in n trials N = number of elements in the population r = number of elements in the population labeled success Hypergeometric Probability Distribution Hypergeometric Probability Function The probability function f(x) on the is usually applicable for values of x = 0, 1, 2, ... n. However, only values of x where: 1) x < r and 2) n - x < N - r are valid. If these two conditions do not hold for a value of x, the corresponding f(x) equals 0. Hypergeometric Probability Distribution Example: Neveready's Batteries Bob Neveready has removed two dead batteries from a flashlight and inadvertently mingled them with the two good batteries he intended as replacements. The four batteries look identical. Bob now randomly selects two of the four batteries. What is the probability he selects the two good batteries? Example: Neveready's Batteries Using the probability function: 𝑓(𝑥)=(■8(𝑟@𝑥))(■8(𝑁−𝑟@𝑛−𝑥))/((■8(𝑁@𝑛)) )=(■8(2@2))(■8(2@0))/((■8(4@2)) )=(2!/2!0!)(2!/0!2!)/((4!/2!2!) )= 1/6= .167 where: x = 2 = number of good batteries selected n = 2 = number of batteries selected N = 4 = number of batteries in total r = 2 = number of good batteries in total Hypergeometric Probability Distribution Mean 𝐸(𝑥)=𝜇=𝑛(𝑟/𝑁)' Variance 𝑉𝑎𝑟(𝑥)=𝜎^2=𝑛(𝑟/𝑁)(1−𝑟/𝑁)((𝑁−𝑛)/(𝑁−1)) Hypergeometric Probability Distribution Example: Neveready's Batteries Mean 𝜇=𝑛(𝑟/𝑁)=2(2/4)= 1 Variance 𝜎^2=2(2/4)(1−2/4)((4−2)/(4−1))=1/3= .333 Hypergeometric Probability Distribution Consider a hypergeometric distribution with n trials and let p = (r/N) denote the probability of a success on the first trial. If the population size is large, the term (N - n)/(N - 1) approaches 1. The expected value and variance can be written E(x) = np and Var(x) = np(1 - p). Note that these are the expressions for the expected value and variance of a binomial distribution. Hypergeometric Probability Distribution When the population size is large, a hypergeometric distribution can be approximated by a binomial distribution with n trials and a probability of success p = (r/N).
What is Intersection of Two Events?
The intersection of events A and B is the set of all sample points that are in both A and B. The intersection of events A and B is denoted by A B. Intersection of A and B Example: Bradley Investments Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable and Collins Mining Profitable M C = {(10, 8), (5, 8)} P(M C) = P(10, 8) + P(5, 8) = .20 + .16 = .36
What is Multiplication Law for Independent Events?
The multiplication law also can be used as a test to see if two events are independent. The law is written as: P(A B) = P(A)P(B) Example: Bradley Investments Event M = Markley Oil Profitable Event C = Collins Mining Profitable Are events M and C independent? Does P(M C) = P(M)P(C) ? We know: P(M C) = .36, P(M) = .70, P(C) = .48 But: P(M)P(C) = (.70)(.48) = .34, not .36 Hence: M and C are not independent.
What is Multiplication Law?
The multiplication law provides a way to compute the probability of the intersection of two events. The law is written as: P(A B) = P(B)P(A|B) or P(A B) = P(A)P(B|A) Example: Bradley Investments Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable and Collins Mining Profitable We know: P(M) = .70, P(C|M) = .5143 Thus: P(M C) = P(M)P(M|C) = (.70)(.5143) = .36 (This result is the same as that obtained earlier using the definition of the probability of an event.)
Explain Normal Probability Distribution?
The normal probability distribution is the most important distribution for describing a continuous random variable. It is widely used in statistical inference. It has been used in a wide variety of applications including: Heights of people Amounts of rainfall Test scores Scientific measurements Abraham de Moivre, a French mathematician, published The Doctrine of Chances in 1733. He derived the normal distribution. Normal Probability Density Function (𝑥)=1/(𝜎√2𝜋) 𝑒^(−1/2((𝑥−𝜇〖)/𝜎)〗^2 ) where: = mean = standard deviation = 3.14159 e = 2.71828 Normal Probability Distribution Characteristics (basis for the empirical rule) 68.26% of values of a normal random variable are within +/- 1 standard deviation of its mean. 95.44% of values of a normal random variable are within +/- 2 standard deviations of its mean. 99.72% of values of a normal random variable are within +/- 3 standard deviations of its mean.
Explain Conditional Probability?
The probability of an event given that another event has occurred is called a conditional probability. The conditional probability of A given B has already occurred is denoted by P(A|B). A conditional probability is computed as follows : 𝑃(𝐴│𝐵)= (𝑃(𝐴∩𝐵))/(𝑃(𝐵)) Example: Bradley Investments Event M = Markley Oil Profitable P(C|M) = Collins Mining Profitable given Markley Oil Profitable We know: P(M C) = .36, P(M) = .70 Event C = Collins Mining Profitable 𝑃(𝐶│𝑀)= (𝑃(𝐶∩𝑀))/(𝑃(𝑀)) = .36/.70= .5143
Explain Union of Two Events?
The union of events A and B is the event containing all sample points that are in A or B or both. The union of events A and B is denoted by A B. Example: Bradley Investments Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable or Collins Mining Profitable (or both) M C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)} P(M C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) + P(0, 8) + P(-20, 8) = .20 + .08 + .16 + .26 + .10 + .02 = .82
What is Variance and Standard Deviation?
The variance summarizes the variability in the values of a random variable. Var(x) = 2 = (x - )2f(x) The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities. The standard deviation, , is defined as the positive square root of the variance. Expected Value Example: JSL Appliances x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40 E(x) = 1.20 = expected number of TVs sold in a day
Explain Some Basic Relationships of Probability?
There are some basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities. Complement of an Event Union of Two Events Intersection of Two Events Mutually Exclusive Events
What is Continuous Probability Distributions?
Uniform Probability Distribution Normal Probability Distribution Exponential Probability Distribution A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval. The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2.
What is Subjective Method?
When economic conditions or a company's circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimate.