DJBruce Posted July 3, 2010 Share Posted July 3, 2010 So awhile back I found some interesting problems from older versions of the Michigan Autumn Take Home Challenge, but sadly they do not post answers to the past exams. While going through a few of them I ran into some that stumped me: 1. Which is greater [math]10!!!!!! [/math] or [math] 10^{10^{10^{10^{10^{10^{10}}}}}}[/math] (Note: In evaluating towers of exponents, precedence rules state that evaluation should be from the top down rather than from the bottom up. For example [math]2^{3^{2}}[/math] should be interpreted as [math]2^{9}[/math] ) I know that factorial growth generally outstrips exponential growth, so I would guess 10!!!!!! is greater, but I really am not confident in that guess. 2. Recall that for a nonempty finite set [math]S={x_{1},x_{2}...x_{3}}[/math] with mean [math]\bar{x}=\frac{x_{1}+x_{2}+...+x_{n}}{n}[/math] the standard deviation of S id defined to be [math]\sigma=\sqrt{\frac{(x_{1}-\bar{x})^{2}+(x_{2}-\bar{x}+...(x_{n}-\bar{x})^{2})^{2}}{n}}[/math] Prove or disprove: for any nonempty finite set S of positive real numbers, the standard deviation of s cannot be larger the the mean of S. My thoughts for this one were to assume that S describes a binomial distribution where n=100, p<<<q. For my counter example I let: [math] n=100, p=1x10^{-7}, q=.9999999[/math] In this situation [math]\bar{x}=np=(100)(1x10^{-7})=1x10^{-5}[/math] [math]\sigma=\sqrt{npq}=\sqrt{(100)(1x10^{-7})(.9999999)}\approx.00316[/math] [math] \sigma=.00316>\bar{x}=1x10^{-5}[/math] So since I have a counterexample their statement is shown to be false. Is this a proper way to prove this or not? Link to comment Share on other sites More sharing options...
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