# HW1 (3).pdf - University of Illinois Urbana-Champaign CS361...

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University of Illinois Urbana-ChampaignCS361 Probability and StatisticsHW #1Question 1AThe minimum of the top possible score is 5.Since we know the median of all the scores is 5, the minimum top possible score mustalso be 5. Since there are five students who took the exam, the top three scores mustall be 5, so the median of the data remains 5.BThe maximum of the top possible score is 10.Since we know the mean of all the scores is 4, we can obtain the sum of all the scoresby multiplying the mean by the number of data points.mean*numberofdatapoints= 4*5 = 20(1)Therefore, the sum of all the scores must be 20. We also know that the median ofthe data has to remain 5. So, the maximum possible score can be 5 points less thanthe total - 20. However, if the possible top score was 15, that would result in themedian score becoming 0, which is incorrect. Therefore, there must be two scoresthat are 5 points each, leaving a remainder of 10 points to be assigned for the toppossible score, and the median remains 5 pointsCThe minimum of standard deviation is 1.265.This standard deviation is calculated over the data set with the lowest possible highscore, which we obtained earlier. For the minimum of standard deviation, split theremaining 5 points over the last two slots to obtain the data set - 5,5,5,2,3.DThe maximum of standard deviation is 3.74.This standard deviation is calculated over the data set with the highest possible highscore, which we obtained earlier. We know that there must be two scores that are 5points each, and the top score is 10, leaving the last two scores to be 0. The data setis - 10,5,5,0,0.

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Term
Fall
Professor
PENG
Tags
University Of Illinois Urbana Champaign
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