347.2010.2.comments

347.2010.2.comments - Comments on Exam 2, Spring 2010...

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Comments on Exam 2, Spring 2010 Problem 1 The distribution is hypergeometric with parameters N = 20, r = 4, and n = 3. Several people said it was binomial. The situation described is sampling WITHOUT REPLACEMENT, which is modeled by a hypergeometric distribution. The binomial models sampling with replacement. A binomial can only be used to approximate a hypergeometric (and thus model sampling without replacement) if the sample size, n , is much smaller than both the number of “defectives,” r , and the number of “non- defectives,” N – r. That is definitely not the case here. Problem 3 In this problem the random variable Y , the # rolls of a fair die required to get a 6, has a geometric distribution with p = 1/6. a. Many evaluated P ( Y = 3) instead of the correct P ( Y 3). b. Some found the value of y so that P ( Y = y ) 0.05 instead of the correct P ( Y > y ) 0.05 (which is equivalent to P ( Y y ) 0.95. Also, many forgot the useful formula that for a geometric distribution as defined in our text, P ( Y > y ) = q y .
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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347.2010.2.comments - Comments on Exam 2, Spring 2010...

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