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MIT6_042JF10_fnl_2008_sol

# MIT6_042JF10_fnl_2008_sol - 6.042/18.062J Mathematics for...

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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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