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Part 1 : 07/27/10 21:34:32 Question 1 - CMA 1289 5-24 - Probability The College Honor Society sells hot pretzels at the home football games. The frequency distribution of the demand for pretzels per game is presented as follows: Unit Sales Volume Probability 2,000 pretzels .10 3,000 pretzels .15 4,000 pretzels .20 5,000 pretzels .35 6,000 pretzels .20 The pretzels are sold for $1.00 each, and the cost per pretzel is $.30. Any unsold pretzels are discarded because they will be stale before the next home game. The conditional profit (loss) per game of having 4,000 pretzels available but being able to sell 5,000 pretzels if they had been available is A. $3,500. B. $2,800. C. $(1,225). D. $4,025. A. It is impossible to sell more products that are supplied, so the conditional profit of having 4,000 pretzels and a demand of 5,000 is the same as having 4,000 pretzels available and a demand of 4,000. B. It is impossible to sell more products than are supplied, so the conditional profit of having 4,000 pretzels and a demand of 5,000 is the same as having 4,000 pretzels available and having a demand of 4,000. If 4,000 are sold at a price of $1.00 each and the cost is $.30 each, the profit is ($1 - $.30) × 4,000 = $2,800. C. This answer is incorrect. See the correct answer for a complete explanation. D. It is impossible to sell more products that are supplied, so the conditional profit of having 4,000 pretzels and a demand of 5,000 is the same as having 4,000 pretzels available and a demand of 4,000. Question 2 - CMA 1287 5-22 - Probability A simple two-state cost investigation model calls for a decision to investigate if the following inequality holds: L(1 í P) > I + C(1 í P) If: L = excess operating cost if out of control P = probability that the process is in control I = cost of investigation C = cost of correction if out of control The critical probability that the process is in control is A. 1 í [I/(L í C)]. B. I/(L í C) í 1. C. Greater than I/C. D. (I + C)/L. A. This inequality would be used to decide whether it was cost-effective to perform an investigation to see if a (c) HOCK international, page 1 Part 1 : 07/27/10 21:34:32 process is out of control. If the cost of leaving the process out of control (L) is less than the cost of investigating to see if it is out of control (I) and fixing it if it turns out to be out of control (C), then it is more beneficial to leave it alone and not even investigate. However, this question is more a test of your ability to simplify an algebraic formula than it is a test of your ability to understand what the inequality is saying. Some algebraic formulas cannot be “solved” in the sense of finding the unique number that a particular variable stands for with the information you have. This is one of those problems. All you can do is simplify this inequality, so that is what you do to answer this question.The problem states that P = the probability that the process is in control. Then, it asks what is the probability that the process is in control. Therefore, to answer the question, you need to solve the inequality for the variable P by simplifying it so that P is on one side of the equation all by itself. Here is the way it works: First, change the inequality to an equation. Then, begin doing the same operations on both sides of the equation until you have the P isolated on one side of the equation. In algebra, when you do the same thing to both sides of an equation, you have not changed the equation. So you could do that forever, and your equation would still be saying the same thing. There are certain other things in a term that you can re-state without changing the term, also. That is what you do here.L(1 í P) = I + C(1 í P) (1) Subtract C(1-P) from both sides: L(1 í P) í C(1 í P) = I (2) Now you have two like terms on the left side. You can transform the left side of the equation from “L(1 í P) – C(1 í P)” to “(L í C)(1 í P)” without changing it: (L - C)(1 - P) = I (3) Divide both sides by (L í C): 1 í P = [I / (L í C)] (4) Subtract 1 from both sides: íP = [I/(L í C)] í 1 (5) Multiply both sides by í1 to change the negative P to a positive P: P = -[I/(L í C) + 1 (6) Rearrange the terms on the right side: P = 1 í [I/(L í C)] And that is the answer to the question. To prove that this has not changed the equation, try using the following numbers for the variables: L = $1,000 P = .90 C = $250 L(1 – P) = I + [C(1 í P)] 1,000(.10) = 75 + (250 × .10) 100 = 100 And: P = 1 í [I/(L í C)] .90 = 1 í [75/(1,000 í 250)] .90 = 1 í (75/750) .90 = 1 - .10 .90 = .90 Both equations are true. B. This answer is incorrect. See the correct answer for a complete explanation. (c) HOCK international, page 2 Part 1 : 07/27/10 21:34:32 C. This answer is incorrect. See the correct answer for a complete explanation. D. This answer is incorrect. See the correct answer for a complete explanation. Question 3 - CMA 1286 5-2 - Probability Decisions are frequently classified as those made under certainty and those made under uncertainty. Certainty exists when A. There is absolutely no doubt that an