Exam1 A_F18_Key-1.pdf

# Typically between 500 and 2800 and the performance of

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(typically between 500 and 2800) and the performance of a player is modeled as a Normal random variable centered on their true rating with standard deviation 285. Supposing that there are only two outcomes, a win or a loss, the probability a player will defeat an opponent with a certain rating is given by the probability of observing a performance higher than the opponent’s rating. Suppose Alice is a chess player with a true skill level of 1400. (a) [5pts.] Using the 68-95-99.7 (Empirical) Rule , what is the approximate probability Alice will defeat a player with a true skill level of 1685? Show all your work needed to justify the final answer. Using the Empirical Rule, P ( X > 1685) = P ( X > μ + σ ) 0 . 16 Form B: P ( X > 2030) = P ( X > μ + 2 σ ) 0 . 025 (b) [5pts.] Using the 68-95-99.7 (Empirical) Rule , Alice said she has, approximately, a 97.5% chance of defeating her next opponent. What is that opponent’s rating? Show all your work needed to justify the final answer. Using the Empirical Rule, P ( X > μ - 2 σ ) . 975 . Then 1400 - 2 × 285 = 830 . Form B: P ( X > μ - σ ) . 84 . Then 1500 - 1 × 265 = 1235 . (c) [5pts.] An alternative rating system uses the same core concept, but has a standard deviation of 25. Suppose Bridget has a rating of 150 in this alternative rating system. What would be more unusual: Alice defeating a player rated 1685 (in the original system) or Bridget losing to a player rated 100 (in the alternative system)? Show all your work needed to justify the final answer. In your answer, be sure to say which is more unusual and why, based on the numerical quantities you calculated. z 1 = 100 - 150 25 = - 2 z 2 = 1685 - 1400 285 = 1 In Form B: z 1 = 90 - 150 20 = - 3 z 2 = 2030 - 1500 265 = 2 Bridget losing is more unusual. Explanation :The z -score for Bridget’s result is further from 0 and both events are concerned with the tail probabilities, implying it’s a more unusual event. Note that if Alice lost or Bridget won, this would not be unusual. -score:

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Exam I Stat 226 Fall 2018 —KEY Page 6 5. Field of Dreams. Ray, who is an excellent farmer, is considering alternative management options. Corn yields for farms in his county are Normally distributed with a mean of 170 bushels per acre and a standard deviation of 15 bushels per acre. In order to properly manage risk for his farm, he needs to consider a few different scenarios. Note 1: Use only Table A to find the required probabilities. DO NOT use the 68-95-99.7 (Empirical) rule. Note 2: Show all your work, including the calculation of the z -score, for full credit. (a) [4pts.] In the previous year, Ray’s farm had a yield of 181 bushels per acre. Assuming this year’s yield is independent of that, what is the probability his farm will have a larger yield this year? Report the probability to 4 decimal places. First, calculate the z-score: z = (181 - 170) / 15 = 0 . 733 ... . This should round down to 0 . 73 . In form B: z = (182 - 165) / 20 = 0 . 85 ... Then compute the probability P ( X > 181) = P ( Z > 0 . 73) = 1 - P ( Z < 0 . 73) = 0 . 2327 ( If choosing z = 0 . 74 , this would be 0 . 2296 ) In form B: P ( X > 182) = P ( Z > 0 . 85) = 1 - P ( Z < 0 . 85) = 0 . 1977 (b) [5pts.] Ray is concerned about the lowest 1% of yields - compute what yield, in bushels per acre, corresponds to that. Round your answer to 2 decimal places.
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