Assignment 3 SOLUTIONS (1206A3w2008)

Assignment 3 SOLUTIONS (1206A3w2008) - MGSC 1206.2...

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MGSC 1206.2 Assignment #3 SOLUTIONS Winter 2008 Due: 1 pm, Friday, February 1 st TOTAL MARKS FOR THIS ASSIGNMENT = 55 Note: Points are deducted when students do not (a) show work including suitable calculations, (b) use proper notation, and (c) answer word problems with words. 8.4 #34 & #36. A company gives prospective workers a 6-question multiple-choice test. Each question has 5 possible answers, so that there is a 1/5 or 20% chance of answering a question correctly just by guessing. #34. Find the probability of getting no correct answers. There is a 26.2% probability of getting no correct answers. #36. Find the probability of getting no more than 3 correct answers. There is a 98.3% probability of getting no more than 3 correct answers. 1 pt 1 pt 1 pt 1 pt 3 pts 1 pt
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8.5 #30. An insurance company has written 100 policies of $10,000, 500 of $5000 and 1000 of $1000 for people of age 20. If experience shows that the probability that a person will die at age 20 is 0.001, how much can the company expect to pay out during the year the policies were written? 3 points for the final answer and following proper calculations… Note: This one can be solved satisfactorily in different ways, all OK… 2 pts for using expected value notation
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8.5 #42. A raffle offers a first prize of $1000, two second prizes of $300 each, and twenty third prizes of $10 each. If 10,000 tickets were sold at 50 cents each, find the expected winnings for a person buying one ticket. Is this a fair game? 1. 3 out of the 10 computers recently purchased by our company from Bob's Computer Emporium have been found to be defective in some respects. Our company has informed Bob that we will no longer purchase from them because of the poor quality of merchandise. Bob responded that the problems were a fluke (highly unlikely) – their sophisticated equipment can be subject to failure, but on average, no more than 10% of Bob’s equipment has defects. a) Show your verification that this is a binomial problem. This problem fits the requirements of the Binomial Distribution each computer is defective or not defective we are counting the number that are defective defects in one computer do not cause defects in any other, so the
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