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Unformatted text preview: Up: Stat 5101 Stat 5131 Final Exam Problem 1 Let X be the number of wins. Then , where n = 100 and p = 18 / 38 = 9 / 19. The number of losses is n X . Let W denote the gambler's net winnings. Then W = X ( n X ) = 2 X n . The mean. By linearity of expectation and the mean of the binomial distribution being n p The standard deviation. By the formula and the variance of the binomial distribution being n p (1  p ) so the standard deviation is . Problem 2 The probability of rolling a six on each roll is 1 / 6 and the rolls are independent, the number of sixes X has a distribution with n = 200 and p = 1 / 6. The mean is E ( X ) = n p = 200 / 6 = 33.33333 The standard deviation is The continuity correction says that we should standardize The normal distribution table gives 0.1210 for the probability of the tail past z . Just for the record (not part of the solution), the exact probability is 0.1223. The normal approximation gives almost three significant figures despite the skewness of the distribution. Without the continuity correction, almost three significant figures despite the skewness of the distribution....
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This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Spring '02 term at Minnesota.
 Spring '02
 Staff

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