event will occur. B. The standard deviation of an event is greater than zero. C. There is more than one outcome for each possible action. D. The probability of the event is less than 1.0. A. When there is no doubt that an event will occur, the probability of the event is 1.0 or 100%, which happens only under a condition of absolute certainty. Under conditions of certainty, conditions are deterministic and known and the probability is either 1.0 or 100% (event occurs) or 0% (event does not occur). B. This answer is incorrect. See the correct answer for a complete explanation. C. More that one outcome is not a characteristic of the conditions of certainty. D. The probability of an event under conditions of certainty is 1.0, not less than 1.0. Question 4 - CMA 1289 5-23 - Probability The College Honor Society sells hot pretzels at the home football games. The frequency distribution of the demand for pretzels per game is presented as follows: Unit Sales Volume Probability 2,000 pretzels .10 3,000 pretzels .15 4,000 pretzels .20 5,000 pretzels .35 6,000 pretzels .20 The pretzels are sold for $1.00 each, and the cost per pretzel is $.30. Any unsold pretzels are discarded because they will be stale before the next home game. The conditional profit per game of having 4,000 pretzels available and selling all 4,000 pretzels is A. $1,200. B. Some amount other than those given. C. $2,800. D. $2,100. A. This is the profit if 4,000 pretzels are available but only 3,000 are sold. See the correct answer for a complete explanation. B. The correct answer is given. See the correct answer for a complete explanation. C. If we assume that all 4,000 pretzels are sold at $1.00 each and the cost per pretzel is $.30, the profit will be (c) HOCK international, page 3 Part 1 : 07/27/10 21:34:32 ($1 í $.30) × 4,000 = $2,800. D. This answer assumes that only 3,000 pretzels are sold at $1.00 each, and the cost of each sold pretzel is $.30 each. However, 4,000 pretzels are sold. Question 5 - CMA 1292 4-21 - Probability A beverage stand can sell either soft drinks or coffee on any given day. If the stand sells soft drinks and the weather is hot, it will make $2,500; if the weather is cold, the profit will be $1,000. If the stand sells coffee and the weather is hot, it will make $1,900; if the weather is cold, the profit will be $2,000. The probability of cold weather on a given day at this time is 60%. The expected payoff for selling coffee is A. $1,360. B. $3,900. C. $2,200. D. $1,960. A. This is the expected payoff for selling soft drinks when the weather is cold multiplied by the probability of cold weather of .60, plus the expected payoff for selling coffee when the weather is hot multiplied by the probability of hot weather of .40. That is not the expected payoff for selling coffee. B. This is not possible, as the maximum the beverage stand can obtain is $2,500 if it sells soft drinks when the weather is hot. C. This is the expected payoff for selling coffee when the weather is cold multiplied by the probability of cold weather of .60, plus the expected payoff for selling soft drinks when the weather is hot multiplied by the probability of hot weather of .40. That is not the expected payoff for selling coffee. D. To solve this problem we have to identify the expected payoff of selling coffee when the weather is either hot or cold. That will be the weighted average of the expected payoffs for serving coffee, weighted according to the probability of cold weather and hot weather. So we will multiply each possible payoff for selling coffee by its corresponding probability. If the stand sells coffee and the weather is hot, it will make $1,900, and the probability of hot weather is 40% (100% í 60%). If the stand sells coffee and the weather is cold, it will make $2,000, and the probability of cold weather is 60%. Thus, the weighted average payoff of selling coffee is ($1,900 × .40) + ($2,000 × .60), which is $1,960. And that is the expected payoff for selling coffee. Question 6 - CIA 1194 III-60 - Probability A company uses two major material inputs in its production. To prepare its manufacturing operations budget, the company has to project the cost changes of these material inputs. The cost changes are independent of one another. The purchasing department provides the following probabilities associated with projected cost changes: Cost Increase Material 1 Material 2 3% .3 .5 5% .5 .4 10% .2 .1 (c) HOCK international, page 4 Part 1 : 07/27/10 21:34:32 The probability of a 3% increase in the cost of both Material 1 and Material 2 is A. 80 percent. B. 40 percent. C. 15 percent. D. 25 percent. A. This answer is incorrect. See the correct answer for a complete explanation. B. This answer is incorrect. See the correct answer for a complete explanation. C. The probability of two events happening simultaneously is calculated by multiplying the chance of each item happening together. In this question, that is 0.3 × 0.5 = 0.15 D. This answer is incorrect. See the correct answer for a complete explanation. Question 7 - CMA 1288 5-14 - Probability Ron Bagley is contemplating whether to