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This question has kinda stumped me:

For a data set, the mean is 5 and the standard deviation is 1. There are 10 values in the data set. Is the range of the data set bigger, smaller or equal to 10?

I don't want the answer, just the technique. Here's what I've tried (admittedly, not much):

$$\sum_{i=0}^{9}a_i / 10= 5$$ $$\sigma = 1$$ Is there a relation between standard deviation and simple mean which I can use? Any help is appreciated. This is not homework, it's a practice qn for the GRE.

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up vote 2 down vote accepted

$\def\implies{\quad\Rightarrow\quad}$If the mean is 5, then you know that

$$\frac{1}{10} \sum_i a_i = 5 \implies \sum_ia_i = 50$$

and if the standard deviation is 1 (and hence the variance is 1) then

$$\frac{1}{10} \sum_i(a_i-5)^2 = 1 \implies \sum_i(a_i-5)^2 = 10$$

To make the range as big as possible, we choose two values $a^+$ and $a^-$ to be as far from the mean as possible (symmetrically) and set the rest of the values to the mean, i.e. $a^+=5+x$ and $a^-=5-x$. What does that tell you about $x$?

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tells me that 2x^2 = 10, which gives x as about 2.24 and -2.24. hence the max possible range is 4.48. is this correct? – OckhamsRazor Jun 1 '12 at 10:53
Precisely.$\,\!\,\!$ – Chris Taylor Jun 1 '12 at 12:14
Thanks, that really helped. – OckhamsRazor Jun 1 '12 at 18:15

Let us define $$\mu=\frac{1}{N}\Sigma_{i=1}^N a_i$$ $$\delta=\max\{|a_i-\mu|:i=1,\ldots,N\}$$ $$\sigma=\sqrt{\frac{1}{N}\Sigma_{i=1}^N (a_i-\mu)^2}.$$ Then we get easily the following estimates $$\sqrt{N}\sigma\geq\delta\geq\sigma.\tag{*}$$ You know $N,\sigma$ and your problem is to estimate $\delta,$ so you can use (*).

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Nice Chris Taylor. I just want to comment about the OPs question when he said "Is there a relation between standard deviation and simple mean which I can use?". In general there is no relationship between a mean and a standard deviation. You can construct families of distributions like the normal where any value can be given to the mean and independently any positive value can be given to the standard deviation. In this case though you are dealing with a sample and you know that the sample mean is 5, the sample standard deviation is 1 and the sample size is known. So you can relate the range here using Chris' trick because he knows how to pick a distribution that maximizes the range. However, be careful, in statistics we often define the sample standard deviation as ∑i(ai−m)$^2$/(n-1) which in your case would give ∑i(ai−5)$^2$/9. If that is what the GRE writers intended when they said the variance is 1 then the correct answer would be 2x$^2$ = 9 or x=√4.5 = 2.121. This could be a problem for you on the GRE because the answer you compute might not be one of the choices.

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ah right, so sample data would produce different results from the whole population data. thanks. – OckhamsRazor Jun 1 '12 at 18:30
Yes If they define the sample variance that way. The MLE which is a slightly biased estimator would have n in the denominator or 10 in your case. But teaching to use the unbiased estimator is so engrained in the way elementary statistics is taught organizations like the ETS would probably use n-1 and think of this as a tricky question. I personally don't like the emphasis on unbiasedness. We should teach students that mean square error measures accuracy and can be more important than bias when the biased estimator has a small bias. – Michael Chernick Jun 1 '12 at 20:00


If the range is $10$, find out how large the standard deviation would have to be (It will be larger than 1)

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