### Module 3 Discussion.

Added on - 23 Sep 2019

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Module 3 DiscussionOver thinking it.Michael Stow posted May 12, 2017 10:51 AMI spent four days on this too and decided that I am not landing a rocket on the moon.Since I am not landing a rocket on the moon I think a quick down and dirty reference forthis question is to use the bell curve.since the standard measured for each curve is based on the standard deviations they arealways the same. Anything that falls between the mean and the first variation is 68%, twovariables is 95.44% and three variables is 99.72%. Regardless each section of the curvemaintains a constant percentage (correct me if I am wrong this never becomes zero).If I drew thisdown on paper real quick and according to the curve:10 would fall between the second and third variable which is 13.59% probability.21 days is over two standard deviations away from the mean giving it a 2.14%probability.These are quick ball park figures, and the link below is for the Pearson explanation.http://www.pearsoncustom.com/mct-comprehensive/asset.php?isbn=1269879944&id=20384Dear Michael,Assuming bell curve for not landing a rocket on the moon does not make sense, also thestandard deviations for each curve if many are drawn cannot be same because of thedifferent values for the mean. The description of the amount of data is incorrectly written,it should be like the amount of data that falls between the mean and the first standarddeviation is 68%, that falls between the mean and the second variation is 95.44%, andthat falls between the mean and the third variation is 99.72%.If the curve is drawn on the curve, then the probability that the value 10 will lie betweensecond value and third value is 13.59%, also mention what does 10 represent and we arediscussing about the probability of 21 days, what does it represent is not defined here.