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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.

Shaily
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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. – hmakholm left over Monica 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? – Emily Feb 11 '13 at 23:04

2 Answers2

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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.

amWhy
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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.

Ross Millikan
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