investigate a labor efficiency variance in the Assembly Department. It will cost $6,000 to undertake the investigation and another $18,000 to correct operations if the department is found to be operating improperly. If the department is operating improperly and Bagley failed to make the investigation, operating costs from the various inefficiencies are expected to amount to $33,000. Bagley would be indifferent between investigating and not investigating the variance if the probability of improper operation is A. 0.29. B. 0.60. C. 0.40. D. 0.71. A. If the probability of improper operations "X" is .29, the investigation is not expected to be cost beneficial. Therefore, Bagley would not be indifferent. B. If the probability of improper operation "X" is 0.60, investigation is expected to be cost beneficial. Therefore, Bagley would not be indifferent. C. In order to determine this percentage, we must determine at what percentage of likelihood of error the expected cost of the investigation is equal to the expected cost of not investigating. In case of investigation, total costs will be equal to the sum of the cost of investigation itself ($6,000) and the expected cost of correction ($18,000X), where "X" is the probability of improper operations. The expected cost of not doing an investigation is $33,000X. Equating both sides would allow us to find X. Our equation is: 33,000X = 6,000 + 18,000X. To solve for X, (1) Subtract 18,000X from both sides of the equation: 15,000X = 6,000 (2) Divide both sides of the equation by 15,000: X = .40 If the probability of an improper operation is 40%, Bagley is indifferent as to whether or not he investigate. (c) HOCK international, page 5 Part 1 : 07/27/10 21:34:32 D. If the probability of improper operation "X" is 0.71, investigation is expected to be cost beneficial. Therefore, Bagley would not be indifferent. Question 8 - CMA 1289 5-22 - Probability The College Honor Society sells hot pretzels at the home football games. The frequency distribution of the demand for pretzels per game is presented as follows: Unit Sales Volume Probability 2,000 pretzels .10 3,000 pretzels .15 4,000 pretzels .20 5,000 pretzels .35 6,000 pretzels .20 The pretzels are sold for $1.00 each, and the cost per pretzel is $.30. Any unsold pretzels are discarded because they will be stale before the next home game. The conditional profit per game of having 4,000 pretzels available but only selling 3,000 pretzels is A. Some amount other than those given. B. $2,100. C. $2,800. D. $1,800. A. The correct answer is given. See the correct answer for a complete explanation. B. This answer is calculated without regard to the unsold pretzels that are discarded because they will be stale before the next home game. Those pretzels are a cost that needs to be included in the calculation of any profit. C. This would be the profit if it were assumed that all 4,000 pretzels were sold. However, only 3,000 pretzels were sold. D. Since unsold pretzels are discarded, to calculate the profit we need to use the cost of all 4,000 pretzels that will be purchased to sell, not only the cost of the 3,000 pretzels that are sold. The cost is $1,200 (4,000 × $.30). The revenue from selling 3,000 pretzels is $3,000 (3,000 × $1). The difference between the revenue of $3,000 and the cost of $1,200 is the profit, which is $1,800. Question 9 - CMA 688 5-25 - Probability A computer store sells four computer models designated as P104, X104, A104 and S104. The store manager has made random number assignments to represent customer choices based on past sales data. The assignments are Model Random Numbers P104 0-1 X104 2-6 A104 7-8 S104 9 The probability that a customer will select model P104 is A. Some probability other than those given. B. 10 percent. (c) HOCK international, page 6 Part 1 : 07/27/10 21:34:32 C. 50 percent. D. 20 percent. A. The probability that a customer will select a P104 is given. B. 10% is the probability that S104 will be selected by a customer. C. 50% is the probability that a customer will select X104. D. In the question 10 random numbers, 0 to 9, are used. 0 and 1 (2 numbers in total) represent customers who choose a P104. There are 10 numbers in total. Therefore, the probability that a customer will select a P104 is 2 out of 10, or 2 ÷ 10, which is 20%. Question 10 - CMA 1280 5-15 - Probability A car rental agency has a policy of replacing the tires on its car fleet as the tires wear out. Management wonders if there would be any cost savings if the tires are periodically replaced at one time on its fleet of 500 cars. The technique the car rental agency would find most useful is A. Statistical sampling. B. Probability analysis. C. Linear programming. D. Learning curve techniques. A. Statistical sampling is used to calculate the condition of a population based on a sample from that population. B. Probability gives a means of measuring numerically how likely it is that an event will occur. It enables us to quantify and analyze uncertainties. Usage of probability technique can provide an estimate of what condition the tires of the fleet are in at any point in time