Sections 5.2 %26 5.3 Recitation Problems

# Sections 5.2 %26 5.3 Recitation Problems - Solutions for...

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Unformatted text preview: Solutions for Recitation Problems from Chapter 5: Problems 14, 20, 10 and 21 (also 19 from HW #6) 14. Let X be the discrete r.v. of profit, in \$K, with values x = -100, 0, 50, 100, 150, and 200. a. To be a valid probability distribution, all the probabilities must sum to one: prob (-100) + prob (0) + prob (50) + prob (100) + prob (150) + prob (200) = 1 But prob(200) is missing. So Prob(200) must be the complement of the sum of all the other probabilities: Prob ( x = 200) = 1 – [ prob ( x = -100) + prob ( x = 0) + prob ( x = 50) + prob ( x = 100) + prob ( x = 150)] = 1 – (0.10 + 0.20 + 0.30 + 0.25 + 0.10) = 1 – 0.95 = 0.05 The probability that MRA will have a \$200,000 profit is 0.05. In other words, there is a 5% chance that MRA will have a \$200,000 profit. The complete probability distribution for profit x is now: x prob(x)-100 0.10 0.20 50 0.30 100 0.25 150 0.10 200 0.05 1.00 b. “Profit” means x > 0, so Chap 5 recitation problems 1 © Harvey Singer 2009 Prob (Profit) = prob ( x > 0) = prob ( x = 50) + prob ( x = 100) + prob ( x = 150) + prob ( x = 200) 200) = 0.30 + 0.25 + 0.10 + 0.05 = 0.70 The probability that MRA will have a profit is 0.70. In other words, there is a 70% chance that MRA will be profitable. c. The phrase “at least \$100,000” means “\$100,000 or more.” The discrete values of x that are “100,000 or more” are x = 100, 150, and 200. Prob (at least 100) = prob ( x ≥ 100) = prob ( x = 100) + prob ( x = 150) + prob ( x = 200) = 0.25 + 0.10 + 0.05 = 0.40 The probability that MRA will have a profit of at least \$100,000 is 0.40. In other words, there is a 40% chance that MRA will have a profit of at least \$100,000. (Another way of stating this: There is a 40% chance that MRA will have a profit of at not less than \$100,000.) ********************************************************* Chap 5 recitation problems 2 © Harvey Singer 2009 New questions: Not asked in the textbook problem, but asked here and now: 1. How much profit should be expected? Calculate expected value of x from the probability distribution: x prob(x) x*prob(x)-100 0.10-10 0.20 50 0.30 15 100 0.25 25 150 0.10 15 200 0.05 10 1.00 55.00 Expect a profit of \$55,000. Of course, the profit may be more or it may be less, but as a first an best guess, expect a profit of \$55K. Notes: a. The expected value is NOT the arithmetic average (mean) of the values of the r.v. x . The arithmetic average of the values x = -100, 0, 50, 100, 150, and 200 is 66.67. But this assumes that all the values are equally likely to occur, so that they make an equal contribution to what’s expected. That’s not true here. These are not equally likely values. Some of the values of x are more likely than others. As a result, the probability of a value, and not just the value itself, also drives what should be expected....
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Sections 5.2 %26 5.3 Recitation Problems - Solutions for...

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