Statistics324_HW2 - Statistics 324 Discussion 311 w Jack...

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Statistics 324 – Discussion 311 w/ Jack Homework 2 – Victoria Yakovleva 1. die1 <- sample(1:6, 120, replace = TRUE) xtabs(~ die1) die1 1 2 3 4 5 6 20 21 21 21 20 17 barchart(xtabs(~ die1)) If the outcomes (1-6) are equally likely, the results should be near a frequency of 20 (1/6 of 120) for each outcome. The above results, in my opinion, indicate equal likelihood, with all the outcomes other than 6 resulting in a frequency of 20 or 21, which is pretty darn good for such a small sample size.
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die2 <- sample(1:6, 120, replace = TRUE) xtabs(~ die2) die2 1 2 3 4 5 6 18 17 23 17 26 19 dsum <- die1 + die2 barchart(xtabs(~ dsum)) The distribution of the sum does not appear to give equal probability to the values 2-12; this is good, because the distribution of the sum isn’t expected to give equal probability to the values. Rather, the (theoretical) distribution of the sum goes from 1/36 (for a sum of 2) to 2/36 (for a sum of 3)… and so on, increasing by increments of 1/36, to a peak of 6/36 (for a sum of 7)… and then back down to 1/36 (for a sum of 12), decreasing by increments of 1/36. Barchart of Minimum
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This note was uploaded on 03/27/2008 for the course STAT 324 taught by Professor Bates during the Fall '06 term at University of Wisconsin.

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Statistics324_HW2 - Statistics 324 Discussion 311 w Jack...

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