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The Obama vacations have cost an average of 4 million and a standard deviation of 5 million. The Bush vacations have cost an average of 3 million and a standard deviation of 2 million. What is more unusual: an Obama vacation that costs 5 million or a Bush vacation that cost 1 million?

What I think: I don't know what "more unusual" means in statistics. I tried to solve this problem by comparing the standard deviation and the z-interval.

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@JacobBlack - Obama seems to get a fair amount of "free" spending money on some of those vacations. It would be interesting if someone could give a distribution having non-negative costs for every vacation and "Obama statistics". –  Mark Bennet Feb 11 '13 at 22:58
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"Please show work" applies to you, the one who's asking a homework question. –  Henning Makholm Feb 11 '13 at 22:59
    
Its a homework review. Should I be using a z-interval to calculate this answer? –  Shaily Feb 11 '13 at 23:02
    
Compute the probability that an Obama vacation costs (at least) \$5 million. Compute the probability that a Bush vacation costs (at least) \$1 million. Which is higher? –  Arkamis Feb 11 '13 at 23:04
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2 Answers

up vote 2 down vote accepted

Bush's vacation costing $1,000,000 is a $1,000,000 less than average, for Bush: a full standard deviation lower than average. Obama's vacation costing $5,000,000 is only 1/5 th of a standard deviation higher than average.

So Bush's vacation cost deviated more from the average, with respect to for Bush, than did Obama's (with respect to Obama's average costs), i.e., Bush's vacation cost was the more unusual in the sense that his vacation costs deviated more than is typical for Bush, than did Obama's vacation cost from what is typical for Obama.

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Ohhhh. Political view exposed. ;-) –  B. S. Feb 25 '13 at 7:04
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A Bush vacation that costs 1M is a full standard deviation low, an Obama one that costs 5M is only 0.2 standard deviation high.

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But are the sample spaces here even large enough for a normal-distribution approximation to make sense? –  Henning Makholm Feb 11 '13 at 23:07
    
@HenningMakholm: without further information, I would have to assume they are. I suspect in reality you are right. I also suspect that whoever posed the problem didn't think about it. –  Ross Millikan Feb 11 '13 at 23:09
    
It has to be normal distribution. We haven't learned anything else yet. –  Shaily Feb 11 '13 at 23:12
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