exam1_2009 - Statistics 403 Exam 1 October 27, 2009...

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Unformatted text preview: Statistics 403 Exam 1 October 27, 2009 Instructions: • You may not use notes, books, formula cards, etc. • You may use a calculator, but only for arithmetic. • If you are asked for a probability that cannot be calculated by hand, you can express your answer either in terms of normal probabilities (e.g. P ( Z < ... ), P ( Z > ... )), or in terms of R commands (e.g. pnorm(...) ). • If you need a quantile, define it as a symbol (e.g. Q . 3 ) and give your answer in terms of the symbol. • Every problem is worth 15 points. • Partial credit will be given. Express you answers clearly and show work where appro- priate. 1. Order the following three distributions in terms of their variances (i.e. state which has the smallest variance, which has the middle variance value, and which has the greatest variance). a. x-1 1 P ( X = x ) 0.2 0.6 0.2 b. x-2 2 P ( X = x ) 0.3 0.4 0.3 c. x-1 1 P ( X = x ) 0.3 0.4 0.3 Solution: You can answer this question numerically – the variance of a, b, and c, respectively, are 0.4, 2.4, and 0.6, so b > c > a. It’s good to think about how these problems can be answered without actually comput- ing the variance (in case you see a problem where it is too tedious to do the calculation). All three distributions have mean zero, so the variances are determined by how far the squared values are likely to be from zero. b and c have the same probabilities, but the two points in b are further from the zero than the points in c. Therefore the variance of b is greater than the variance of c. 1 a and c have the same sample space, but in a there is a higher probability of observing a zero, so the variance of c is greater than the variance of a....
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This note was uploaded on 02/06/2012 for the course STAT 403 taught by Professor Kerbyshedden during the Winter '12 term at University of Michigan-Dearborn.

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exam1_2009 - Statistics 403 Exam 1 October 27, 2009...

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