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MIT6_042JF10_fnl_2008_sol

MIT6_042JF10_fnl_2008_sol - 6.042/18.062J Mathematics for...

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6.042/18.062J Mathematics for Computer Science December 17, 2008 Tom Leighton and Marten van Dijk Final Exam Problem 1. [25 points] The Final Breakdown Suppose the 6.042 final consists of: 36 true/false questions worth 1 point each. 1 induction problem worth 15 points. 1 giant problem that combines everything from the semester, worth 49 points. Grading goes as follows: The TAs choose to grade the easy true/false questions. For each individual point, they flip a fair coin. If it comes up heads, the student gets the point. Marten and Brooke split the task of grading the induction problem. With 1/3 probability, Marten grades the problem. His grading policy is as follows: Either he gets exasperated by the improper use of math symbols and gives 0 points (which happens with 2/5 probability), or he finds the answer satisfactory and gives 15 points (which happens with 3/5 probability). With 2/3 probability, Brooke grades the problem. Her grading policy is as follows: She selects a random integer point value from the range from 0 to 15, inclusive, with uniform probability. • Finally, Tom grades the giant problem. He rolls two fair seven -sided dice (which have values from 1 to 7, inclusive), takes their product, and subtracts it from 49 to determine the score. (Example: Tom rolls a 3 and a 4. The score is then 49 3 4 = · 37.) Assume all random choices during the grading process are mutually independent. The problem parts start on the next page. Show your work to receive partial credit. (a) [7 pts] What is the expected score on the exam?
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2 Final Exam Solution. 36/2 + ( 1/3 ) 15 ( 3/5 ) + ( 2/3 ) 15/2 + ( 49 ( 4 4 )) = 18 + 3 + 5 + 33 = 59 The expected score on the exam is the sum of the expected scores on the individual problems. Ex ( test score ) = Ex ( mc score ) + Ex ( ind score ) + Ex ( giant score ) • The expected multiple choice score is just the sum of the expectations on 36 coin tosses. Since the coin is fair, the expected number of heads on each flip is 1/2. There- fore: Ex ( mc score ) = 36 1/2 = 18 · • The expected induction score is a weighted sum of the expectations on problems graded by Marten and Brooke. Let M be the event that Marten graded the problem, and B be the event that Brooke graded the problem. Therefore: Ex ( ind score | M ) = 0 15 · 2/5 + 15 · 3/5 = 9 Ex ( ind score B ) = ( i 1/16 ) = 15/2 · | i = 0 Ex ( ind score ) = Ex ( ind score | M ) Pr { M } + Ex ( ind score | B ) Pr { B } = 9 1/3 + 15/2 2/3 = 8 · · The expected giant problem score is the expectation on 49 minus the product of rolls of two fair seven-sided dice. We can pull out the constant 49, and (since the dice are independent) the expectation on the product of the rolls becomes the product of the expectations on the rolls, which is 4 in this case (average of numbers 1 to 7). Therefore, if R is the random variable representing the roll of a single die: Ex ( giant score ) = Ex 49 R 2 = 49 Ex ( R ) 2 = 49 ( 4 ) 2 = 49 16 = 33 Therefore, the overall expectation is: Ex ( test score ) = 18 + 8 + 33 = 59 (b) [5 pts] What is the variance on the 36 true/false questions?
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