DSC330_hw1_solutions

DSC330_hw1_solutions - DSC 330(Spring 2011 HW1 SOLUTIONS 1...

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Unformatted text preview: DSC 330 (Spring 2011) HW1 SOLUTIONS 1. Suppose a random variable, u1D44C 1 is from the population u1D441 (1 , 9). Calculate the proba- bility u1D443u1D45F ( u1D44C 1 1) and u1D443u1D45F ( u1D44C 1 4). Solutions : Since u1D44C 1 is symmetric about 1, u1D443u1D45F ( u1D44C 1 1) = 0 . 5. u1D707 = 1 and u1D70E = 3, so u1D443u1D45F ( u1D44C 1 4) = u1D443u1D45F ( u1D44C 1 1 3 4 1 3 ) = u1D443u1D45F ( u1D44D 1) = 0 . 5 . 3413 = . 1587 2. Textbook 1.38(a) (This question asks you to calculate the probability u1D443u1D45F ( u1D465 6) ) Solutions : For this problem, u1D707 u1D465 = u1D707 = 4 . 29 and u1D70E u1D465 = u1D70E u1D45B = 2 . 95 50 = . 4172 by the Central Limit Theorem. a. u1D443u1D45F ( u1D465 6) = u1D443u1D45F ( u1D465 4 . 29 . 4172 6 4 . 29 . 4172 ) = u1D443u1D45F ( u1D44D 4 . 10) = 0 . 5 . 5 = 0. Since the probability of observing a sample mean CAHS score of 6 or higher is so small ( u1D45D is essentially 0), we would not expect to see a sample mean of 6 or higher.essentially 0), we would not expect to see a sample mean of 6 or higher....
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This note was uploaded on 01/29/2012 for the course DSC 330 taught by Professor Staff during the Spring '08 term at Oregon.

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DSC330_hw1_solutions - DSC 330(Spring 2011 HW1 SOLUTIONS 1...

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