and the expected value of replacing all tires at once versus replacing them as they wear out. C. Linear programming is used to either maximize or minimize some quantity (called the objective function) in the presence of constraints, or restrictions, such as limited quantities of labor or materials. D. Learning curves describe the fact that the more experience people have with something, the more efficient they become in doing that task. Higher costs per unit early in production are part of the start-up costs. Usually new products and production processes experience a period of low productivity followed by increased productivity. However, the rate of productivity improvement declines over time until it reaches a level where it remains, until another change in production occurs. Question 11 - CMA 1289 5-21 - Probability The College Honor Society sells hot pretzels at the home football games. The frequency distribution of the demand for pretzels per game is presented as follows: United Sales Volume Probability 2,000 pretzels .10 3,000 pretzels .15 4,000 pretzels .20 5,000 pretzels .35 6,000 pretzels .20 The pretzels are sold for $1.00 each, and the cost per pretzel is $.30. Any unsold pretzels are discarded because they will be stale before the next home game. (c) HOCK international, page 7 Part 1 : 07/27/10 21:34:32 The estimated demand for pretzels at the next home football game using a deterministic approach based on the most likely outcome is A. 5,000 pretzels. B. 4,400 pretzels. C. 6,000 pretzels. D. 4,000 pretzels. A. The deterministic approach assumes that the volume of sales will be the amount of sales that corresponds to the most likely outcome, or the one with the highest probability. In this case the highest probability is 35% which corresponds to 5,000 pretzels. B. 4,400 pretzels is calculated using an expected value approach. C. This is simply the greatest possible volume of sales. D. 4,000 pretzels is an average unit sales volume giving equal weight to each possible outcome. Question 12 - CMA 1287 5-21 - Probability A simple two-state cost investigation model calls for a decision to investigate if the following inequality holds: L(1 - P) > I + C(1 - P) If: L = excess operating cost if out of control P = probability that the process is in control I = cost of investigation C = cost of correction if out of control The value (1 - P) must be A. The probability that the process is out of control. B. The inverse of the probability matrix. C. A measure of the seriousness of the consequences. D. The transpose of the probability matrix. A. If P represents the probability that the process is in control, then (1 í P) must represent the probability that the process is out of control. The problem tells us that the model is a two-state model. The process can be either in control or out of control, and there can't be any other probabilities. The sum of all probabilities is always equal to 1. Therefore, when P is the probability that the process is in control, (1 í P) must be the probability that it is out of control. B. This answer is incorrect. See the correct answer for a complete explanation. C. This answer is incorrect. See the correct answer for a complete explanation. D. This answer is incorrect. See the correct answer for a complete explanation. Question 13 - CMA 683 5-8 - Probability A company is simulating the actions of a government agency in which 50% of the time a recall of a product is required, 40% of the time only notification of the buyer about a potential defect is required, and 10% of the time no action on its part is required. Random numbers of 1 to 100 are being used. An appropriate assignment of random numbers for the recall category would be (c) HOCK international, page 8 Part 1 : 07/27/10 21:34:32 A. 40-90. B. 1-40. C. 61-100. D. 11-60. A. Since 50% of time a recall of a product is required, 50 of the numbers 1-100 need to be assigned to that alternative. The number of numbers in an interval is calculated as the highest number minus the lowest number plus 1. The numbers 40 - 90 represent 51 random numbers, not 50. B. Since 50% of time a recall of a product is required, 50 of the numbers 1-100 need to be assigned to that alternative. The number of numbers in an interval is calculated as the highest number minus the lowest number plus 1. The numbers 1 - 40 represent 40 random numbers, not 50. C. Since 50% of time a recall of a product is required, 50 of the numbers 1-100 need to be assigned to that alternative. The number of numbers in an interval is calculated as the highest number minus the lowest number plus 1. The numbers 61 - 100 represent 40 random numbers, not 50. D. Since 50% of time a recall of a product is required, 50 of the numbers 1-100 need to be assigned to that alternative. The number of numbers in an interval is calculated as the highest number minus the lowest number plus 1. Numbers 11 - 60 represent 50 random numbers, and this is the only answer choice from among the given possibilities that contains 50 numbers in the interval. (c) HOCK international, page 9

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